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Electric Field and Coulomb's Law — JEE Previous Year Questions

Every Electric Field and Coulomb's Law question asked in JEE Main and JEE Advanced across the last 186 papers — 182 questions, each with its correct answer. Free to read, no account needed.

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182

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72%

All 182 Electric Field and Coulomb's Law questions

Most recent papers first.

Q1·PhysicsMultiple correctJEE Advanced 2026
Consider an electric dipole comprising two charges +q+q+q and −q-q−q each with mass mmm, separated by a fixed distance ddd and initially at rest with its dipole moment pointing along i^\hat{i}i^. A uniform electric field Ej^E\hat{j}Ej^​ is turned on at time t=0t = 0t=0 and it is turned off at t=tft = t_{f}t=tf​, when the dipole moment makes an angle θf\theta_{f}θf​ with i^\hat{i}i^. Neglecting any sources of energy loss, correct option(s) is/are:
  1. (A)The center of mass of the dipole is deflected towards j^\hat{j}j^​ in the presence of the field.
  2. (B)If the magnitude of the final angular velocity ωf=2qEmd\omega_{f} = \sqrt{\frac{2qE}{md}}ωf​=md2qE​​, then θf=π6\theta_{f} = \frac{\pi}{6}θf​=6π​.
  3. (C)If θf=π/3\theta_{f} = \pi/3θf​=π/3, then the change in kinetic energy of the dipole is given by 23 qEd2\sqrt{3}\,qEd23​qEd.
  4. (D)For θf=π/4\theta_{f} = \pi/4θf​=π/4, the dipole rotates around its center of mass with a constant angular velocity after t>tft > t_{f}t>tf​.

Correct answer: (B), (D)

Step-by-step solution →
Q2·PhysicsSingle correctJEE Main 2026
Two point charges q1=3 μCq_1 = 3\,\mu Cq1​=3μC and q2=−4 μCq_2 = -4\,\mu Cq2​=−4μC are placed at points (2i^+3j^+3k^)(2\hat{i}+3\hat{j}+3\hat{k})(2i^+3j^​+3k^) and (i^+j^+k^)(\hat{i}+\hat{j}+\hat{k})(i^+j^​+k^) respectively. Force on charge q2q_2q2​ is ________ N. (Take 14πϵ0=9×109 SI Units)\left(\text{Take } \dfrac{1}{4\pi\epsilon_0} = 9\times 10^{9}\text{ SI Units}\right)(Take 4πϵ0​1​=9×109 SI Units)
  1. (A)(12i^+24j^+24k^)×10−3(12\hat{i}+24\hat{j}+24\hat{k})\times 10^{-3}(12i^+24j^​+24k^)×10−3
  2. (B)(4i^+8j^+8k^)×10−3(4\hat{i}+8\hat{j}+8\hat{k})\times 10^{-3}(4i^+8j^​+8k^)×10−3
  3. (C)(3i^+6j^+6k^)×10−3(3\hat{i}+6\hat{j}+6\hat{k})\times 10^{-3}(3i^+6j^​+6k^)×10−3
  4. (D)(−4i^−8j^−8k^)×10−3(-4\hat{i}-8\hat{j}-8\hat{k})\times 10^{-3}(−4i^−8j^​−8k^)×10−3

Correct answer: (B)

Step-by-step solution →
Q3·PhysicsSingle correctJEE Main 2026
A thin half ring of radius 35 cm is uniformly charged with a total charge of Q coulomb. If the magnitude of the electric field at centre of the half ring is 100 V/m, then the value of Q is ________ nC. (ε₀ = 8.85 × 10−12^{-12}−12 C²/Nm² and π = 3.14)
  1. (A)2.14
  2. (B)2.44
  3. (C)3.25
  4. (D)0.7

Correct answer: (A)

Step-by-step solution →
Q4·PhysicsSingle correctJEE Main 2026
A rigid dipole undergoes a simple harmonic motion about its centre in the presence of an electric field E⃗1=E0x^\vec{E}_{1} = E_{0}\hat{x}E1​=E0​x^. If another electric field E⃗2=2E0(y^+z^)\vec{E}_{2} = 2E_{0}(\hat{y} + \hat{z})E2​=2E0​(y^​+z^) is introduced to the system, what will be the percentage change in the frequency of the oscillation (approximate)?
  1. (A)73%
  2. (B)63%
  3. (C)83%
  4. (D)53%

Correct answer: (A)

Step-by-step solution →
Q5·PhysicsSingle correctJEE Main 2026
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A: In electrostatics, a conductor does not store any net charge inside. Reason R: Inside the capacitor (with no dielectric medium), the free charge carriers, if placed between the plates of capacitor, experience force and drift. Choose the correct answer from the options given below
  1. (A)Both A and R are true and R is the correct explanation of A
  2. (B)Both A and R are true but R is NOT the correct explanation of A
  3. (C)A is true but R is false
  4. (D)A is false but R is true

Correct answer: (B)

Step-by-step solution →
Q6·PhysicsSingle correctJEE Main 2026
Two short electric dipoles AAA and BBB having dipole moment p1p_{1}p1​ and p2p_{2}p2​ respectively are placed with their axis mutually perpendicular as shown in the figure. The resultant electric field at a point xxx is making an angle of 60° with the line joining points OOO and xxx. The ratio of the dipole moments p2/p1p_{2}/p_{1}p2​/p1​ is ______.
  1. (A)32\frac{\sqrt{3}}{2}23​​
  2. (B)232\sqrt{3}23​
  3. (C)13\frac{1}{\sqrt{3}}3​1​
  4. (D)3\sqrt{3}3​

Correct answer: (B)

Step-by-step solution →
Q7·PhysicsSingle correctJEE Main 2026
Two point charges 8 μC8\,\mu C8μC and −2 μC-2\,\mu C−2μC are located at x=2x = 2x=2 cm and x=4x = 4x=4 cm, respectively on the xxx-axis. The ratio of electric flux due to these charges through two spheres of radii 3 cm and 5 cm with their centers at the origin is ________.
  1. (A)4:14 : 14:1
  2. (B)3:43 : 43:4
  3. (C)4:34 : 34:3
  4. (D)4:54 : 54:5

Correct answer: (C)

Step-by-step solution →
Q8·PhysicsSingle correctJEE Main 2026
Identify the correct statements : A. Electrostatic field lines form closed loops. B. The electric field lines point radially outward when charge is greater than zero. C. The Gauss-Law is valid only for inverse-square force. D. The workdone in moving a charged particle in a static electric field around a closed path is zero. E. The motion of a particle under Coulomb's force must take place in a plane. Choose the correct answer from the options given below :
  1. (A)A, B, D, E Only
  2. (B)A, B, C, D Only
  3. (C)B, C, D, E Only
  4. (D)A, C, E Only

Correct answer: (C)

Step-by-step solution →
Q9·PhysicsSingle correctJEE Main 2026
Three charges +2q. +3q and –4q are situated at (0,–3a), (2a, 0) and (–2a,0) respectively in the xy plane. The resultant dipole moment about origin is____
  1. (A)2qa(3j^−i^)2qa\left(3\hat{j}-\hat{i}\right)2qa(3j^​−i^)
  2. (B)2qa(3i^−7j^)2qa\left(3\hat{i}-7\hat{j}\right)2qa(3i^−7j^​)
  3. (C)2qa(7i^−3j^)2qa\left(7\hat{i}-3\hat{j}\right)2qa(7i^−3j^​)
  4. (D)2qa(3j^−7i^)2qa\left(3\hat{j}-7\hat{i}\right)2qa(3j^​−7i^)

Correct answer: (C)

Step-by-step solution →
Q10·PhysicsSingle correctJEE Main 2026
The electrostatic potential in a charged spherical region of radius r varies as V=ar3+bV = ar^{3} + bV=ar3+b, where a and b are constants. The total charge in the sphere of unit radius is α×πa∈0\alpha \times \pi a \in_{0}α×πa∈0​. The value of α\alphaα is ______. (permittivity of vacuum is ∈0\in_{0}∈0​)
  1. (A)–12
  2. (B)–6
  3. (C)–9
  4. (D)–8

Correct answer: (A)

Step-by-step solution →
Q11·PhysicsNumericalJEE Main 2026
A point charge q = 1 μ\muμC is located at a distance 2 cm from one end of a thin insulating wire of length 10 cm having a charge QQQ = 24 μ\muμC, distributed uniformly along its length, as shown in figure. Force between q and wire is ____ N. (Use 14πϵ0=9×109 N.m2/C2\frac{1}{4\pi\epsilon_0} = 9\times10^{9}\,\text{N.m}^{2}/\text{C}^{2}4πϵ0​1​=9×109N.m2/C2)

Correct answer: 90

Step-by-step solution →
Q12·PhysicsSingle correctJEE Main 2026
Two shorts dipoles (A,B)(A, B)(A,B), AAA having charges ± 2μC and length 1 cm and BBB having charges ± 4μC and length 1 cm are placed with their centres 80 cm apart as shown in the figure. The electric field at a point PPP, equi-distant from the centres of both dipoles is_____N/C.
  1. (A)9162×105\frac{9}{16}\sqrt{2}\times 10^{5}169​2​×105
  2. (B)4.52×1044.5\sqrt{2}\times 10^{4}4.52​×104
  3. (C)92×1049\sqrt{2}\times 10^{4}92​×104
  4. (D)9162×104\frac{9}{16}\sqrt{2}\times 10^{4}169​2​×104

Correct answer: (D)

Step-by-step solution →
Q13·PhysicsSingle correctJEE Main 2026
Two point charges 2q and q are placed at vertex A and centre of face CDEF of the cube as shown in figure. The electric flux passing through the cube is :
  1. (A)3qε0\frac{3q}{\varepsilon_0}ε0​3q​
  2. (B)qε0\frac{q}{\varepsilon_0}ε0​q​
  3. (C)3q2ε0\frac{3q}{2\varepsilon_0}2ε0​3q​
  4. (D)3q4ε0\frac{3q}{4\varepsilon_0}4ε0​3q​

Correct answer: (D)

Step-by-step solution →
Q14·PhysicsSingle correctJEE Main 2026
A simple pendulum has a bob with mass mmm and charge q. The pendulum string has negligible mass. When a uniform and horizontal electric field E⃗\vec{E}E is applied, the tension in the string changes. The final tension in the string, when pendulum attains an equilibrium position is ______. (g : acceleration due to gravity)
  1. (A)mg−qEmg - qEmg−qE
  2. (B)mg+qEmg + qEmg+qE
  3. (C)m2g2+q2E2\sqrt{m^2g^2 + q^2E^2}m2g2+q2E2​
  4. (D)m2g2−q2E2\sqrt{m^2g^2 - q^2E^2}m2g2−q2E2​

Correct answer: (C)

Step-by-step solution →
Q15·PhysicsSingle correctJEE Main 2026
Six point charges are kept 60° apart from each other on the circumference of a circle of radius R as shown in figure. The net electric field at the centre of the circle is _________. (∈o\in_{o}∈o​ is permittivity of free space)
  1. (A)−5Q8π∈oR2(i^+3j^)-\frac{5Q}{8\pi \in_{o} R^{2}}(\hat{i}+\sqrt{3}\hat{j})−8π∈o​R25Q​(i^+3​j^​)
  2. (B)−Q4π∈oR2(3 i^−j^)-\frac{Q}{4\pi \in_{o} R^{2}}\left(\sqrt{3}\,\hat{i}-\hat{j}\right)−4π∈o​R2Q​(3​i^−j^​)
  3. (C)−(5Q8π∈oR2)(i^−3j^)-\left(\frac{5Q}{8\pi \in_{o} R^{2}}\right)\left(\hat{i}-3\hat{j}\right)−(8π∈o​R25Q​)(i^−3j^​)
  4. (D)Q4π∈oR2(3 i^−j^)\frac{Q}{4\pi \in_{o} R^{2}}\left(\sqrt{3}\,\hat{i}-\hat{j}\right)4π∈o​R2Q​(3​i^−j^​)

Correct answer: (B)

Step-by-step solution →
Q16·PhysicsSingle correctJEE Main 2026
Five positive charges each having charge qqq are placed at the vertices of a pentagon as shown in the figure. The electric potential (V)(V)(V) and the electric field (E⃗)(\vec{E})(E) at the center O of the pentagon due to these five positive charges are :
  1. (A)V=5q4πε0rV = \dfrac{5q}{4\pi\varepsilon_0 r}V=4πε0​r5q​ and E⃗=0\vec{E} = 0E=0
  2. (B)V=5q4πε0rV = \dfrac{5q}{4\pi\varepsilon_0 r}V=4πε0​r5q​ and E⃗=53q8πε0r2 r^\vec{E} = \dfrac{5\sqrt{3}q}{8\pi\varepsilon_0 r^2}\,\hat{r}E=8πε0​r253​q​r^
  3. (C)V=5q4πε0rV = \dfrac{5q}{4\pi\varepsilon_0 r}V=4πε0​r5q​ and E⃗=5q4πε0r2 r^\vec{E} = \dfrac{5q}{4\pi\varepsilon_0 r^2}\,\hat{r}E=4πε0​r25q​r^
  4. (D)V=0V = 0V=0 and E⃗=0\vec{E} = 0E=0

Correct answer: (A)

Step-by-step solution →
Q17·PhysicsSingle correctJEE Advanced 2025
Two co-axial conducting cylinders of same length ℓ\ellℓ with radii 2R\sqrt{2}R2​R and 2R are kept, as shown in Fig. 1. The charge on the inner cylinder is Q and the outer cylinder is grounded. The annular region between the cylinders is filled with a material of dielectric constant κ =5. Consider an imaginary plane of the same length ℓ\ellℓ at a distance R from the common axis of the cylinders. This plane is parallel to the axis of the cylinders. The cross-sectional view of this arrangement is shown in Fig. 2. Ignoring edge effects, the flux of the electric field through the plane is ( ε0\varepsilon_0ε0​ is the permittivity of free space):
  1. (A)Q30ε0\frac{Q}{30\varepsilon_0}30ε0​Q​
  2. (B)Q15ε0\frac{Q}{15\varepsilon_0}15ε0​Q​
  3. (C)Q60ε0\frac{Q}{60\varepsilon_0}60ε0​Q​
  4. (D)Q120ε0\frac{Q}{120\varepsilon_0}120ε0​Q​

Correct answer: (C)

Step-by-step solution →
Q18·PhysicsMultiple correctJEE Advanced 2025
Six infinitely large and thin non-conducting sheets are fixed in configurations I and II. As shown in the figure, the sheets carry uniform surface charge densities which are indicated in terms of σ0\sigma_0σ0​. The separation between any two consecutive sheets is 1 μm. The various regions between the sheets are denoted as 1, 2, 3, 4 and 5. If σ0=9\sigma_0 = 9σ0​=9 μC/m2^22, then which of the following statements is/are correct: (Take permittivity of free space ε0=9×10−12\varepsilon_0 = 9\times 10^{-12}ε0​=9×10−12 F/m)
  1. (A)In region 4 of the configuration I, the magnitude of the electric field is zero.
  2. (B)In region 3 of the configuration II, the magnitude of the electric field is σ0ε0\frac{\sigma_0}{\varepsilon_0}ε0​σ0​​ .
  3. (C)Potential difference between the first and the last sheets of the configuration I is 5 V .
  4. (D)Potential difference between the first and the last sheets of the configuration II is zero.

Correct answer: (A)

Step-by-step solution →
Q19·PhysicsSingle correctJEE Advanced 2025
List-I shows four configurations, each consisting of a pair of ideal electric dipoles. Each dipole has a dipole moment of magnitude p, oriented as marked by arrows in the figures. In all the configurations the dipoles are fixed such that they are at a distance 2r apart along the x direction. The midpoint of the line joining the two dipoles is X. The possible resultant electric fields E⃗\vec{E}E at X are given in List-II. Choose the option that describes the correct match between the entries in List-I to those in List-II.
List-IList-II
P.see figure1.E⃗=0\vec{E} = 0E=0
Q.see figure2.E⃗=−p2πε0r3j^\vec{E} = -\frac{p}{2\pi\varepsilon_{0}r^{3}}\hat{j}E=−2πε0​r3p​j^​
R.see figure3.E⃗=−p4πε0r3(i^−j^)\vec{E} = -\frac{p}{4\pi\varepsilon_{0}r^{3}}(\hat{i} - \hat{j})E=−4πε0​r3p​(i^−j^​)
S.see figure4.E⃗=p4πε0r3(2i^−j^)\vec{E} = \frac{p}{4\pi\varepsilon_{0}r^{3}}(2\hat{i} - \hat{j})E=4πε0​r3p​(2i^−j^​)
5.E⃗=pπε0r3i^\vec{E} = \frac{p}{\pi\varepsilon_{0}r^{3}}\hat{i}E=πε0​r3p​i^
  1. (A)(P) → (3), (Q) → (1), (R) → (2), (S) → (4)
  2. (B)(P) → (4), (Q) → (5), (R) → (3), (S) → (1)
  3. (C)(P) → (2), (Q) → (1), (R) → (4), (S) → (5)
  4. (D)(P) → (2), (Q) → (1), (R) → (3), (S) → (5)

Correct answer: (C)

Step-by-step solution →
Q20·PhysicsSingle correctJEE Main 2025
Electric charge is transferred to an irregular metallic disk as shown in figure. If σ1\sigma_1σ1​, σ2\sigma_2σ2​, σ3\sigma_3σ3​ and σ4\sigma_4σ4​ are charge densities at given points then, choose the correct answer from the options given below: (A) σ1>σ3\sigma_1>\sigma_3σ1​>σ3​ ; σ2=σ4\sigma_2=\sigma_4σ2​=σ4​ (B) σ1>σ2\sigma_1>\sigma_2σ1​>σ2​ ; σ3>σ4\sigma_3>\sigma_4σ3​>σ4​ (C) σ1>σ3>σ2=σ4\sigma_1>\sigma_3>\sigma_2=\sigma_4σ1​>σ3​>σ2​=σ4​ (D) σ1<σ3<σ2=σ4\sigma_1<\sigma_3<\sigma_2=\sigma_4σ1​<σ3​<σ2​=σ4​ (E) σ1=σ2=σ3=σ4\sigma_1=\sigma_2=\sigma_3=\sigma_4σ1​=σ2​=σ3​=σ4​
  1. (A)A, B and C Only
  2. (B)A and C Only
  3. (C)D and E Only
  4. (D)B and C Only

Correct answer: (A)

Step-by-step solution →
Q21·PhysicsSingle correctJEE Main 2025
An infinitely long wire has uniform linear charge density λ=2 nC/m\lambda=2\ nC/mλ=2 nC/m. The net flux through a Gaussian cube of side length 3\sqrt33​ cm, if the wire passes through any two corners of the cube, that are maximally displaced from each other, would be x Nm2C−1x\ Nm^2C^{-1}x Nm2C−1, where xxx is: [Neglect any edge effects and use 14πε0=9×109\dfrac{1}{4\pi\varepsilon_0}=9\times10^94πε0​1​=9×109 SI units]
  1. (A)0.72π0.72\pi0.72π
  2. (B)1.44π1.44\pi1.44π
  3. (C)6.48π6.48\pi6.48π
  4. (D)2.16π2.16\pi2.16π

Correct answer: (D)

Step-by-step solution →
Q22·PhysicsSingle correctJEE Main 2025
Two metal spheres of radius RRR and 3R3R3R have same surface charge density σ\sigmaσ. If they are brought in contact and then separated, the surface charge density on smaller and bigger sphere becomes σ1\sigma_1σ1​ and σ2\sigma_2σ2​, respectively. The ratio σ1σ2\dfrac{\sigma_1}{\sigma_2}σ2​σ1​​ is:
  1. (A)19\dfrac{1}{9}91​
  2. (B)999
  3. (C)13\dfrac{1}{3}31​
  4. (D)333

Correct answer: (D)

Step-by-step solution →
Q23·PhysicsIntegerJEE Main 2025
The electric field in a region is given by E⃗=(2i^+4j^+6k^)×103\vec{E}=\left(2\hat{i}+4\hat{j}+6\hat{k}\right)\times10^{3}E=(2i^+4j^​+6k^)×103 N/C. The flux of the field through a rectangular surface parallel to the x-z plane is 6.06.06.0 Nm2^22C−1^{-1}−1. The area of the surface is __________ cm2^22.

Correct answer: 15

Step-by-step solution →
Q24·PhysicsSingle correctJEE Main 2025
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): The outer body of an aircraft is made of metal which protects persons sitting inside from lightning-strikes. Reason (R): The electric field inside the cavity enclosed by a conductor is zero. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)Both (A) and (R) are correct and (R) is the correct explanation of (A)
  2. (B)(A) is correct but (R) is not correct
  3. (C)Both (A) and (R) are correct but (R) is not correct explanation of (A)
  4. (D)(A) is not correct but (R) is correct

Correct answer: (A)

Step-by-step solution →
Q25·PhysicsSingle correctJEE Main 2025
A dipole with two electric charges of 2 μC magnitude each, with separation distance 0.5 μm, is placed between the plates of a capacitor such that its axis is parallel to an electric field established between the plates with a potential difference of 5 V applied. Separation between the plates is 0.5 mm. If the dipole is rotated by 30∘30^\circ30∘ from the axis, it tends to realign in the direction due to a torque. The value of torque is:
  1. (A)5×10−95\times10^{-9}5×10−9 Nm
  2. (B)5×10−35\times10^{-3}5×10−3 Nm
  3. (C)2.5×10−122.5\times10^{-12}2.5×10−12 Nm
  4. (D)2.5×10−92.5\times10^{-9}2.5×10−9 Nm

Correct answer: (A)

Step-by-step solution →
Q26·PhysicsSingle correctJEE Main 2025
Two infinite identical charged sheets and a charged spherical body of charge density ρ\rhoρ are arranged as shown in figure. Then the correct relation between the electrical fields at A, B, C and D points is:
  1. (A)E⃗A=E⃗B\vec{E}_A=\vec{E}_BEA​=EB​; E⃗C=E⃗D\vec{E}_C=\vec{E}_DEC​=ED​
  2. (B)E⃗A>E⃗B\vec{E}_A>\vec{E}_BEA​>EB​; E⃗C=E⃗D\vec{E}_C=\vec{E}_DEC​=ED​
  3. (C)E⃗C≠E⃗D\vec{E}_C\ne\vec{E}_DEC​=ED​; E⃗A>E⃗B\vec{E}_A>\vec{E}_BEA​>EB​
  4. (D)∣E⃗A∣=∣E⃗B∣|\vec{E}_A|=|\vec{E}_B|∣EA​∣=∣EB​∣; E⃗C>E⃗D\vec{E}_C>\vec{E}_DEC​>ED​

Correct answer: (C)

Step-by-step solution →
Q27·PhysicsSingle correctJEE Main 2025
Two small spherical balls of mass 10 g each with charges −2 μC-2\,\mu C−2μC and 2 μC2\,\mu C2μC, are attached to two ends of very light rigid rod of length 20 cm. The arrangement is now placed near an infinite non-conducting charge sheet with uniform charge density of 100 μC/m2100\,\mu C/m^2100μC/m2 such that length of rod makes an angle of 30∘30^\circ30∘ with electric field generated by charge sheet. Net torque acting on the rod is: (Take ε0:8.85×10−12\varepsilon_0:8.85\times10^{-12}ε0​:8.85×10−12 C2^22/Nm2^22)
  1. (A)112 Nm
  2. (B)1.12 Nm
  3. (C)2.24 Nm
  4. (D)11.2 Nm

Correct answer: (B)

Step-by-step solution →
Q28·PhysicsSingle correctJEE Main 2025
A metallic ring is uniformly charged as shown in the figure. AC and BD are two mutually perpendicular diameters. Electric field due to arc AB to "O" is "E" is magnitude. What would be the magnitude of electric field at "O" due to arc ABC?
  1. (A)2E2E2E
  2. (B)2 E\sqrt2\,E2​E
  3. (C)E2\dfrac{E}{2}2E​
  4. (D)Zero

Correct answer: (B)

Step-by-step solution →
Q29·PhysicsSingle correctJEE Main 2025
Consider a circular loop that is uniformly charged and has a radius a2a\sqrt2a2​. Find the position along the positive z-axis of the cartesian coordinate system where the electric field is maximum, if the ring is assumed to be placed in the xy-plane at the origin:
  1. (A)a2\dfrac{a}{\sqrt2}2​a​
  2. (B)a2\dfrac{a}{2}2a​
  3. (C)aaa
  4. (D)000

Correct answer: (C)

Step-by-step solution →
Q30·PhysicsSingle correctJEE Main 2025
A small bob of mass 100 mg and charge +10 μC+10\ \mu C+10 μC is connected to an insulating string of length 1 m. It is brought near to an infinitely long non-conducting sheet of charge density ‘σ\sigmaσ’ as shown in figure. If the string subtends an angle of 45∘45^\circ45∘ with the sheet at equilibrium, the charge density of the sheet will be: (Given ε0=8.85×10−12\varepsilon_0=8.85\times10^{-12}ε0​=8.85×10−12 F/m and acceleration due to gravity g=10g=10g=10 m/s2^22)
  1. (A)0.885 nC/m2^22
  2. (B)17.7 nC/m2^22
  3. (C)885 nC/m2^22
  4. (D)1.77 nC/m2^22

Correct answer: (D)

Step-by-step solution →
Q31·PhysicsSingle correctJEE Main 2025
Consider two infinitely large plane parallel conducting plates as shown below. The plates are uniformly charged with a surface charge density +σ+\sigma+σ and −2σ-2\sigma−2σ. The force experienced by a point charge +q+q+q placed at the mid point between two plates will be:
  1. (A)σq4ε0\dfrac{\sigma q}{4\varepsilon_0}4ε0​σq​
  2. (B)3σq2ε0\dfrac{3\sigma q}{2\varepsilon_0}2ε0​3σq​
  3. (C)3σq4ε0\dfrac{3\sigma q}{4\varepsilon_0}4ε0​3σq​
  4. (D)σq2ε0\dfrac{\sigma q}{2\varepsilon_0}2ε0​σq​

Correct answer: (B)

Step-by-step solution →
Q32·PhysicsSingle correctJEE Main 2025
A point charge +q+q+q is placed at the origin. A second point charge +9q+9q+9q is placed at (d,0,0)(d,0,0)(d,0,0) in Cartesian coordinate system. The point in between them where the electric field vanishes is:
  1. (A)(4d3,0,0)\left(\dfrac{4d}{3},0,0\right)(34d​,0,0)
  2. (B)(d4,0,0)\left(\dfrac{d}{4},0,0\right)(4d​,0,0)
  3. (C)(3d4,0,0)\left(\dfrac{3d}{4},0,0\right)(43d​,0,0)
  4. (D)(d3,0,0)\left(\dfrac{d}{3},0,0\right)(3d​,0,0)

Correct answer: (B)

Step-by-step solution →
Q33·PhysicsSingle correctJEE Main 2025
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Net dipole moment of a polar linear isotropic dielectric substance is not zero even in the absence of an external electric field. Reason (R): In absence of an external electric field, the different permanent dipoles of a polar dielectric substance are oriented in random directions. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)(A) is correct but (R) is not correct
  2. (B)Both (A) and (R) are correct but (R) is not the correct explanation of (A)
  3. (C)Both (A) and (R) are correct and (R) is the correct explanation of (A)
  4. (D)(A) is not correct but (R) is correct

Correct answer: (D)

Step-by-step solution →
Q34·PhysicsSingle correctJEE Main 2025
A point charge causes an electric flux of −2×104-2\times 10^4−2×104 Nm²C⁻¹ to pass through a spherical Gaussian surface of 8.0 cm radius, centred on the charge. The value of the point charge is: (Given ε0=8.85×10−12\varepsilon_0=8.85\times 10^{-12}ε0​=8.85×10−12 C²N⁻¹m⁻²)
  1. (A)−17.7×10−8-17.7\times 10^{-8}−17.7×10−8 C
  2. (B)−15.7×10−8-15.7\times 10^{-8}−15.7×10−8 C
  3. (C)17.7×10−817.7\times 10^{-8}17.7×10−8 C
  4. (D)15.7×10−815.7\times 10^{-8}15.7×10−8 C

Correct answer: (A)

Step-by-step solution →
Q35·PhysicsSingle correctJEE Main 2025
An electric dipole of mass m, charge q, and length lll is placed in a uniform electric field E⃗=E0i^\vec{E}=E_0\hat{i}E=E0​i^. When the dipole is rotated slightly from its equilibrium position and released, the time period of its oscillations will be:
  1. (A)12π2mlqE0\dfrac{1}{2\pi}\sqrt{\dfrac{2ml}{qE_0}}2π1​qE0​2ml​​
  2. (B)2πmlqE02\pi\sqrt{\dfrac{ml}{qE_0}}2πqE0​ml​​
  3. (C)12πml2qE0\dfrac{1}{2\pi}\sqrt{\dfrac{ml}{2qE_0}}2π1​2qE0​ml​​
  4. (D)2πml2qE02\pi\sqrt{\dfrac{ml}{2qE_0}}2π2qE0​ml​​

Correct answer: (D)

Step-by-step solution →
Q36·PhysicsSingle correctJEE Main 2025
An electric dipole is placed at a distance of 2 cm from an infinite plane sheet having positive charge density σ0\sigma_0σ0​ (orientation as shown in the figure). Choose the correct option from the following.
  1. (A)Torque on dipole is zero and net force is directed away from the sheet.
  2. (B)Torque on dipole is zero and net force acts towards the sheet.
  3. (C)Potential energy of dipole is minimum and torque is zero.
  4. (D)Potential energy and torque both are maximum.

Correct answer: (C)

Step-by-step solution →
Q37·PhysicsSingle correctJEE Main 2025
Match List-I with List-II. Choose the correct answer from the options given below:
List-IList-II
A.Electric field inside (distance r > 0 from center) of a uniformly charged spherical shell with surface charge density σ, radius RI.σε0\dfrac{\sigma}{\varepsilon_0}ε0​σ​
B.Electric field at distance r > 0 from a uniformly charged infinite plane sheet with surface charge density σII.σ2ε0\dfrac{\sigma}{2\varepsilon_0}2ε0​σ​
C.Electric field outside (distance r > 0 from center) of a uniformly charged spherical shell with surface charge density σ and radius RIII.000
D.Electric field between 2 oppositely charged infinite plane parallel sheets with uniform surface charge density σIV.σR2ε0r2\dfrac{\sigma R^2}{\varepsilon_0 r^2}ε0​r2σR2​
  1. (A)(A)-(IV), (B)-(II), (C)-(III), (D)-(II)
  2. (B)(A)-(IV), (B)-(II), (C)-(III), (D)-(I)
  3. (C)(A)-(III), (B)-(II), (C)-(IV), (D)-(I)
  4. (D)(A)-(III), (B)-(II), (C)-(IV), (D)-(II)

Correct answer: (C)

Step-by-step solution →
Q38·PhysicsIntegerJEE Main 2025
An electric dipole of dipole moment 6×10−66\times10^{-6}6×10−6 Cm is placed in uniform electric field of magnitude 10610^6106 V/m. Initially, the dipole moment is parallel to electric field. The work that needs to be done on the dipole to make its dipole moment opposite to the field, will be ______ J.

Correct answer: 12

Step-by-step solution →
Q39·PhysicsSingle correctJEE Main 2025
A particle of mass "m" and charge "q" is fastened to one end "A" of a massless string having equilibrium length ℓ\ellℓ, whose other end is fixed at point "O". The whole system is placed on a frictionless horizontal plane and is initially at rest. If uniform electric field is switched on along the direction as shown in figure, then the speed of the particle when it crosses the x-axis is
  1. (A)2qEℓm\sqrt{\frac{2qE\ell}{m}}m2qEℓ​​
  2. (B)qEℓ4m\sqrt{\frac{qE\ell}{4m}}4mqEℓ​​
  3. (C)qEℓm\sqrt{\frac{qE\ell}{m}}mqEℓ​​
  4. (D)qEℓ2m\sqrt{\frac{qE\ell}{2m}}2mqEℓ​​

Correct answer: (C)

Step-by-step solution →
Q40·PhysicsSingle correctJEE Main 2025
A small uncharged conducting sphere is placed in contact with an identical sphere but having 4×10−84\times10^{-8}4×10−8 C charge and then removed to a distance such that the force of repulsion between them is 9×10−39\times10^{-3}9×10−3 N. The distance between them is (Take 14πε0\dfrac{1}{4\pi\varepsilon_0}4πε0​1​ as 9×1099\times10^99×109 in SI units):
  1. (A)2 cm
  2. (B)3 cm
  3. (C)4 cm
  4. (D)1 cm

Correct answer: (A)

Step-by-step solution →
Q41·PhysicsIntegerJEE Main 2025
A square loop of sides a = 1 m is held normally in front of a point charge q = 1C. The flux of the electric field through the shaded region is 5p×1ε0\dfrac{5}{p}\times\dfrac{1}{\varepsilon_0}p5​×ε0​1​ Nm2^22/C, where the value of p is _______.

Correct answer: 48

Step-by-step solution →
Q42·PhysicsSingle correctJEE Main 2025
Two point charges −4 μc-4\,\mu c−4μc and 4 μc4\,\mu c4μc, constituting an electric dipole, are placed at (−9,0,0)(-9, 0, 0)(−9,0,0) cm and (9,0,0)(9, 0, 0)(9,0,0) cm in a uniform electric field of strength 104 NC−110^4\ \mathrm{NC^{-1}}104 NC−1. The work done on the dipole in rotating it from the equilibrium through 180∘180^\circ180∘ is :
  1. (A)14.414.414.4 mJ
  2. (B)18.418.418.4 mJ
  3. (C)12.412.412.4 mJ
  4. (D)16.416.416.4 mJ

Correct answer: (A)

Step-by-step solution →
Q43·PhysicsSingle correctJEE Main 2025
The electric flux is ϕ=ασ+βλ\phi=\alpha\sigma+\beta\lambdaϕ=ασ+βλ where λ\lambdaλ and σ\sigmaσ are linear and surface charge density, respectively. (αβ)\left(\dfrac{\alpha}{\beta}\right)(βα​) represents
  1. (A)charge
  2. (B)electric field
  3. (C)displacement
  4. (D)area

Correct answer: (C)

Step-by-step solution →
Q44·PhysicsSingle correctJEE Main 2025
A point particle of charge Q is located at P along the axis of an electric dipole 1 at a distance r as shown in the figure. The point P is also on the equatorial plane of a second electric dipole 2 at a distance r. The dipoles are made of opposite charges q separated by a distance 2a. For the charge particle at P not to experience any net force, which of the following correctly describes the situation?
  1. (A)ar∼20\dfrac{a}{r}\sim 20ra​∼20
  2. (B)ar∼10\dfrac{a}{r}\sim 10ra​∼10
  3. (C)ar∼0.5\dfrac{a}{r}\sim 0.5ra​∼0.5
  4. (D)ar∼3\dfrac{a}{r}\sim 3ra​∼3

Correct answer: (D)

Step-by-step solution →
Q45·PhysicsSingle correctJEE Main 2025
A line charge of length a/2 is kept at the centre of an edge BC of a cube ABCDEFGH having edge length a (see figure). If the linear charge density of the line is λ C per unit length, then the total electric flux through all the faces of the cube is: (take ε₀ = free-space permittivity)
  1. (A)λa/8ε₀
  2. (B)λa/16ε₀
  3. (C)λa/2ε₀
  4. (D)λa/4ε₀

Correct answer: (A)

Step-by-step solution →
Q46·PhysicsSingle correctJEE Main 2025
For a short dipole placed at origin O, the dipole moment P is along x-axis, as shown in the figure. If the electric potential and electric field at A are V0V_0V0​ and E0E_0E0​, respectively, then the correct combination of the electric potential and electric field, respectively, at point B on the y-axis is given by
  1. (A)V02\frac{V_0}{2}2V0​​ and E016\frac{E_0}{16}16E0​​
  2. (B)zero and E08\frac{E_0}{8}8E0​​
  3. (C)zero and E016\frac{E_0}{16}16E0​​
  4. (D)V0V_0V0​ and E04\frac{E_0}{4}4E0​​

Correct answer: (C)

Step-by-step solution →
Q47·PhysicsNumericalJEE Advanced 2024
A charge is kept at the central point P of a cylindrical region. The two edges subtend a half-angle θ at P, as shown in the figure. When θ = 30°, then the electric flux through the curved surface of the cylinder is Φ. If θ = 60°, then the electric flux through the curved surface becomes ϕn\frac{\phi}{\sqrt{n}}n​ϕ​, where the value of n is ______

Correct answer: 3

Step-by-step solution →
Q48·PhysicsMultiple correctJEE Advanced 2024
A small electric dipole p⃗0\vec{p}_0p​0​ , having a moment of inertia I about its center, is kept at a distance r from the center of a spherical shell of radius R . The surface charge density σ is uniformly distributed on the spherical shell. The dipole is initially oriented at a small angle θ as shown in the figure. While staying at a distance r , the dipole is free to rotate about its center. If released from rest, then which of the following statement(s) is(are) correct? [ε₀ is the permittivity of free space.]
  1. (A)The dipole will undergo small oscillations at any finite value of r .
  2. (B)The dipole will undergo small oscillations at any finite value of r > R .
  3. (C)The dipole will undergo small oscillations with an angular frequency of 2σp0ε0I\sqrt{\frac{2\sigma p_0}{\varepsilon_0 I}}ε0​I2σp0​​​ at r = 2R .
  4. (D)The dipole will undergo small oscillations with an angular frequency σp0100ε0I\sqrt{\frac{\sigma p_0}{100\varepsilon_0 I}}100ε0​Iσp0​​​ at r = 10R .

Correct answer: (B), (D)

Step-by-step solution →
Q49·PhysicsSingle correctJEE Main 2024
Five charges +q+q+q, +5q+5q+5q, −2q-2q−2q, +3q+3q+3q and −4q-4q−4q are situated as shown in the figure. The electric flux due to this configuration through the surface S is:
  1. (A)5qε0\dfrac{5q}{\varepsilon_{0}}ε0​5q​
  2. (B)4qε0\dfrac{4q}{\varepsilon_{0}}ε0​4q​
  3. (C)3qε0\dfrac{3q}{\varepsilon_{0}}ε0​3q​
  4. (D)qε0\dfrac{q}{\varepsilon_{0}}ε0​q​

Correct answer: (B)

Step-by-step solution →
Q50·PhysicsNumericalJEE Main 2024
At the centre of a half ring of radius R=10R=10R=10 cm and linear charge density 4n C m−14n\ \mathrm{C\ m^{-1}}4n C m−1, the potential is x π Vx\,\pi\ \mathrm{V}xπ V. The value of xxx is ______.

Correct answer: 36

Step-by-step solution →
Q51·PhysicsNumericalJEE Main 2024
An electric field E⃗=(2xi^)\vec{E}=(2x\hat{i})E=(2xi^) N/C exists in space. A cube of side 2 m is placed in the space with its edges parallel to the coordinate axes and one corner at the origin. The electric flux through the cube is _______ N m2^{2}2/C.

Correct answer: 16

Step-by-step solution →
Q52·PhysicsNumericalJEE Main 2024
An electric field, E⃗=2i^+6j^+8k^6\vec E=\dfrac{2\hat i+6\hat j+8\hat k}{\sqrt6}E=6​2i^+6j^​+8k^​ passes through the surface of 4 m24\,m^24m2 area having unit vector n^=(2i^+j^+k^6)\hat n=\left(\dfrac{2\hat i+\hat j+\hat k}{\sqrt6}\right)n^=(6​2i^+j^​+k^​). The electric flux for that surface is ___ V m.

Correct answer: 12

Step-by-step solution →
Q53·PhysicsNumericalJEE Main 2024
If the net electric field at point P along the Y axis is zero, then the ratio of q2q3\dfrac{q_2}{q_3}q3​q2​​ is 85x\dfrac{8}{5x}5x8​, where x = ________ .

Correct answer: 5

Step-by-step solution →
Q54·PhysicsNumericalJEE Main 2024
Three infinitely long charged thin sheets are placed as shown in the figure. The magnitude of the electric field at the point PPP is xσε0\tfrac{x\sigma}{\varepsilon_{0}}ε0​xσ​. The value of xxx is _______. (all quantities are measured in SI units)

Correct answer: 2

Step-by-step solution →
Q55·PhysicsSingle correctJEE Main 2024
σ\sigmaσ is the uniform surface charge density of a thin spherical shell of radius RRR. The electric field at any point on the surface of the spherical shell is:
  1. (A)σε0R\tfrac{\sigma}{\varepsilon_{0}}Rε0​σ​R
  2. (B)σ2ε0\tfrac{\sigma}{2\varepsilon_{0}}2ε0​σ​
  3. (C)σε0\tfrac{\sigma}{\varepsilon_{0}}ε0​σ​
  4. (D)σ4ε0\tfrac{\sigma}{4\varepsilon_{0}}4ε0​σ​

Correct answer: (C)

Step-by-step solution →
Q56·PhysicsSingle correctJEE Main 2024
Two identical conducting spheres P and S with charge Q on each, repel each other with a force 16N. A third identical uncharged conducting sphere R is successively brought in contact with the two spheres. The new force of repulsion between P and S is :
  1. (A)4 N
  2. (B)6 N
  3. (C)1 N
  4. (D)12 N

Correct answer: (B)

Step-by-step solution →
Q57·PhysicsNumericalJEE Main 2024
The electric field at a point P, at a distance rrr on the axial line of a short electric dipole, is E. The electric field at a point R, at a distance 2r2r2r on the equatorial line of the same dipole, will be Ex\dfrac{E}{x}xE​. The value of xxx is __________.

Correct answer: 16

Step-by-step solution →
Q58·PhysicsSingle correctJEE Main 2024
The vehicles carrying inflammable fluids usually have metallic chains touching the ground:
  1. (A)To conduct excess charge due to air friction to the ground and prevent sparking.
  2. (B)To alert other vehicles.
  3. (C)To protect tyres from catching dirt from the ground.
  4. (D)It is a custom.

Correct answer: (A)

Step-by-step solution →
Q59·PhysicsNumericalJEE Main 2024
An infinite plane sheet of charge having uniform surface charge density +σs C/m2+\sigma_s\,C/m^2+σs​C/m2 is placed on x-y plane. Another infinitely long line charge having uniform linear charge density +λs C/m+\lambda_s\,C/m+λs​C/m is placed at z=4 mz=4\,mz=4m plane and parallel to y-axis. If the magnitude when ∣σs∣=2∣λs∣|\sigma_s|=2|\lambda_s|∣σs​∣=2∣λs​∣ then at point (0,0,2)(0,0,2)(0,0,2), the ratio of magnitudes of electric field values due to sheet charge to that of line charge is πn:1\pi\sqrt n:1πn​:1. The value of nnn is ___

Correct answer: 16

Step-by-step solution →
Q60·PhysicsSingle correctJEE Main 2024
An infinitely long positively charged straight thread has a linear charge density λ Cm−1\lambda\,Cm^{-1}λCm−1. An electron revolves along a circular path having axis along the length of the wire. The graph that correctly represents the variation of the kinetic energy of electron as a function of radius of circular path from the wire is:
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (B)

Step-by-step solution →
Q61·PhysicsSingle correctJEE Main 2024
A charge qqq is placed at the center of one of the surface of a cube. The flux linked with the cube is
  1. (A)q4ε0\tfrac{q}{4\varepsilon_0}4ε0​q​
  2. (B)q2ε0\tfrac{q}{2\varepsilon_0}2ε0​q​
  3. (C)q8ε0\tfrac{q}{8\varepsilon_0}8ε0​q​
  4. (D)Zero

Correct answer: (B)

Step-by-step solution →
Q62·PhysicsSingle correctJEE Main 2024
C1C_1C1​ and C2C_2C2​ are two hollow concentric cubes enclosing charges 2Q2Q2Q and 3Q3Q3Q respectively as shown in figure. The ratio of electric flux passing through C1C_1C1​ and C2C_2C2​ is:
  1. (A)2:5
  2. (B)5:2
  3. (C)2:3
  4. (D)3:2

Correct answer: (A)

Step-by-step solution →
Q63·PhysicsNumericalJEE Main 2024
Suppose a uniformly charged wall provides a uniform electric field of 2×1042\times 10^42×104 N/C normally. A charged particle of mass 2 g being suspended through a silk thread of length 20 cm and remain stayed at a distance of 10 cm from the wall. Then the charge on the particle will be 1x μ\dfrac{1}{\sqrt x}\,\mux​1​μC where x=x=x= __________. [use g=10g=10g=10 m/s2^22]

Correct answer: 3

Step-by-step solution →
Q64·PhysicsSingle correctJEE Main 2024
Two charges q and 3q are separated by a distance 'r' in air. At a distance x from charge q, the resultant electric field is zero. The value of x is :
  1. (A)(1+3)r\dfrac{(1+\sqrt{3})}{r}r(1+3​)​
  2. (B)r3(1+3)\dfrac{r}{3(1+\sqrt{3})}3(1+3​)r​
  3. (C)r(1+3)\dfrac{r}{(1+\sqrt{3})}(1+3​)r​
  4. (D)r(1+3)r(1+\sqrt{3})r(1+3​)

Correct answer: (C)

Step-by-step solution →
Q65·PhysicsSingle correctJEE Main 2024
A particle of charge −q-q−q and mass mmm moves in a circle of radius rrr around an infinitely long line charge of linear density +λ+\lambda+λ. Then time period will be given as: (Consider kkk as Coulomb's constant)
  1. (A)T2=4π2m2kλqr3T^2=\dfrac{4\pi^2 m}{2k\lambda q}r^3T2=2kλq4π2m​r3
  2. (B)T=2πrm2kλqT=2\pi r\sqrt{\dfrac{m}{2k\lambda q}}T=2πr2kλqm​​
  3. (C)T=12πrm2kλqT=\dfrac{1}{2\pi r}\sqrt{\dfrac{m}{2k\lambda q}}T=2πr1​2kλqm​​
  4. (D)T=12π2kλqmT=\dfrac{1}{2\pi}\sqrt{\dfrac{2k\lambda q}{m}}T=2π1​m2kλq​​

Correct answer: (B)

Step-by-step solution →
Q66·PhysicsSingle correctJEE Main 2024
Two charges of 5Q5Q5Q and −2Q-2Q−2Q are situated at the points (3a,0)(3a,0)(3a,0) and (−5a,0)(-5a,0)(−5a,0) respectively. The electric flux through a sphere of radius 4a4a4a having center at origin is :
  1. (A)2Qϵ0\dfrac{2Q}{\epsilon_0}ϵ0​2Q​
  2. (B)5Qϵ0\dfrac{5Q}{\epsilon_0}ϵ0​5Q​
  3. (C)7Qϵ0\dfrac{7Q}{\epsilon_0}ϵ0​7Q​
  4. (D)3Qϵ0\dfrac{3Q}{\epsilon_0}ϵ0​3Q​

Correct answer: (B)

Step-by-step solution →
Q67·PhysicsNumericalJEE Main 2024
An electron is moving under the influence of the electric field of a uniformly charged infinite plane sheet S having surface charge density +σ+\sigma+σ. The electron at t=0t=0t=0 is at a distance of 1 m from S and has a speed of 1 m/s. The maximum value of σ\sigmaσ if the electron strikes S at t=1t=1t=1 s is α[mϵ0e]Cm2\alpha\left[\dfrac{m\epsilon_0}{e}\right]\dfrac{C}{m^2}α[emϵ0​​]m2C​ the value of α\alphaα is ______ .

Correct answer: 8

Step-by-step solution →
Q68·PhysicsSingle correctJEE Main 2024
An electric field is given by (6i^+5j^+3k^)(6\hat{i}+5\hat{j}+3\hat{k})(6i^+5j^​+3k^) N/C. The electric flux through a surface area 30i^30\hat{i}30i^ m2^{2}2 lying in YZ-plane (in SI unit) is:
  1. (A)90
  2. (B)150
  3. (C)180
  4. (D)60

Correct answer: (C)

Step-by-step solution →
Q69·PhysicsNumericalJEE Main 2024
Two charges of −4 μ-4\,\mu−4μC and +4 μ+4\,\mu+4μC are placed at the points A(1,0,4) m and B(2,-1,5) m located in an electric field E⃗=0.20 i^\vec E=0.20\,\hat iE=0.20i^ V/cm. The magnitude of the torque acting on the dipole is 8α×10−58\sqrt\alpha\times10^{-5}8α​×10−5 Nm. Where α\alphaα is ________.

Correct answer: 2.00

Step-by-step solution →
Q70·PhysicsNumericalJEE Main 2024
A thin metallic wire having cross sectional area of 10−410^{-4}10−4 m2^22 is used to make a ring of radius 30 cm. A positive charge of 2π2\pi2π C is uniformly distributed over the ring, while another positive charge of 30 pC is kept at the centre of the ring. The tension in the ring is __________ N; provided that the ring does not get deformed (neglect the influence of gravity). (given, 14πϵ0=9×109\dfrac{1}{4\pi\epsilon_0}=9\times 10^94πϵ0​1​=9×109 SI units).

Correct answer: 3

Step-by-step solution →
Q71·PhysicsSingle correctJEE Main 2023
The electric field due to a short electric dipole at a large distance (r)(r)(r) from center of dipole on the equatorial plane varies with distance as :
  1. (A)rrr
  2. (B)1r\frac{1}{r}r1​
  3. (C)1r3\frac{1}{r^{3}}r31​
  4. (D)1r2\frac{1}{r^{2}}r21​

Correct answer: (C)

Step-by-step solution →
Q72·PhysicsSingle correctJEE Main 2023
A 10 μ\muμC charge is divided into two parts and placed at 1 cm distance so that the repulsive force between them is maximum. The charges of the two parts are:
  1. (A)9 μ\muμC, 1 μ\muμC
  2. (B)5 μ\muμC, 5 μ\muμC
  3. (C)7 μ\muμC, 3 μ\muμC
  4. (D)8 μ\muμC, 2 μ\muμC

Correct answer: (B)

Step-by-step solution →
Q73·PhysicsNumericalJEE Main 2023
A thin infinite sheet charge and an infinite line charge of respective charge densities +σ+\sigma+σ and +λ+\lambda+λ are placed parallel at 5 m5\,m5m distance from each other. Points 'P' and 'Q' are at 3π m\dfrac{3}{\pi}\,mπ3​m and 4π m\dfrac{4}{\pi}\,mπ4​m perpendicular distance from line charge towards sheet charge, respectively. ′EP′'E_{P}'′EP′​ and ′EQ′'E_{Q}'′EQ′​ are the magnitudes of resultant electric field intensities at point 'P' and 'Q', respectively. If EPEQ=4a\dfrac{E_{P}}{E_{Q}}=\dfrac{4}{a}EQ​EP​​=a4​ for 2∣σ∣=∣λ∣2|\sigma|=|\lambda|2∣σ∣=∣λ∣. Then the value of aaa is _____.

Correct answer: 6

Step-by-step solution →
Q74·PhysicsNumericalJEE Main 2023
Three point charges qqq, −2q-2q−2q and 2q2q2q are placed on x-axis at a distance x=0x = 0x=0, x=34Rx = \dfrac{3}{4}Rx=43​R and x=Rx = Rx=R respectively from origin as shown. If q=2×10−6q = 2 \times 10^{-6}q=2×10−6 C and R = 2 cm, the magnitude of net force experienced by the charge −2q-2q−2q is ____ N.

Correct answer: 5440

Step-by-step solution →
Q75·PhysicsSingle correctJEE Main 2023
Two charges each of magnitude 0.01 C0.01\,C0.01C and separated by a distance of 0.4 m0.4\,m0.4m constitute an electric dipole. If the dipole is placed in an uniform electric field E⃗\vec EE of 10 dyne/C10\,dyne/C10dyne/C making 30∘30^\circ30∘ angle with E⃗\vec EE, the magnitude of torque acting on the dipole is:
  1. (A)4.0×10−5 Nm4.0\times10^{-5}\,Nm4.0×10−5Nm
  2. (B)2.0×10−5 Nm2.0\times10^{-5}\,Nm2.0×10−5Nm
  3. (C)4.0×10−8 Nm4.0\times10^{-8}\,Nm4.0×10−8Nm
  4. (D)1.5×10−5 Nm1.5\times10^{-5}\,Nm1.5×10−5Nm

Correct answer: (B)

Step-by-step solution →
Q76·PhysicsNumericalJEE Main 2023
As shown in the figure. a configuration of two equal point charges (q0=+2μq_0 = +2\muq0​=+2μ C) is placed on an inclined plane. Mass of each point charge is 20 g. Assume that there is no friction between charge and plane. For the system of two point charges to be in equilibrium (at rest) the height h=x×10−3h = x \times 10^{-3}h=x×10−3 m The value of x is ____. (Take 14πε0=9×109 N m2C−2,g=10 ms−1\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \text{ N m}^2\text{C}^{-2}, g = 10 \text{ ms}^{-1}4πε0​1​=9×109 N m2C−2,g=10 ms−1)

Correct answer: 300

Step-by-step solution →
Q77·PhysicsNumericalJEE Main 2023
An electron revolves around an infinite cylindrical wire having uniform linear charge density 2×10−82\times10^{-8}2×10−8 Cm−1^{-1}−1 in a circular path under the influence of the attractive electrostatic field as shown in the figure. The velocity with which it is revolving is 9×10x9\times10^{x}9×10x m/s. (Given mass of electron =9×10−31=9\times10^{-31}=9×10−31 kg). The value of x is _______ .

Correct answer: 8

Step-by-step solution →
Q78·PhysicsNumericalJEE Main 2023
An electric dipole of dipole moment is 6.0×10−6 Cm6.0\times10^{-6}\,\text{Cm}6.0×10−6Cm placed in a uniform electric field of 1.5×103 NC−11.5\times10^{3}\,\text{NC}^{-1}1.5×103NC−1 in such a way that dipole moment is along electric field. The work done in rotating dipole by 180∘180^\circ180∘ in this field will be ______ mJ\text{mJ}mJ

Correct answer: 18

Step-by-step solution →
Q79·PhysicsSingle correctJEE Main 2023
Graphical variation of electric field due to a uniformly charged insulating solid sphere of radius R, with distance r from the centre O is represented by:
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (A)

Step-by-step solution →
Q80·PhysicsSingle correctJEE Main 2023
A dipole comprises of two charged particles of identical magnitude qqq and opposite in nature. The mass 'mmm' of the positive charged particle is half of the mass of the negative charged particle. The two charges are separated by a distance 'lll'. If the dipole is placed in a uniform electric field 'E⃗\vec{E}E' in such a way that dipole axis makes a very small angle with the electric field 'E⃗\vec{E}E'. The angular frequency of the oscillations of the dipole when released is given by:
  1. (A)8qE3ml\sqrt{\dfrac{8qE}{3ml}}3ml8qE​​
  2. (B)4qEml\sqrt{\dfrac{4qE}{ml}}ml4qE​​
  3. (C)4qE3ml\sqrt{\dfrac{4qE}{3ml}}3ml4qE​​
  4. (D)8qEml\sqrt{\dfrac{8qE}{ml}}ml8qE​​

Correct answer: (C)

Step-by-step solution →
Q81·PhysicsNumericalJEE Main 2023
A cubical volume is bounded by the surfaces x=0, x=a, y=0, y=a, z=0, z=ax=0,\ x=a,\ y=0,\ y=a,\ z=0,\ z=ax=0, x=a, y=0, y=a, z=0, z=a. The electric field in the region is given by E⃗=E0xi^\vec E=E_0 x\hat iE=E0​xi^, where E0=4×104 NC−1m−1E_0=4\times10^4\,NC^{-1}m^{-1}E0​=4×104NC−1m−1. If a=2a=2a=2 cm, the charge contained in the cubical volume is Q×10−14Q\times10^{-14}Q×10−14 C. The value of Q is _________. (Take ε0=9×10−12 C2/Nm2\varepsilon_0=9\times10^{-12}\,C^2/Nm^2ε0​=9×10−12C2/Nm2)

Correct answer: 288

Step-by-step solution →
Q82·PhysicsNumericalJEE Main 2023
Two equal positive point charges are separated by a distance 2a2a2a. The distance of a point from the centre of the line joining two charges on the equatorial line (perpendicular bisector) at which force experienced by a test charge q0q_0q0​ becomes maximum is ax\tfrac{a}{\sqrt{x}}x​a​. The value of xxx is _______ .

Correct answer: 2

Step-by-step solution →
Q83·PhysicsSingle correctJEE Main 2023
Let σ\sigmaσ be the uniform surface charge density of two infinite thin plane sheets shown in the figure. Then the electric fields in three different regions EIE_IEI​, EIIE_{II}EII​ and EIIIE_{III}EIII​ are:
  1. (A)E⃗I=2σε0n^, E⃗II=0, E⃗III=2σε0n^\vec{E}_I=\tfrac{2\sigma}{\varepsilon_0}\hat{n},\ \vec{E}_{II}=0,\ \vec{E}_{III}=\tfrac{2\sigma}{\varepsilon_0}\hat{n}EI​=ε0​2σ​n^, EII​=0, EIII​=ε0​2σ​n^
  2. (B)E⃗I=σ2ε0n^, E⃗II=0, E⃗III=σ2ε0n^\vec{E}_I=\tfrac{\sigma}{2\varepsilon_0}\hat{n},\ \vec{E}_{II}=0,\ \vec{E}_{III}=\tfrac{\sigma}{2\varepsilon_0}\hat{n}EI​=2ε0​σ​n^, EII​=0, EIII​=2ε0​σ​n^
  3. (C)E⃗I=−σε0n^, E⃗II=0, E⃗III=σε0n^\vec{E}_I=-\tfrac{\sigma}{\varepsilon_0}\hat{n},\ \vec{E}_{II}=0,\ \vec{E}_{III}=\tfrac{\sigma}{\varepsilon_0}\hat{n}EI​=−ε0​σ​n^, EII​=0, EIII​=ε0​σ​n^
  4. (D)E⃗I=0, E⃗II=σε0n^, E⃗III=0\vec{E}_I=0,\ \vec{E}_{II}=\tfrac{\sigma}{\varepsilon_0}\hat{n},\ \vec{E}_{III}=0EI​=0, EII​=ε0​σ​n^, EIII​=0

Correct answer: (C)

Step-by-step solution →
Q84·PhysicsSingle correctJEE Main 2023
Considering a group of positive charges, which of the following statements is correct?
  1. (A)both the net potential and the net electric field cannot be zero at a point.
  2. (B)net potential of the system at a point can be zero but net electric field can't be zero at that point.
  3. (C)net potential of the system cannot be zero at a point but net electric field can be zero at that point.
  4. (D)Both the net potential and the net field can be zero at a point.

Correct answer: (C)

Step-by-step solution →
Q85·PhysicsNumericalJEE Main 2023
Expression for an electric field is given by E⃗=4000 x2 i^ Vm\vec{E} = 4000\,x^2\,\hat{i}\,\frac{\text{V}}{\text{m}}E=4000x2i^mV​. The electric flux through the cube of side 202020 cm when placed in electric field (as shown in the figure) is __________ V cm.

Correct answer: 640

Step-by-step solution →
Q86·PhysicsSingle correctJEE Main 2023
Electric field in a certain region is given by E⃗=(Ax2i^+By3j^)\vec E=\left(\dfrac{A}{x^2}\hat i+\dfrac{B}{y^3}\hat j\right)E=(x2A​i^+y3B​j^​). The SI unit of A and B are:
  1. (A)Nm3C−1; Nm2C−1Nm^3C^{-1};\ Nm^2C^{-1}Nm3C−1; Nm2C−1
  2. (B)Nm2C−1; Nm3C−1Nm^2C^{-1};\ Nm^3C^{-1}Nm2C−1; Nm3C−1
  3. (C)Nm3C; Nm2CNm^3C;\ Nm^2CNm3C; Nm2C
  4. (D)Nm2C; Nm3CNm^2C;\ Nm^3CNm2C; Nm3C

Correct answer: (B)

Step-by-step solution →
Q87·PhysicsSingle correctJEE Main 2023
A point charge QQQ is placed at the centre of a conducting spherical shell of inner radius aaa and outer radius bbb. The electric field due to charge QQQ in the three regions I (r<ar<ar<a), II (a<r<ba<r<ba<r<b) and III (r>br>br>b) is given by:
  1. (A)EI=0, EII=0, EIII=0E_I=0,\ E_{II}=0,\ E_{III}=0EI​=0, EII​=0, EIII​=0
  2. (B)EI=0, EII=0, EIII≠0E_I=0,\ E_{II}=0,\ E_{III}\neq0EI​=0, EII​=0, EIII​=0
  3. (C)EI≠0, EII=0, EIII≠0E_I\neq0,\ E_{II}=0,\ E_{III}\neq0EI​=0, EII​=0, EIII​=0
  4. (D)EI≠0, EII=0, EIII=0E_I\neq0,\ E_{II}=0,\ E_{III}=0EI​=0, EII​=0, EIII​=0

Correct answer: (C)

Step-by-step solution →
Q88·PhysicsNumericalJEE Main 2023
A cuboid with one corner at the origin and edges of length 1 m, 2 m and 3 m along the x, y and z axes respectively lies in a region with electric field E⃗=2x2i^−4yj^+6k^\vec{E}=2x^{2}\hat{i}-4y\hat{j}+6\hat{k}E=2x2i^−4yj^​+6k^ N/C. The magnitude of charge within the cuboid is nε0n\varepsilon_0nε0​ C. The value of nnn is _______.

Correct answer: 12

Step-by-step solution →
Q89·PhysicsNumericalJEE Main 2023
A point charge q1=4q0q_1 = 4q_0q1​=4q0​ is placed at the origin. Another point charge q2=−q0q_2 = -q_0q2​=−q0​ is placed at x=12x = 12x=12 cm. The charge of a proton is q0q_0q0​. The proton is placed on the x-axis so that the electrostatic force on the proton is zero. In this situation, the position of the proton from the origin is ________ cm.

Correct answer: 24

Step-by-step solution →
Q90·PhysicsSingle correctJEE Main 2023
In a cuboid of dimension 2L×2L×L2L \times 2L \times L2L×2L×L, a charge qqq is placed at the center of the surface 'SSS' having area 4L24L^24L2. The flux through the opposite surface to 'SSS' is given by:
  1. (A)q12ε0\dfrac{q}{12\varepsilon_0}12ε0​q​
  2. (B)q6ε0\dfrac{q}{6\varepsilon_0}6ε0​q​
  3. (C)q3ε0\dfrac{q}{3\varepsilon_0}3ε0​q​
  4. (D)q2ε0\dfrac{q}{2\varepsilon_0}2ε0​q​

Correct answer: (B)

Step-by-step solution →
Q91·PhysicsNumericalJEE Main 2023
For a charged spherical ball, electrostatic potential inside the ball varies with rrr as V=2ar2+bV=2ar^2+bV=2ar2+b. Here, aaa and bbb are constant and rrr is the distance from the center. The volume charge density inside the ball is −λaε-\lambda a\varepsilon−λaε. The value of λ\lambdaλ is _____. ε=\varepsilon=ε= permittivity of the medium

Correct answer: 12

Step-by-step solution →
Q92·PhysicsNumericalJEE Main 2023
A uniform electric field of 10 N/C is created between two parallel charged plates. An electron enters the field symmetrically between the plates with a kinetic energy 0.5 eV. The length of each plate is 10 cm. The angle (θ\thetaθ) of deviation of the path of the electron as it comes out of the field is _______ (in degree).

Correct answer: 45

Step-by-step solution →
Q93·PhysicsSingle correctJEE Main 2023
A point charge of 10 μC10\,\mu C10μC is placed at the origin. At what location on the X-axis should a point charge of 40 μC40\,\mu C40μC be placed so that the net electric field is zero at x=2x = 2x=2 cm on the X-axis?
  1. (A)x=−4x = -4x=−4 cm
  2. (B)x=6x = 6x=6 cm
  3. (C)x=4x = 4x=4 cm
  4. (D)x=8x = 8x=8 cm

Correct answer: (B)

Step-by-step solution →
Q94·PhysicsSingle correctJEE Main 2023
If two charges q1q_1q1​ and q2q_2q2​ are separated with distance ′d′'d'′d′ and placed in a medium of dielectric constant K. What will be the equivalent distance between charges in air for the same electrostatic force?
  1. (A)2dK2d\sqrt{K}2dK​
  2. (B)1.5 dK1.5\,d\sqrt{K}1.5dK​
  3. (C)dKd\sqrt{K}dK​
  4. (D)KdK\sqrt{d}Kd​

Correct answer: (C)

Step-by-step solution →
Q95·PhysicsNumericalJEE Main 2023
A stream of positively charged particles having qm=2×1011 Ckg\tfrac{q}{m}=2\times10^{11}\,\tfrac{\text{C}}{\text{kg}}mq​=2×1011kgC​ and velocity v⃗0=3×107 i^\vec{v}_0=3\times10^7\,\hat{i}v0​=3×107i^ m/s is deflected by an electric field 1.8 j^1.8\,\hat{j}1.8j^​ kV/m. The electric field exists in a region of 10 cm along xxx direction. Due to the electric field, the deflection of the charge particles in the yyy direction is _______ mm.

Correct answer: 2

Step-by-step solution →
Q96·PhysicsMultiple correctJEE Advanced 2022
Six charges are placed around a regular hexagon of side length a as shown in the figure. Five of them have charge q , and the remaining one has charge x . The perpendicular from each charge to the nearest hexagon side passes through the center O of the hexagon and is bisected by the side. Which of the following statement(s) is(are) correct in SI units?
  1. (A)When x = q , the magnitude of the electric field at O is zero.
  2. (B)When x = −-−q , the magnitude of the electric field at O is q6πϵ0a2\frac{q}{6\pi \epsilon_0 a^2}6πϵ0​a2q​
  3. (C)When x = 2q, the potential at O is 7q43πϵ0a\frac{7q}{4\sqrt{3}\pi \epsilon_0 a}43​πϵ0​a7q​ .
  4. (D)When x = −-− 3q , the potential at O is −3q43πϵ0a-\frac{3q}{4\sqrt{3}\pi \epsilon_0 a}−43​πϵ0​a3q​

Correct answer: (A), (B), (C)

Step-by-step solution →
Q97·PhysicsIntegerJEE Advanced 2022
A charge q is surrounded by a closed surface consisting of an inverted cone of height h and base radius R , and a hemisphere of radius R as shown in the figure. The electric flux through the conical surface is nq6ϵ0\frac{nq}{6\epsilon_0}6ϵ0​nq​ (in SI units). The value of n is_________

Correct answer: 3

Step-by-step solution →
Q98·PhysicsMultiple correctJEE Advanced 2022
In the figure, the inner (shaded) region AAA represents a sphere of radius rA=1r_A = 1rA​=1, within which the electrostatic charge density varies with the radial distance rrr from the center as ρA=kr\rho_A = krρA​=kr, where kkk is positive. In the spherical shell BBB of outer radius rBr_BrB​, the electrostatic charge density varies as ρB=2kr\rho_B = \frac{2k}{r}ρB​=r2k​. Assume that dimensions are taken care of. All physical quantities are in their SI units. Which of the following statement(s) is(are) correct?
  1. (A)If rB=32r_B = \sqrt{\frac{3}{2}}rB​=23​​, then the electric field is zero everywhere outside BBB.
  2. (B)If rB=32r_B = \frac{3}{2}rB​=23​, then the electric potential just outside BBB is kϵ0\frac{k}{\epsilon_0}ϵ0​k​.
  3. (C)If rB=2r_B = 2rB​=2, then the total charge of the configuration is 15πk15\pi k15πk.
  4. (D)If rB=52r_B = \frac{5}{2}rB​=25​, then the magnitude of the electric field just outside BBB is 13πkϵ0\frac{13\pi k}{\epsilon_0}ϵ0​13πk​.

Correct answer: (B)

Step-by-step solution →
Q99·PhysicsSingle correctJEE Main 2022
A spherically symmetric charge distribution is considered with charge density varying as ρ(r)={ρ0(34−rR)for r≤RZerofor r>R\rho(r) = \begin{cases} \rho_{0}\left(\frac{3}{4} - \frac{r}{R}\right) & \text{for } r \leq R \\ \text{Zero} & \text{for } r > R \end{cases}ρ(r)={ρ0​(43​−Rr​)Zero​for r≤Rfor r>R​ Where, r(r < R) is the distance from the centre O (as shown in figure). The electric field at point P will be :
  1. (A)ρ0r4ε0(34−rR)\frac{\rho_{0}r}{4\varepsilon_{0}}\left(\frac{3}{4} - \frac{r}{R}\right)4ε0​ρ0​r​(43​−Rr​)
  2. (B)ρ0r3ε0(34−rR)\frac{\rho_{0}r}{3\varepsilon_{0}}\left(\frac{3}{4} - \frac{r}{R}\right)3ε0​ρ0​r​(43​−Rr​)
  3. (C)ρ0r4ε0(1−rR)\frac{\rho_{0}r}{4\varepsilon_{0}}\left(1 - \frac{r}{R}\right)4ε0​ρ0​r​(1−Rr​)
  4. (D)ρ0r5ε0(1−rR)\frac{\rho_{0}r}{5\varepsilon_{0}}\left(1 - \frac{r}{R}\right)5ε0​ρ0​r​(1−Rr​)

Correct answer: (C)

Step-by-step solution →
Q100·PhysicsSingle correctJEE Main 2022
Two identical metallic spheres A and B when placed at certain distance in air repel each other with a force of F. Another identical uncharged sphere C is first placed in contact with A and then in contact with B and finally placed at midpoint between spheres A and B. The force experienced by sphere C will be :
  1. (A)3F/2
  2. (B)3F/4
  3. (C)F
  4. (D)2F

Correct answer: (B)

Step-by-step solution →
Q101·PhysicsNumericalJEE Main 2022
Two electric dipoles of dipole moments 1.2 × 10−30^{-30}−30 cm and 2.4 × 10−30^{-30}−30 cm are placed in two difference uniform electric fields of strengths 5 × 104^{4}4 NC−1^{-1}−1 and 15 × 104^{4}4 NC−1^{-1}−1 respectively. The ratio of maximum torque experienced by the electric dipoles will be 1x\frac{1}{x}x1​. The value of x is ________.

Correct answer: 6

Step-by-step solution →
Q102·PhysicsSingle correctJEE Main 2022
A uniform electric field E = (8m/e) V/m is created between two parallel plates of length 1m as shown in figure, (where m = mass of electron and e = charge of electron). An electron enters the field symmetrically between the plates with a speed of 2m/s. The angle of the deviation (θ\thetaθ) of the path of the electron as it comes out of the field will be ……….
  1. (A)tan⁡−1(4)\tan^{-1}(4)tan−1(4)
  2. (B)tan⁡−1(2)\tan^{-1}(2)tan−1(2)
  3. (C)tan⁡−1(13)\tan^{-1}\left(\dfrac{1}{3}\right)tan−1(31​)
  4. (D)tan⁡−1(3)\tan^{-1}(3)tan−1(3)

Correct answer: (B)

Step-by-step solution →
Q103·PhysicsSingle correctJEE Main 2022
A charge of 4 μC is to be divided into two. The distance between the two divided charges is constant. The magnitude of the divided charges so that the force between them is maximum, will be:
  1. (A)1 μC and 3 μC
  2. (B)2 μC and 2 μC
  3. (C)0 and 4 μC
  4. (D)1.5 μC and 2.5 μC

Correct answer: (B)

Step-by-step solution →
Q104·PhysicsNumericalJEE Main 2022
A long cylindrical volume contains a uniformly distributed charge of density ρ\rhoρ Cm−3\mathrm{Cm}^{-3}Cm−3. The electric field inside the cylindrical volume at a distance x=2ε0ρx = \frac{2\varepsilon_0}{\rho}x=ρ2ε0​​ m from its axis is ______ Vm−1\mathrm{Vm}^{-1}Vm−1

Correct answer: 1

Step-by-step solution →
Q105·PhysicsNumericalJEE Main 2022
Three point charges of magnitude 5μC5\mu C5μC, 0.16μC0.16\mu C0.16μC and 0.3μC0.3\mu C0.3μC are located at the vertices A, B, C of a right angled triangle whose sides are AB = 3cm, BC=32BC = 3\sqrt{2}BC=32​ cm and CA=3 cm and point A is the right angle corner. Charge at point A experiences ________ N of electrostatic force due to the other two charges.

Correct answer: 17

Step-by-step solution →
Q106·PhysicsNumericalJEE Main 2022
The volume charge density of a sphere of radius 6 m is 2 μC\mu CμC cm−3cm^{-3}cm−3. The number of lines of force per unit surface area coming out from the surface of the sphere is _______ ×1010\times 10^{10}×1010 NC−1NC^{-1}NC−1. [Given : Permittivity of vacuum ∈0=8.85×10−12\in_0 = 8.85 \times 10^{-12}∈0​=8.85×10−12 C2C^2C2 N−1−m−2N^{-1} - m^{-2}N−1−m−2 ]

Correct answer: 45

Step-by-step solution →
Q107·PhysicsSingle correctJEE Main 2022
A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength 1×1051 \times 10^51×105 NC−1^{-1}−1. If the charge on the particle is 40 μC and the initial velocity is 200 ms−1^{-1}−1, how much distance it will travel before coming to the rest momentarily :
  1. (A)1 m
  2. (B)5 m
  3. (C)10 m
  4. (D)0.5 m

Correct answer: (D)

Step-by-step solution →
Q108·PhysicsSingle correctJEE Main 2022
Two point charges Q each are placed at a distance d apart. A third point charge q is placed at a distance x from mid-point on the perpendicular bisector. The value of x at which charge q will experience the maximum Coulomb's force is :
  1. (A)x=dx = dx=d
  2. (B)x=d2x = \frac{d}{2}x=2d​
  3. (C)x=d2x = \frac{d}{\sqrt{2}}x=2​d​
  4. (D)x=d22x = \frac{d}{2\sqrt{2}}x=22​d​

Correct answer: (D)

Step-by-step solution →
Q109·PhysicsSingle correctJEE Main 2022
The three charges q/2, q and q/2 are placed at the corners A, B and C of a square of side 'a' as shown in figure. The magnitude of electric field (E) at the comer D of the square, is :
  1. (A)q4πϵ0a2(12+12)\frac{q}{4\pi\epsilon_{0}a^{2}}\left(\frac{1}{\sqrt{2}} + \frac{1}{2}\right)4πϵ0​a2q​(2​1​+21​)
  2. (B)q4πϵ0a2(1+12)\frac{q}{4\pi\epsilon_{0}a^{2}}\left(1 + \frac{1}{\sqrt{2}}\right)4πϵ0​a2q​(1+2​1​)
  3. (C)q4πϵ0a2(1−12)\frac{q}{4\pi\epsilon_{0}a^{2}}\left(1 - \frac{1}{\sqrt{2}}\right)4πϵ0​a2q​(1−2​1​)
  4. (D)q4πϵ0a2(12−12)\frac{q}{4\pi\epsilon_{0}a^{2}}\left(\frac{1}{\sqrt{2}} - \frac{1}{2}\right)4πϵ0​a2q​(2​1​−21​)

Correct answer: (A)

Step-by-step solution →
Q110·PhysicsSingle correctJEE Main 2022
Given below are two statements : Statement-I : A point charge is brought in an electric field. The value of electric field at a point near to the charge may increase if the charge is positive. Statement-II : An electric dipole is placed in a non-uniform electric field. The net electric force on the dipole will not be zero. Choose the correct answer from the options given below :
  1. (A)Both statement-I and statement-II are true.
  2. (B)Both statement-I and statement-I are false.
  3. (C)Statement-I is true but statement-II is false.
  4. (D)Statement-I is false but statement-II is true.

Correct answer: (A)

Step-by-step solution →
Q111·PhysicsSingle correctJEE Main 2022
Two point charges A and B of magnitude +8×10−6+8\times10^{-6}+8×10−6 C and −8×10−6-8\times10^{-6}−8×10−6 C respectively are placed at a distance d apart. The electric field at the middle point O between the charges is 6.4×1046.4\times10^46.4×104 NC−1^{-1}−1. The distance 'd' between the point charges A and B is:
  1. (A)2.0 m
  2. (B)3.0 m
  3. (C)1.0 m
  4. (D)4.0 m

Correct answer: (B)

Step-by-step solution →
Q112·PhysicsSingle correctJEE Main 2022
If a charge q is placed at the centre of a closed hemispherical non-conducting surface, the total flux passing through the flat surface would be :
  1. (A)qε0\dfrac{q}{\varepsilon_{0}}ε0​q​
  2. (B)q2ε0\dfrac{q}{2\varepsilon_{0}}2ε0​q​
  3. (C)q4ε0\dfrac{q}{4\varepsilon_{0}}4ε0​q​
  4. (D)q2πε0\dfrac{q}{2\pi\varepsilon_{0}}2πε0​q​

Correct answer: (B)

Step-by-step solution →
Q113·PhysicsSingle correctJEE Main 2022
Three identical charged balls each of charge 2C are suspended from a common point P by silk threads of 2m each (as shown in figure). They form an equilateral triangle of side 1m. The ratio of net force on a charged ball to the force between any two charged balls will be :
  1. (A)1 : 1
  2. (B)1 : 4
  3. (C)3\sqrt{3}3​ : 2
  4. (D)3\sqrt{3}3​ : 1

Correct answer: (D)

Step-by-step solution →
Q114·PhysicsSingle correctJEE Main 2022
Sixty four conducting drops each of radius 0.02 m and each carrying a charge of 5 μC are combined to form a bigger drop. The ratio of surface density of bigger drop to the smaller drop will be :
  1. (A)1 : 4
  2. (B)4 : 1
  3. (C)1 : 8
  4. (D)8 : 1

Correct answer: (B)

Step-by-step solution →
Q115·PhysicsSingle correctJEE Main 2022
In the figure, a very large plane sheet of positive charge is shown. P1P_1P1​ and P2P_2P2​ are two points at distance lll and 2l2l2l from the charge distribution. If σ is the surface charge density, then the magnitude of electric fields E1E_1E1​ and E2E_2E2​ at P1P_1P1​ and P2P_2P2​ respectively are :
  1. (A)E1=σ/ε0E_1=\sigma/\varepsilon_0E1​=σ/ε0​, E2=σ/2ε0E_2=\sigma/2\varepsilon_0E2​=σ/2ε0​
  2. (B)E1=2σ/ε0E_1=2\sigma/\varepsilon_0E1​=2σ/ε0​, E2=σ/ε0E_2=\sigma/\varepsilon_0E2​=σ/ε0​
  3. (C)E1=E2=σ/2ε0E_1=E_2=\sigma/2\varepsilon_0E1​=E2​=σ/2ε0​
  4. (D)E1=E2=σ/ε0E_1=E_2=\sigma/\varepsilon_0E1​=E2​=σ/ε0​

Correct answer: (C)

Step-by-step solution →
Q116·PhysicsSingle correctJEE Main 2022
Two identical charged particles each having a mass 10 g and charge 2.0×10−72.0\times10^{-7}2.0×10−7 C area placed on a horizontal table with a separation of L between then such that they stay in limited equilibrium. If the coefficient of friction between each particle and the table is 0.25, find the value of L. [Use g = 10 ms−2^{-2}−2]
  1. (A)12 cm
  2. (B)10 cm
  3. (C)8 cm
  4. (D)5 cm

Correct answer: (A)

Step-by-step solution →
Q117·PhysicsSingle correctJEE Main 2022
A long cylindrical volume contains a uniformly distributed charge of density ρ\rhoρ. The radius of cylindrical volume is R. A charge particle (q) revolves around the cylinder in a circular path. The kinetic of the particle is :
  1. (A)ρqR24ε0\dfrac{\rho qR^{2}}{4\varepsilon_{0}}4ε0​ρqR2​
  2. (B)ρqR22ε0\dfrac{\rho qR^{2}}{2\varepsilon_{0}}2ε0​ρqR2​
  3. (C)qρ4ε0R2\dfrac{q\rho}{4\varepsilon_{0}R^{2}}4ε0​R2qρ​
  4. (D)4ε0R2qρ\dfrac{4\varepsilon_{0}R^{2}}{q\rho}qρ4ε0​R2​

Correct answer: (A)

Step-by-step solution →
Q118·PhysicsSingle correctJEE Main 2022
A vertical electric field of magnitude 4.9×1054.9\times10^{5}4.9×105 N/C just prevents a water droplet of a mass 0.1 g from falling. The value of charge on the droplet will be : (Given g = 9.8 m/s2^{2}2)
  1. (A)1.6×10−91.6\times10^{-9}1.6×10−9 C
  2. (B)2.0×10−92.0\times10^{-9}2.0×10−9 C
  3. (C)3.2×10−93.2\times10^{-9}3.2×10−9 C
  4. (D)0.5×10−90.5\times10^{-9}0.5×10−9 C

Correct answer: (B)

Step-by-step solution →
Q119·PhysicsSingle correctJEE Main 2021
A cube is placed inside an electric field, E⃗=150y2j^\vec{E} = 150y^{2}\hat{j}E=150y2j^​. The side of the cube is 0.5 m and is placed in the field as shown in the given figure. The charge inside the cube is :
  1. (A)3.8 × 10−11^{-11}−11 C
  2. (B)8.3 × 10−11^{-11}−11 C
  3. (C)3.8 × 10−12^{-12}−12 C
  4. (D)8.3 × 10−12^{-12}−12 C

Correct answer: (B)

Step-by-step solution →
Q120·PhysicsSingle correctJEE Main 2021
Two particles A and B having charges 20 μ\muμC and −5-5−5 μ\muμC respectively are held fixed with a separation of 5 cm. At what position a third charged particle should be placed so that it does not experience a net electric force?
  1. (A)At 5 cm from 20 μ\muμC on the left side of system
  2. (B)At 5 cm from −-− 5 μ\muμC on the right side
  3. (C)At 1.25 cm from −-− 5 μ\muμC between two charges
  4. (D)At midpoint between two charges

Correct answer: (B)

Step-by-step solution →
Q121·PhysicsSingle correctJEE Main 2021
Choose the incorrect statement : (a) The electric lines of force entering into a Gaussian surface provide negative flux. (b) A charge 'q' is placed at the centre of a cube. The flux through all the faces will be the same. (c) In a uniform electric field net flux through a closed Gaussian surface containing no net charge, is zero. (d) When electric field is parallel to a Gaussian surface, it provides a finite non-zero flux. Choose the most appropriate answer from the options given below
  1. (A)(c) and (d) only
  2. (B)(b) and (d) only
  3. (C)(d) only
  4. (D)(a) and (c) only

Correct answer: (C)

Step-by-step solution →
Q122·PhysicsSingle correctJEE Main 2021
A uniformly charged disc of radius R having surface charge density σ\sigmaσ is placed in the xy plane with its center at the origin. Find the electric field intensity along the z-axis at a distance Z from origin :-
  1. (A)E=σ2ε0(1−Z(Z2+R2)1/2)E = \frac{\sigma}{2\varepsilon_0}\left(1 - \frac{Z}{(Z^2 + R^2)^{1/2}}\right)E=2ε0​σ​(1−(Z2+R2)1/2Z​)
  2. (B)E=σ2ε0(1+Z(Z2+R2)1/2)E = \frac{\sigma}{2\varepsilon_0}\left(1 + \frac{Z}{(Z^2 + R^2)^{1/2}}\right)E=2ε0​σ​(1+(Z2+R2)1/2Z​)
  3. (C)E=2ε0σ(1(Z2+R2)1/2+Z)E = \frac{2\varepsilon_0}{\sigma}\left(\frac{1}{(Z^2 + R^2)^{1/2}} + Z\right)E=σ2ε0​​((Z2+R2)1/21​+Z)
  4. (D)E=σ2ε0(1(Z2+R2)+1Z2)E = \frac{\sigma}{2\varepsilon_0}\left(\frac{1}{(Z^2 + R^2)} + \frac{1}{Z^2}\right)E=2ε0​σ​((Z2+R2)1​+Z21​)

Correct answer: (A)

Step-by-step solution →
Q123·PhysicsSingle correctJEE Main 2021
Figure shows a rod AB, which is bent in a 120° circular arc of radius R. A charge (−Q) is uniformly distributed over rod AB. What is the electric field E⃗\vec{E}E at the centre of curvature O ?
  1. (A)33 Q8πε0R2(i^)\frac{3\sqrt{3}\,Q}{8\pi\varepsilon_{0}R^{2}}(\hat{i})8πε0​R233​Q​(i^)
  2. (B)33 Q8π2ε0R2(i^)\frac{3\sqrt{3}\,Q}{8\pi^{2}\varepsilon_{0}R^{2}}(\hat{i})8π2ε0​R233​Q​(i^)
  3. (C)33 Q16π2ε0R2(i^)\frac{3\sqrt{3}\,Q}{16\pi^{2}\varepsilon_{0}R^{2}}(\hat{i})16π2ε0​R233​Q​(i^)
  4. (D)33 Q8π2ε0R2(−i^)\frac{3\sqrt{3}\,Q}{8\pi^{2}\varepsilon_{0}R^{2}}(-\hat{i})8π2ε0​R233​Q​(−i^)

Correct answer: (B)

Step-by-step solution →
Q124·PhysicsSingle correctJEE Main 2021
Two identical tennis balls each having mass 'm' and charge 'q' are suspended from a fixed point by threads of length 'ℓ'. What is the equilibrium separation when each thread makes a small angle 'θ' with the vertical ?
  1. (A)x=(q2ℓ2πε0mg)1/2x = \left( \frac{q^2 \ell}{2 \pi \varepsilon_0 mg} \right)^{1/2}x=(2πε0​mgq2ℓ​)1/2
  2. (B)x=(q2ℓ2πε0mg)1/3x = \left( \frac{q^2 \ell}{2 \pi \varepsilon_0 mg} \right)^{1/3}x=(2πε0​mgq2ℓ​)1/3
  3. (C)x=(q2ℓ22πε0m2g2)1/3x = \left( \frac{q^2 \ell^2}{2 \pi \varepsilon_0 m^2 g^2} \right)^{1/3}x=(2πε0​m2g2q2ℓ2​)1/3
  4. (D)x=(q2ℓ22πε0m2g)1/3x = \left( \frac{q^2 \ell^2}{2 \pi \varepsilon_0 m^2 g} \right)^{1/3}x=(2πε0​m2gq2ℓ2​)1/3

Correct answer: (B)

Step-by-step solution →
Q125·PhysicsSingle correctJEE Main 2021
Two ideal electric dipoles A and B, having their dipole moment p1p_1p1​ and p2p_2p2​ respectively are placed on a plane with their centres at O as shown in the figure. At point C on the axis of dipole A, the resultant electric field is making an angle of 37∘37^\circ37∘ with the axis. The ratio of the dipole moment of A and B, p1p2\frac{p_1}{p_2}p2​p1​​ is : (take sin⁡37∘=35\sin 37^\circ = \frac{3}{5}sin37∘=53​)
  1. (A)38\frac{3}{8}83​
  2. (B)32\frac{3}{2}23​
  3. (C)43\frac{4}{3}34​
  4. (D)23\frac{2}{3}32​

Correct answer: (D)

Step-by-step solution →
Q126·PhysicsNumericalJEE Main 2021
A particle of mass 1 mg and charge q is lying at the mid-point of two stationary particles kept at a distance '2 m' when each is carrying same change 'q'. If the free charged particle is displaced from its equilibrium position through distance 'x' (x << 1m). The particle executes SHM. Its angular frequency of oscillation will be ________ ×105\times 10^5×105 rad /s if q2=10C2q^2 = 10C^2q2=10C2.

Correct answer: 6000

Step-by-step solution →
Q127·PhysicsSingle correctJEE Main 2021
An electric dipole is placed on x −-− axis in proximity to a line charge of linear density 3.0×10−63.0\times10^{-6}3.0×10−6 C / m. Line charge is placed on z −-−axis and positive and negative charge of dipole is at a distance of 10 mm and 12 mm from the origin respectively. If total force of 4 N is exerted on the dipole, find out the amount of positive charge of the dipole.
  1. (A)0.485 mC
  2. (B)8.8 μC
  3. (C)815.1 nC
  4. (D)4.44 μC

Correct answer: (D)

Step-by-step solution →
Q128·PhysicsNumericalJEE Main 2021
The total charge enclosed in an incremental volume of 2×10−9 m32 \times 10^{-9}\,\mathrm{m^3}2×10−9m3 located at the origin is ________ nC, if electric flux density of its field is found as D=e−xsin⁡y i^−e−xcos⁡y j^+2z k^D = e^{-x}\sin y\,\hat{i} - e^{-x}\cos y\,\hat{j} + 2z\,\hat{k}D=e−xsinyi^−e−xcosyj^​+2zk^ C/m2^22.

Correct answer: 4

Step-by-step solution →
Q129·PhysicsSingle correctJEE Main 2021
A certain charge Q is divided into parts q and (Q − q). How should the charges Q and q be divided so that q and (Q − q) placed at a certain distance apart experience maximum electrostatic repulsion ?
  1. (A)Q=q2Q = \frac{q}{2}Q=2q​
  2. (B)Q=4qQ = 4qQ=4q
  3. (C)Q=2qQ = 2qQ=2q
  4. (D)Q=3qQ = 3qQ=3q

Correct answer: (C)

Step-by-step solution →
Q130·PhysicsNumericalJEE Main 2021
A body having specific charge 8 μC / g is resting on a frictionless plane at a distance 10 cm from the wall (as shown in the figure). It starts moving towards the wall when a uniform electric field of 100 V / m is applied horizontally towards the wall. If the collision of the body with the wall is perfectly elastic, then the time period of the motion will be ____________ s.

Correct answer: 1

Step-by-step solution →
Q131·PhysicsNumericalJEE Main 2021
An infinite number of point charges, each carrying 1 μC charge, are placed along the y-axis at y = 1 m, 2 m, 4 m, 8 m.................. The total force on a 1 C point charge, placed at the origin, is x × 103^{3}3 N. The value of x, to the nearest integer, is________. [Take 14πϵ0\frac{1}{4\pi \epsilon_{0}}4πϵ0​1​ = 9 × 109^{9}9 Nm2^{2}2/C2^{2}2]

Correct answer: 12

Step-by-step solution →
Q132·PhysicsSingle correctJEE Main 2021
An oil drop of radius 2 mm with a density 3g cm−3^{-3}−3 is held stationary under a constant electric field 3.55 × 105^{5}5 V m−1^{-1}−1 in the Millikan's oil drop experiment. What is the number of excess electrons that the oil drop will possess ? (consider g = 9.81 m/s2^{2}2)
  1. (A)48.8 × 1011^{11}11
  2. (B)1.73 × 1010^{10}10
  3. (C)17.3 × 1010^{10}10
  4. (D)1.73 × 1012^{12}12

Correct answer: (B)

Step-by-step solution →
Q133·PhysicsNumericalJEE Main 2021
The electric field in a region is given by E⃗=25E0i^+35E0j^\vec{E} = \frac{2}{5}E_{0}\hat{i} + \frac{3}{5}E_{0}\hat{j}E=52​E0​i^+53​E0​j^​ with E0_{0}0​ = 4.0 × 103^{3}3 NC\frac{N}{C}CN​ .The flux of this field through a rectangular surface area 0.4 m2^{2}2 parallel to the Y − Z plane is _________Nm2^{2}2C−1^{-1}−1.

Correct answer: 640

Step-by-step solution →
Q134·PhysicsSingle correctJEE Main 2021
Find out the surface charge density at the intersection of point x = 3 m plane and x-axis, in the region of uniform line charge of 8 nC/ m lying along the z-axis in free space.
  1. (A)0.424 nC m−2^{-2}−2
  2. (B)47.88 C/m
  3. (C)0.07 nC m−2^{-2}−2
  4. (D)4.0 nC m−2^{-2}−2

Correct answer: (A)

Step-by-step solution →
Q135·PhysicsSingle correctJEE Main 2021
Find the electric field at point P (as shown in figure) on the perpendicular bisector of a uniformly charged thin wire of length L carrying a charge Q. The distance of the point P from the centre of the rod is a = 32\frac{\sqrt{3}}{2}23​​L .
  1. (A)Q23πε0L2\frac{Q}{2\sqrt{3}\pi\varepsilon_0 L^{2}}23​πε0​L2Q​
  2. (B)3Q4πε0L2\frac{\sqrt{3}Q}{4\pi\varepsilon_0 L^{2}}4πε0​L23​Q​
  3. (C)Q3πε0L2\frac{Q}{3\pi\varepsilon_0 L^{2}}3πε0​L2Q​
  4. (D)Q4πε0L2\frac{Q}{4\pi\varepsilon_0 L^{2}}4πε0​L2Q​

Correct answer: (A)

Step-by-step solution →
Q136·PhysicsSingle correctJEE Main 2021
Given below are two statements : Statement – I : An electric dipole is placed at the centre of a hollow sphere. The flux of electric field through the sphere is zero but the electric field is not zero anywhere in the sphere. Statement – II : If R is the radius of a solid metallic sphere and Q be the total charge on it. The electric field at any point on the spherical surface of radius r ( < R) is zero but theelectric flux passing through this closed spherical surface of radius r is not zero. In the light of the above statements. Choose the correct answerfrom the option given below :
  1. (A)Statement I is true but Statement II is false
  2. (B)Statement I is false but Statement II is true
  3. (C)Both Statement I and Statement II are true
  4. (D)Both Statement I and Statement II are false

Correct answer: (A)

Step-by-step solution →
Q137·PhysicsNumericalJEE Main 2021
The electric field in a region is given by E⃗=(35E0i^+45E0j^)NC\vec{E} = \left(\dfrac{3}{5}E_0\hat{i} + \dfrac{4}{5}E_0\hat{j}\right)\dfrac{N}{C}E=(53​E0​i^+54​E0​j^​)CN​. The ratio of flux of reported field through the rectangular surface of area 0.2 m2m^2m2 (parallel to y-z plane) to that of the surface of area 0.3 m2m^2m2 (parallel to x-z plane) is a : b, where a = __________ [Here i^\hat{i}i^, j^\hat{j}j^​ and k^\hat{k}k^ are unit vectors along x, y and z-axes respectively]

Correct answer: 1

Step-by-step solution →
Q138·PhysicsNumericalJEE Main 2021
Two identical conducting spheres with negligible volume have 2.1nC and -0.1nC charges, respectively. They are brought into contact and then separated by a distance of 0.5 m. The electrostatic force acting between the spheres is _______× 10−9^{-9}−9 N. 1 [Given : 4πε0_{0}0​=9×109_{9}9​ SI unit]

Correct answer: 36

Step-by-step solution →
Q139·PhysicsSingle correctJEE Main 2021
An electron with kinetic energy K1 enters between parallel plates of a capacitor at an angle 'α’ with the plates. It leaves the plates at angle 'β' with kinetic energy K2. Then the ratio of kinetic energies K1 : K2 will be:
  1. (A)cossin^{2}^{2}βα
  2. (B)coscos^{2}^{2}αβ
  3. (C)cossinαβ
  4. (D)coscosαβ

Correct answer: (B)

Step-by-step solution →
Q140·PhysicsSingle correctJEE Main 2021
A charge 'q' is placed at one corner of a cube as shown in figure. The flux of electrostatic field E though the shaded area is: Z qY X qqqq
  1. (A)q48ε0\dfrac{q}{48\varepsilon_{0}}48ε0​q​
  2. (B)q8ε0\dfrac{q}{8\varepsilon_{0}}8ε0​q​
  3. (C)q24ε0\dfrac{q}{24\varepsilon_{0}}24ε0​q​
  4. (D)q4ε0\dfrac{q}{4\varepsilon_{0}}4ε0​q​

Correct answer: (C)

Step-by-step solution →
Q141·PhysicsNumericalJEE Main 2021
Two small spheres each of mass 10 mg are suspended from a point by threads 0.5 m long. They are equally charged and repel each other to a distance of 0.20 m . Then charge on each of the sphere is a×10−^{-}− 8^{8}8C. The value of 'a' will be________. 21

Correct answer: 20

Step-by-step solution →
Q142·PhysicsSingle correctJEE Main 2021
Two electrons each are fixed at a distance ‘2d’. A third charge proton placed at the midpoint is displaced slightly by a distance x (x<<d) perpendicular to the line joining the two fixed charges. Proton will execute simple harmonic motion having angular frequency: (m = mass of charged particle)
  1. (A)(q22πε0md3)12\left(\dfrac{q^{2}}{2\pi\varepsilon_{0}md^{3}}\right)^{\frac{1}{2}}(2πε0​md3q2​)21​
  2. (B)(πε0md32q2)12\left(\dfrac{\pi\varepsilon_{0}md^{3}}{2q^{2}}\right)^{\frac{1}{2}}(2q2πε0​md3​)21​
  3. (C)(2πε0md3q2)12\left(\dfrac{2\pi\varepsilon_{0}md^{3}}{q^{2}}\right)^{\frac{1}{2}}(q22πε0​md3​)21​
  4. (D)(2q2πε0md3)12\left(\dfrac{2q^{2}}{\pi\varepsilon_{0}md^{3}}\right)^{\frac{1}{2}}(πε0​md32q2​)21​

Correct answer: (A)

Step-by-step solution →
Q143·PhysicsSingle correctJEE Main 2021
A cube of side 'a' has point charges +Q located at each of its vertices except at the origin where the charge is –Q. The electric field at the centre of cube is :
  1. (A)2Q33πε0a2(x^+y^+z^)\frac{2Q}{3\sqrt{3}\pi\varepsilon_0 a^2}\left(\hat{x}+\hat{y}+\hat{z}\right)33​πε0​a22Q​(x^+y^​+z^)
  2. (B)Q33πε0a2(x^+y^+z^)\frac{Q}{3\sqrt{3}\pi\varepsilon_0 a^2}\left(\hat{x}+\hat{y}+\hat{z}\right)33​πε0​a2Q​(x^+y^​+z^)
  3. (C)−2Q33πε0a2(x^+y^+z^)\frac{-2Q}{3\sqrt{3}\pi\varepsilon_0 a^2}\left(\hat{x}+\hat{y}+\hat{z}\right)33​πε0​a2−2Q​(x^+y^​+z^)
  4. (D)−Q33πε0a2(x^+y^+z^)\frac{-Q}{3\sqrt{3}\pi\varepsilon_0 a^2}\left(\hat{x}+\hat{y}+\hat{z}\right)33​πε0​a2−Q​(x^+y^​+z^)

Correct answer: (C)

Step-by-step solution →
Q144·PhysicsNumericalJEE Main 2021
A point charge of +12 μ\muμC is at a distance 6 cm vertically above the centre of a square of side 12 cm as shown in figure. The magnitude of the electric flux through the square will be ______ ×103\times 10^{3}×103 Nm2^{2}2/C.

Correct answer: 226

Step-by-step solution →
Q145·PhysicsMultiple correctJEE Advanced 2020
Two identical non-conducting solid spheres of same mass and charge are suspended in air from a common point by two non-conducting, massless strings of same length. At equilibrium, the angle between the strings is α\alphaα. The spheres are now immersed in a dielectric liquid of density 800 kg m−3800\ \text{kg m}^{-3}800 kg m−3 and dielectric constant 21. If the angle between the strings remains the same after the immersion, then
  1. (A)electric force between the spheres remains unchanged
  2. (B)electric force between the spheres reduces
  3. (C)mass density of the spheres is 840 kg m−3840\ \text{kg m}^{-3}840 kg m−3
  4. (D)the tension in the strings holding the spheres remains unchanged

Correct answer: (B), (C)

Step-by-step solution →
Q146·PhysicsIntegerJEE Advanced 2020
Two large circular discs separated by a distance of 0.01 m are connected to a battery via a switch as shown in the figure. Charged oil drops of density 900 kg m−3900\ \text{kg m}^{-3}900 kg m−3 are released through a tiny hole at the center of the top disc. Once some oil drops achieve terminal velocity, the switch is closed to apply a voltage of 200 V across the discs. As a result, an oil drop of radius 8×10−78 \times 10^{-7}8×10−7 m stops moving vertically and floats between the discs. The number of electrons present in this oil drop is _________. (neglect the buoyancy force, take acceleration due to gravity =10 ms−2= 10\ \text{ms}^{-2}=10 ms−2 and charge on an electron (e) =1.6×10−19= 1.6 \times 10^{-19}=1.6×10−19 C)

Correct answer: 6

Step-by-step solution →
Q147·PhysicsNumericalJEE Advanced 2020
A circular disc of radius RRR carries surface charge density σ(r)=σ0(1−rR)\sigma(r) = \sigma_0\left(1 - \frac{r}{R}\right)σ(r)=σ0​(1−Rr​), where σ0\sigma_0σ0​ is a constant and rrr is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is ϕ0\phi_0ϕ0​. Electric flux through another spherical surface of radius R4\frac{R}{4}4R​ and concentric with the disc is ϕ\phiϕ. Then the ratio ϕ0ϕ\frac{\phi_0}{\phi}ϕϕ0​​ is_________.

Correct answer: 6.40

Step-by-step solution →
Q148·PhysicsMultiple correctJEE Advanced 2020
A uniform electric field, E⃗=−4003y^\vec{E} = -400\sqrt{3}\hat{y}E=−4003​y^​ NC−1NC^{-1}NC−1 is applied in a region. A charged particle of mass m carrying positive charge q is projected in this region with an initial speed of 210×1062\sqrt{10} \times 10^{6}210​×106 ms−1ms^{-1}ms−1. This particle is aimed to hit a target T, which is 5 m away from its entry point into the field as shown schematically in the figure. Take qm=1010\frac{q}{m} = 10^{10}mq​=1010 Ckg−1Ckg^{-1}Ckg−1. Then-
  1. (A)the particle will hit T if projected at an angle 45º from the horizontal
  2. (B)the particle will hit T if projected either at an angle 30º or 60º from the horizontal
  3. (C)time taken by the particle to hit T could be 56\sqrt{\frac{5}{6}}65​​ μs as well as 52\sqrt{\frac{5}{2}}25​​ μs
  4. (D)time taken by the particle to hit T is 53\sqrt{\frac{5}{3}}35​​ μs

Correct answer: (B), (C)

Step-by-step solution →
Q149·PhysicsSingle correctJEE Main 2020
Charge Q1Q_1Q1​ and Q2Q_2Q2​ are at point A and B of a right angle triangle OAB (see figure). The resultant electric field at point O is perpendicular to the hypotenuse, then Q1Q2\dfrac{Q_1}{Q_2}Q2​Q1​​ is proportional to:
  1. (A)x13x23\dfrac{x_1^{3}}{x_2^{3}}x23​x13​​
  2. (B)x2x1\dfrac{x_2}{x_1}x1​x2​​
  3. (C)x1x2\dfrac{x_1}{x_2}x2​x1​​
  4. (D)x22x12\dfrac{x_2^{2}}{x_1^{2}}x12​x22​​

Correct answer: (C)

Step-by-step solution →
Q150·PhysicsSingle correctJEE Main 2020
Consider the force F on a charge 'q' due to a uniformly charged spherical shell of radius R carrying charge Q distributed uniformly over it. Which one of the following statements is true for F, if 'q' is placed at distance r from the centre of the shell?
  1. (A)F=14π∈0QqR2F = \frac{1}{4\pi ∈_{0}}\frac{Qq}{R^{2}}F=4π∈0​1​R2Qq​ for r<Rr < Rr<R
  2. (B)14π∈0qQR2>F>0\frac{1}{4\pi ∈_{0}}\frac{qQ}{R^{2}} > F > 04π∈0​1​R2qQ​>F>0 for r<Rr < Rr<R
  3. (C)F=14π∈0Qqr2F = \frac{1}{4\pi ∈_{0}}\frac{Qq}{r^{2}}F=4π∈0​1​r2Qq​ for r>Rr > Rr>R
  4. (D)F=14π∈0Qqr2F = \frac{1}{4\pi ∈_{0}}\frac{Qq}{r^{2}}F=4π∈0​1​r2Qq​ for all r

Correct answer: (C)

Step-by-step solution →
Q151·PhysicsSingle correctJEE Main 2020
A particle of charge q and mass m is subjected to an electric field E=E0(1−ax2)E = E_0(1 - ax^{2})E=E0​(1−ax2) in the x-direction, where a and E0E_0E0​ are constants. Initially the particle was at rest at x=0x = 0x=0. Other then the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is:
  1. (A)3a\sqrt{\frac{3}{a}}a3​​
  2. (B)1a\sqrt{\frac{1}{a}}a1​​
  3. (C)a
  4. (D)2a\sqrt{\frac{2}{a}}a2​​

Correct answer: (A)

Step-by-step solution →
Q152·PhysicsSingle correctJEE Main 2020
Two charged thin infinite plane sheets of uniform surface charge density σ+\sigma_+σ+​ and σ−\sigma_-σ−​, where ∣σ+∣>∣σ−∣|\sigma_+| > |\sigma_-|∣σ+​∣>∣σ−​∣, intersect at right angle. Which of the following best represents the electric field lines for this system.
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (D)

Step-by-step solution →
Q153·PhysicsSingle correctJEE Main 2020
Consider a sphere of radius R which carries a uniform charge density ρ\rhoρ. If a sphere of radius R2\dfrac{R}{2}2R​ is carved out of it, as shown the ratio ∣E⃗A∣∣E⃗B∣\dfrac{|\vec{E}_{A}|}{|\vec{E}_{B}|}∣EB​∣∣EA​∣​ of magnitude of electric field E⃗A\vec{E}_{A}EA​ and E⃗B\vec{E}_{B}EB​, respectively, at point A and B due to the remaining portion is:
  1. (A)1834\dfrac{18}{34}3418​
  2. (B)1754\dfrac{17}{54}5417​
  3. (C)1854\dfrac{18}{54}5418​
  4. (D)2134\dfrac{21}{34}3421​

Correct answer: (A)

Step-by-step solution →
Q154·PhysicsNumericalJEE Main 2020
An electric field E⃗=4xi^−(y2+1)j^\vec{E}=4x\hat{i}-(y^{2}+1)\hat{j}E=4xi^−(y2+1)j^​ N/C passes through the box shown in figure. The flux of the electric field through surfaces ABCD and BCGF are marked as ϕI\phi_{I}ϕI​ and ϕII\phi_{II}ϕII​ respectively. The difference between (ϕI−ϕII\phi_{I}-\phi_{II}ϕI​−ϕII​) is (in N m2^{2}2/C) __________.

Correct answer: -48

Step-by-step solution →
Q155·PhysicsSingle correctJEE Main 2020
An electric dipole of moment p⃗=(−i^−3j^+2k^)×10−29\vec{p} = (-\hat{i} - 3\hat{j} + 2\hat{k}) \times 10^{-29}p​=(−i^−3j^​+2k^)×10−29 cm is at the origin (0, 0, 0). The electric field due to this dipole at r⃗=+i^+3j^+5k^\vec{r} = +\hat{i} + 3\hat{j} + 5\hat{k}r=+i^+3j^​+5k^ (note that r⃗⋅p⃗=0\vec{r} \cdot \vec{p} = 0r⋅p​=0) is parallel to:
  1. (A)(−i^−3j^+2k^)(-\hat{i} - 3\hat{j} + 2\hat{k})(−i^−3j^​+2k^)
  2. (B)(+i^−3j^−2k^)(+\hat{i} - 3\hat{j} - 2\hat{k})(+i^−3j^​−2k^)
  3. (C)(−i^+3j^−2k^)(-\hat{i} + 3\hat{j} - 2\hat{k})(−i^+3j^​−2k^)
  4. (D)(+i^+3j^−2k^)(+\hat{i} + 3\hat{j} - 2\hat{k})(+i^+3j^​−2k^)

Correct answer: (D)

Step-by-step solution →
Q156·PhysicsSingle correctJEE Main 2020
A particle of mass m and charge q is released from rest in a uniform electric field. If there is no other force on the particle, the dependence of its speed v on the distance x travelled by it is correctly given by (graphs are schematic and not drawn to scale)
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (D)

Step-by-step solution →
Q157·PhysicsSingle correctJEE Main 2020
Three charged particles A, B and C with charges −4q-4q−4q, 2q2q2q and −2q-2q−2q are present on the circumference of a circle of radius d. The charged particles A, C and centre O of the circle formed an equilateral triangle as shown in figure. Electric field at O along x-direction is:
  1. (A)23qπε0d2\dfrac{2\sqrt{3}q}{\pi\varepsilon_{0}d^{2}}πε0​d223​q​
  2. (B)3qπε0d2\dfrac{\sqrt{3}q}{\pi\varepsilon_{0}d^{2}}πε0​d23​q​
  3. (C)33q4πε0d2\dfrac{3\sqrt{3}q}{4\pi\varepsilon_{0}d^{2}}4πε0​d233​q​
  4. (D)3q4πε0d2\dfrac{\sqrt{3}q}{4\pi\varepsilon_{0}d^{2}}4πε0​d23​q​

Correct answer: (B)

Step-by-step solution →
Q158·PhysicsSingle correctJEE Main 2020
In finding the electric field using Gauss law the formula ∣E⃗∣=qencε0∣A∣|\vec{E}| = \dfrac{q_{enc}}{\varepsilon_{0}|A|}∣E∣=ε0​∣A∣qenc​​ is applicable. In the formula ε0\varepsilon_0ε0​ is permittivity of free space, A is the area of Gaussain surface and qencq_{enc}qenc​ is charge enclosed by the Gaussian surface. This equation can be used in which of the following situation? Only when the Gaussian surface is an
  1. (A)equipotential surface and ∣E⃗∣|\vec{E}|∣E∣ is constant on the surface.
  2. (B)For any choice of Gaussian surface.
  3. (C)Only when ∣E⃗∣|\vec{E}|∣E∣ = constant on the surface.
  4. (D)Only when the Gaussian surface is an equipotential surface.

Correct answer: (A)

Step-by-step solution →
Q159·PhysicsSingle correctJEE Main 2020
Two infinite planes each with uniform surface charge density +σ+\sigma+σ are kept in such a way that the angle between them is 30°. The electric field in the region shown between them is given by:
  1. (A)σ2ε0[(1−32)y^−x^2]\dfrac{\sigma}{2\varepsilon_0}\left[\left(1-\dfrac{\sqrt{3}}{2}\right)\hat{y}-\dfrac{\hat{x}}{2}\right]2ε0​σ​[(1−23​​)y^​−2x^​]
  2. (B)σ2ε0[(1+3)y^−x^2]\dfrac{\sigma}{2\varepsilon_0}\left[\left(1+\sqrt{3}\right)\hat{y}-\dfrac{\hat{x}}{2}\right]2ε0​σ​[(1+3​)y^​−2x^​]
  3. (C)σ2ε0[(1+3)y^+x^2]\dfrac{\sigma}{2\varepsilon_0}\left[\left(1+\sqrt{3}\right)\hat{y}+\dfrac{\hat{x}}{2}\right]2ε0​σ​[(1+3​)y^​+2x^​]
  4. (D)σε0[(1+32)y^+x^2]\dfrac{\sigma}{\varepsilon_0}\left[\left(1+\dfrac{\sqrt{3}}{2}\right)\hat{y}+\dfrac{\hat{x}}{2}\right]ε0​σ​[(1+23​​)y^​+2x^​]

Correct answer: (A)

Step-by-step solution →
Q160·PhysicsMultiple correctJEE Advanced 2019
A charged shell of radius R carries a total charge Q. Given φ as the flux of electric field through a closed cylindrical surface of height h, radius r and with its center same as that of the shell. Here, center of the cylinder is a point on the axis of the cylinder which is equidistant from its top and bottom surfaces. Which of the following option(s) is/are correct? [ε0\varepsilon_0ε0​ is permittivity of free space]
  1. (A)If h < 8R/5 and r = 3R/5 then φ = 0
  2. (B)If h > 2R and r > R then φ = Q/ε0\varepsilon_0ε0​
  3. (C)If h > 2R and r = 4R/5 then φ = Q/5ε05\varepsilon_05ε0​
  4. (D)If h > 2R and r = 3R/5 then φ = Q/5ε05\varepsilon_05ε0​

Correct answer: (A), (B), (D)

Step-by-step solution →
Q161·PhysicsSingle correctJEE Main 2019
Let a total charge 2Q be distributed in a sphere of radius R, with the charge density given by r(r) = kr, where r is the distance from the centre. Two charges A and B, of −Q each, are placed on diametrically opposite points, at equal distance, a, from the centre. If A and B do not experience any force, then :
  1. (A)a=3R21/4a = \dfrac{3R}{2^{1/4}}a=21/43R​
  2. (B)a=2−1/4Ra = 2^{-1/4}Ra=2−1/4R
  3. (C)a=8−1/4Ra = 8^{-1/4}Ra=8−1/4R
  4. (D)a=R/3a = R/\sqrt{3}a=R/3​

Correct answer: (C)

Step-by-step solution →
Q162·PhysicsSingle correctJEE Main 2019
Shown in the figure is a shell made of a conductor. It has inner radius a and outer radius b, and carries charge Q. At its centre is a dipole P⃗\vec{P}P as shown. In this case :
  1. (A)Surface charge density on the inner surface of the shell is zero everywhere.
  2. (B)Electric field outside the shell is the same as that of a point charge at the centre of the shell.
  3. (C)Surface charge density on the inner surface is uniform and equal to (Q/2)4πa2\dfrac{(Q/2)}{4\pi a^{2}}4πa2(Q/2)​
  4. (D)Surface charge density on the outer surface depends on ∣P⃗∣|\vec{P}|∣P∣

Correct answer: (B)

Step-by-step solution →
Q163·PhysicsSingle correctJEE Main 2019
Four point charges −q-q−q, +q+q+q, +q+q+q and −q-q−q are placed on y axis at y = -2d, y = -d, y = +d and y = +2d, respectively. The magnitude of the electric field E at a point on the x −-− axis at x = D, with D >>>>>> d, will vary as:
  1. (A)E∝1DE \propto \dfrac{1}{D}E∝D1​
  2. (B)E∝1D3E \propto \dfrac{1}{D^{3}}E∝D31​
  3. (C)E∝1D2E \propto \dfrac{1}{D^{2}}E∝D21​
  4. (D)E∝1D4E \propto \dfrac{1}{D^{4}}E∝D41​

Correct answer: (D)

Step-by-step solution →
Q164·PhysicsSingle correctJEE Main 2019
The bob of a simple pendulum has mass 2g and a charge of 5.0μC5.0\mu C5.0μC. It is at rest in a uniform horizontal electric field of intensity 2000Vm2000\dfrac{V}{m}2000mV​. At equilibrium, the angle that the pendulum makes with the vertical is: (take g=10ms2g=10\dfrac{m}{s^2}g=10s2m​)
  1. (A)tan⁡−1(2.0)\tan^{-1}(2.0)tan−1(2.0)
  2. (B)tan⁡−1(0.2)\tan^{-1}(0.2)tan−1(0.2)
  3. (C)tan⁡−1(5.0)\tan^{-1}(5.0)tan−1(5.0)
  4. (D)tan⁡−1(0.5)\tan^{-1}(0.5)tan−1(0.5)

Correct answer: (D)

Step-by-step solution →
Q165·PhysicsSingle correctJEE Main 2019
Determine the electric dipole moment of the system of three charges, placed on the vertices of an equilateral triangle, as shown in the figure:
  1. (A)3 qℓj^−i^2\sqrt{3}\, q\ell \frac{\hat{j} - \hat{i}}{\sqrt{2}}3​qℓ2​j^​−i^​
  2. (B)(qℓ)i^+j^2(q\ell)\frac{\hat{i} + \hat{j}}{\sqrt{2}}(qℓ)2​i^+j^​​
  3. (C)2qℓj^2q\ell \hat{j}2qℓj^​
  4. (D)−3 qℓj^-\sqrt{3}\, q\ell \hat{j}−3​qℓj^​

Correct answer: (D)

Step-by-step solution →
Q166·PhysicsSingle correctJEE Main 2019
Three charges +Q, q, +Q are placed respectively, at distance, 0, d/2 and d from the origin, on the x-axis. If the net force experienced by +Q, placed at x = 0, is zero, then value of q is
  1. (A)-Q/4
  2. (B)+Q/2
  3. (C)+Q/4
  4. (D)-Q/2

Correct answer: (A)

Step-by-step solution →
Q167·PhysicsSingle correctJEE Main 2019
Two point charges q1(10 μC)q_1(\sqrt{10}\ \mu C)q1​(10​ μC) and q2(−25 μC)q_2(-25\ \mu C)q2​(−25 μC) are placed on the x-axis at x = 1 m and x = 4 m respectively. The electric field (in V/m) at a point y = 3 m on y-axis is, [take 14πε0=9×109 Nm2C−2]\left[\text{take } \frac{1}{4\pi\varepsilon_0} = 9 \times 10^{9}\ Nm^{2}C^{-2}\right][take 4πε0​1​=9×109 Nm2C−2]
  1. (A)(63i^−27i^)×102(63\hat{i} - 27\hat{i}) \times 10^{2}(63i^−27i^)×102
  2. (B)(−63i^+27i^)×102(-63\hat{i} + 27\hat{i}) \times 10^{2}(−63i^+27i^)×102
  3. (C)(81i^−81i^)×102(81\hat{i} - 81\hat{i}) \times 10^{2}(81i^−81i^)×102
  4. (D)(−81i^+81i^)×102(-81\hat{i} + 81\hat{i}) \times 10^{2}(−81i^+81i^)×102

Correct answer: (A)

Step-by-step solution →
Q168·PhysicsSingle correctJEE Main 2019
For a uniformly charged ring of radius R, the electric field on its axis has the largest magnitude at a distance h from its centre. Then value of h is:
  1. (A)R5\dfrac{R}{\sqrt{5}}5​R​
  2. (B)R2\dfrac{R}{\sqrt{2}}2​R​
  3. (C)R
  4. (D)R2R\sqrt{2}R2​

Correct answer: (B)

Step-by-step solution →
Q169·PhysicsSingle correctJEE Advanced 2018
The electric field E is measured at a point P(0, 0, d) generated due to various charge distributions and the dependence of E on d is found to be different for different charge distributions. List-I contains different relations between EEE and d. List-II describes different electric charge distributions, along with their locations. Match the functions in List-I with the related charge distributions in List-II.
LIST-ILIST-II
P.E is independent of d1.A point charge Q at the origin
Q.E∝1dE \propto \dfrac{1}{d}E∝d1​2.A small dipole with point charges Q at (0,0,ℓ)(0, 0, \ell)(0,0,ℓ) and −Q-Q−Q at (0,0,−ℓ)(0, 0, -\ell)(0,0,−ℓ). Take 2ℓ≪d2\ell \ll d2ℓ≪d
R.E∝1d2E \propto \dfrac{1}{d^{2}}E∝d21​3.An infinite line charge coincident with the x-axis, with uniform linear charge density λ\lambdaλ
S.E∝1d3E \propto \dfrac{1}{d^{3}}E∝d31​4.Two infinite wires carrying uniform linear charge density parallel to the x- axis. The one along (y=0,z=ℓ)(y = 0, z = \ell)(y=0,z=ℓ) has a charge density +λ+\lambda+λ and the one along (y=0,z=−ℓ)(y = 0, z = -\ell)(y=0,z=−ℓ) has a charge density −λ-\lambda−λ. Take 2ℓ<<d2\ell << d2ℓ<<d
5.Infinite plane charge coincident with the xy-plane with uniform surface charge density
  1. (A)P →\rightarrow→ 5; Q →\rightarrow→ 3, 4; R →\rightarrow→ 1; S →\rightarrow→ 2
  2. (B)P →\rightarrow→ 5; Q →\rightarrow→ 3; R →\rightarrow→ 1, 4; S →\rightarrow→ 2
  3. (C)P →\rightarrow→ 5; Q →\rightarrow→ 3; R →\rightarrow→ 1, 2; S →\rightarrow→ 4
  4. (D)P →\rightarrow→ 4; Q →\rightarrow→ 2, 3; R →\rightarrow→ 1; S →\rightarrow→ 5

Correct answer: (B)

Step-by-step solution →
Q170·PhysicsNumericalJEE Advanced 2018
A particle, of mass 10−310^{-3}10−3 kg and charge 1.0 C, is initially at rest. At time t = 0, the particle comes under the influence of an electric field E⃗(t)=E0sin⁡ωt i^\vec{E}(t) = E_{0}\sin\omega t\,\hat{i}E(t)=E0​sinωti^, where E0=1.0E_{0} = 1.0E0​=1.0 NC−1^{-1}−1 and ω=103\omega = 10^{3}ω=103 rad s−1^{-1}−1. Consider the effect of only the electrical force on the particle. Then the maximum speed, in m s−1^{-1}−1, attained by the particle at subsequent times is ____________.

Correct answer: 2.00

Step-by-step solution →
Q171·PhysicsMultiple correctJEE Advanced 2018
An infinitely long thin non-conducting wire is parallel to the z-axis and carries a uniform line charge density λ\lambdaλ. It pierces a thin non-conducting spherical shell of radius RRR in such a way that the arc PQPQPQ subtends an angle 120∘120^{\circ}120∘ at the centre OOO of the spherical shell, as shown in the figure. The permittivity of free space is ε0\varepsilon_{0}ε0​. Which of the following statements is (are) true?
  1. (A)The electric flux through the shell is 3Rλ/ε0\sqrt{3}R\lambda/\varepsilon_{0}3​Rλ/ε0​
  2. (B)The z-component of the electric field is zero at all the points on the surface of the shell
  3. (C)The electric flux through the shell is 2Rλ/ε0\sqrt{2}R\lambda/\varepsilon_{0}2​Rλ/ε0​
  4. (D)The electric field is normal to the surface of the shell at all points

Correct answer: (A), (B)

Step-by-step solution →
Q172·PhysicsMultiple correctJEE Advanced 2017
A point charge +Q+Q+Q is placed just outside an imaginary hemispherical surface of radius R as shown in the figure. Which of the following statements is/are correct?
  1. (A)The electric flux passing through the *curved* surface of the hemisphere is −Q2ε0(1−12)-\dfrac{Q}{2\varepsilon_{0}}\left(1 - \dfrac{1}{\sqrt{2}}\right)−2ε0​Q​(1−2​1​)
  2. (B)Total flux through the curved and the flat surfaces is Qε0\dfrac{Q}{\varepsilon_{0}}ε0​Q​
  3. (C)The component of the electric field normal to the flat surface is constant over the surface
  4. (D)The circumference of the flat surface is an equipotential

Correct answer: (A), (D)

Step-by-step solution →
Q173·PhysicsSingle correctJEE Advanced 2016
Consider an evacuated cylindrical chamber of height h having rigid conducting plates at the ends and an insulating curved surface as shown in the figure. A number of spherical balls made of a light weight and soft material and coated with a conducting material are placed on the bottom plate. The balls have a radius r << h. Now a high voltage source (HV) is connected across the conducting plates such that the bottom plate is at +V0+V_{0}+V0​ and the top plate at −V0-V_{0}−V0​. Due to their conducting surface, the balls will get charged, will become equipotential with the plate and are repelled by it. The balls will eventually collide with the top plate, where the coefficient of restitution can be taken to be zero due to the soft nature of the material of the balls. The electric field in the chamber can be considered to be that of a parallel plate capacitor. Assume that there are no collisions between the balls and the interaction between them is negligible. (Ignore gravity) Which one of the following statements is correct?
  1. (A)The balls will bounce back to the bottom plate carrying the opposite charge they went up with
  2. (B)The balls will execute simple harmonic motion between the two plates
  3. (C)The balls will bounce back to the bottom plate carrying the same charge they went up with
  4. (D)The balls will stick to the top plate and remain there

Correct answer: (A)

Step-by-step solution →
Q174·PhysicsSingle correctJEE Advanced 2016
Consider an evacuated cylindrical chamber of height h having rigid conducting plates at the ends and an insulating curved surface as shown in the figure. A number of spherical balls made of a light weight and soft material and coated with a conducting material are placed on the bottom plate. The balls have a radius r << h. Now a high voltage source (HV) is connected across the conducting plates such that the bottom plate is at +V0+V_{0}+V0​ and the top plate at −V0-V_{0}−V0​. Due to their conducting surface, the balls will get charged, will become equipotential with the plate and are repelled by it. The balls will eventually collide with the top plate, where the coefficient of restitution can be taken to be zero due to the soft nature of the material of the balls. The electric field in the chamber can be considered to be that of a parallel plate capacitor. Assume that there are no collisions between the balls and the interaction between them is negligible. (Ignore gravity) The average current in the steady state registered by the ammeter in the circuit will be
  1. (A)proportional to V01/2V_{0}^{1/2}V01/2​
  2. (B)proportional to V02V_{0}^{2}V02​
  3. (C)proportional to the potential V0V_{0}V0​
  4. (D)zero

Correct answer: (B)

Step-by-step solution →
Q175·PhysicsMultiple correctJEE Advanced 2015
Consider a uniform spherical charge distribution of radius R1R_{1}R1​ centred at the origin O. In this distribution, a spherical cavity of radius R2R_{2}R2​, centred at P with distance OP=a=R1−R2OP = a = R_{1} - R_{2}OP=a=R1​−R2​ (see figure) is made. If the electric field inside the cavity at position r⃗\vec{r}r is E⃗(r⃗)\vec{E}(\vec{r})E(r), then the correct statement(s) is(are)
  1. (A)E⃗\vec{E}E is uniform, its magnitude is independent of R2R_{2}R2​ but its direction depends on r⃗\vec{r}r
  2. (B)E⃗\vec{E}E is uniform, its magnitude depends on R2R_{2}R2​ and its direction depends on r⃗\vec{r}r
  3. (C)E⃗\vec{E}E is uniform, its magnitude is independent of aaa but its direction depends on a⃗\vec{a}a
  4. (D)E⃗\vec{E}E is uniform and both its magnitude and direction depend on a⃗\vec{a}a

Correct answer: (D)

Step-by-step solution →
Q176·PhysicsMultiple correctJEE Advanced 2015
The figures below depict two situations in which two infinitely long static line charges of constant positive line charge density λ\lambdaλ are kept parallel to each other. In their resulting electric field, point charges qqq and −q-q−q are kept in equilibrium between them. The point charges are confined to move in the x direction only. If they are given a small displacement about their equilibrium positions, then the correct statement(s) is(are)
  1. (A)Both charges execute simple harmonic motion.
  2. (B)Both charges will continue moving in the direction of their displacement.
  3. (C)Charge +q+q+q executes simple harmonic motion while charge −q-q−q continues moving in the direction of its displacement.
  4. (D)Charge −q-q−q executes simple harmonic motion while charge +q+q+q continues moving in the direction of its displacement.

Correct answer: (C)

Step-by-step solution →
Q177·PhysicsIntegerJEE Advanced 2015
An infinitely long uniform line charge distribution of charge per unit length λ\lambdaλ lies parallel to the y-axis in the y-z plane at z=32az = \dfrac{\sqrt{3}}{2}az=23​​a (see figure). If the magnitude of the flux of the electric field through the rectangular surface ABCD lying in the x-y plane with its center at the origin is λLnε0\dfrac{\lambda L}{n\varepsilon_{0}}nε0​λL​ (ε0\varepsilon_{0}ε0​ = permittivity of free space), then the value of nnn is

Correct answer: 6

Step-by-step solution →
Q178·PhysicsMultiple correctJEE Advanced 2014
Let E1(r)E_1(r)E1​(r), E2(r)E_2(r)E2​(r) and E3(r)E_3(r)E3​(r) be the respective electric fields at a distance rrr from a point charge QQQ, an infinitely long wire with constant linear charge density λ\lambdaλ, and an infinite plane with uniform surface charge density σ\sigmaσ. If E1(r0)=E2(r0)=E3(r0)E_1(r_0) = E_2(r_0) = E_3(r_0)E1​(r0​)=E2​(r0​)=E3​(r0​) at a given distance r0r_0r0​, then
  1. (A)Q=4σπr02Q = 4\sigma\pi r_0^2Q=4σπr02​
  2. (B)r0=λ2πσr_0 = \frac{\lambda}{2\pi\sigma}r0​=2πσλ​
  3. (C)E1(r0/2)=2E2(r0/2)E_1(r_0/2) = 2E_2(r_0/2)E1​(r0​/2)=2E2​(r0​/2)
  4. (D)E2(r0/2)=4E3(r0/2)E_2(r_0/2) = 4E_3(r_0/2)E2​(r0​/2)=4E3​(r0​/2)

Correct answer: (C)

Step-by-step solution →
Q179·PhysicsSingle correctJEE Advanced 2014
Charges QQQ, 2Q2Q2Q and 4Q4Q4Q are uniformly distributed in three dielectric solid spheres 1, 2 and 3 of radii R/2R/2R/2, RRR and 2R2R2R respectively, as shown in figure. If magnitudes of the electric fields at point P at a distance RRR from the centre of spheres 1, 2 and 3 are E1E_{1}E1​, E2E_{2}E2​ and E3E_{3}E3​ respectively, then
  1. (A)E1>E2>E3E_{1} > E_{2} > E_{3}E1​>E2​>E3​
  2. (B)E3>E1>E2E_{3} > E_{1} > E_{2}E3​>E1​>E2​
  3. (C)E2>E1>E3E_{2} > E_{1} > E_{3}E2​>E1​>E3​
  4. (D)E3>E2>E1E_{3} > E_{2} > E_{1}E3​>E2​>E1​

Correct answer: (C)

Step-by-step solution →
Q180·PhysicsSingle correctJEE Advanced 2014
Four charges Q1Q_{1}Q1​, Q2Q_{2}Q2​, Q3Q_{3}Q3​ and Q4Q_{4}Q4​ of same magnitude are fixed along the xxx axis at x=−2ax = -2ax=−2a, −a-a−a, +a+a+a and +2a+2a+2a, respectively. A positive charge qqq is placed on the positive yyy axis at a distance b>0b > 0b>0. Four options of the signs of these charges are given in List I. The direction of the forces on the charge qqq is given in List II. Match List I with List II and select the correct answer using the code given below the lists.
List IList II
P.Q1Q_{1}Q1​, Q2Q_{2}Q2​, Q3Q_{3}Q3​ Q4Q_{4}Q4​ all positive1.+x+x+x
Q.Q1Q_{1}Q1​, Q2Q_{2}Q2​ positive; Q3Q_{3}Q3​, Q4Q_{4}Q4​ negative2.−x-x−x
R.Q1Q_{1}Q1​, Q4Q_{4}Q4​ positive ; Q2Q_{2}Q2​, Q3Q_{3}Q3​ negative3.+y+y+y
S.Q1Q_{1}Q1​, Q3Q_{3}Q3​ positive; Q2Q_{2}Q2​, Q4Q_{4}Q4​ negative4.−y-y−y
  1. (A)P-3, Q-1, R-4, S-2
  2. (B)P-4, Q-2, R-3, S-1
  3. (C)P-3, Q-1, R-2, S-4
  4. (D)P-4, Q-2, R-1, S-3

Correct answer: (A)

Step-by-step solution →
Q181·PhysicsMultiple correctJEE Advanced 2013
Two non-conducting solid spheres of radii R and 2R, having uniform volume charge densities ρ1\rho_{1}ρ1​ and ρ2\rho_{2}ρ2​ respectively, touch each other. The net electric field at a distance 2R from the centre of the smaller sphere, along the line joining the centre of the spheres is zero. The ratio ρ1/ρ2\rho_{1}/\rho_{2}ρ1​/ρ2​ can be
  1. (A)−4-4−4
  2. (B)−3225-\frac{32}{25}−2532​
  3. (C)3225\frac{32}{25}2532​
  4. (D)444

Correct answer: (B), (D)

Step-by-step solution →
Q182·PhysicsMultiple correctJEE Advanced 2013
Two non-conducting spheres of radii R1R_{1}R1​ and R2R_{2}R2​ and carrying uniform volume charge densities +ρ+\rho+ρ and −ρ-\rho−ρ, respectively, are placed such that they partially overlap, as shown in the figure. At all points in the overlapping region,
  1. (A)the electrostatic field is zero
  2. (B)the electrostatic potential is constant
  3. (C)the electrostatic field is constant in magnitude
  4. (D)the electrostatic field has same direction

Correct answer: (C), (D)

Step-by-step solution →

Electric Field and Coulomb's Law — frequently asked

How many questions from Electric Field and Coulomb's Law appear in JEE?

Electric Field and Coulomb's Law has appeared in 133 of the last 186 JEE Main and JEE Advanced papers — about 72% of them — contributing 182 questions in total across those papers.

Is Electric Field and Coulomb's Law an important chapter for JEE?

Judged by how often it is actually tested, it appears in roughly 72% of papers. Chapters above about 50% are effectively guaranteed to show up every session, so they repay thorough preparation; lower-frequency chapters are better treated as targeted revision.

Where do these Electric Field and Coulomb's Law questions come from?

Every question is from an official JEE Main or JEE Advanced paper, transcribed from the original paper and tagged to this chapter. Answers follow the official answer key.

Other Physics chapters

  • Properties of Solids and Liquids 344
  • Current Electricity 309
  • Rotational Motion 266
  • Geometrical Optics 259
  • Kinematics 255
  • Magnetic Field of Current 211
  • Electromagnetic Waves 208
  • Thermodynamics 201

All 28 Physics chapters →

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