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Electric Potential — JEE Previous Year Questions

Every Electric Potential question asked in JEE Main and JEE Advanced across the last 186 papers — 70 questions, each with its correct answer. Free to read, no account needed.

Questions

70

Papers it appeared in

63/186

Appearance rate

34%

All 70 Electric Potential questions

Most recent papers first.

Q1·PhysicsMultiple correctJEE Advanced 2026
Two charges Q1=qQ_{1} = qQ1​=q and Q2=mqQ_{2} = mqQ2​=mq are placed at the points P1(a,b)P_{1}(a, b)P1​(a,b) and P2(ma,mb)P_{2}(ma, mb)P2​(ma,mb), respectively, in the XYXYXY plane, where a,b≠0a, b \neq 0a,b=0 and m≠0,1m \neq 0, 1m=0,1. If V1V_{1}V1​ is the potential at a point in the XYXYXY plane due to charge Q1Q_{1}Q1​ and V2V_{2}V2​ is the potential at that point due to charge Q2Q_{2}Q2​. Correct statement(s) for the points at which ∣V1∣=∣V2∣|V_{1}| = |V_{2}|∣V1​∣=∣V2​∣ is/are:
  1. (A)For m=−1m = -1m=−1, locus of these points is ax+by=0ax + by = 0ax+by=0.
  2. (B)For m=2m = 2m=2, the locus of these points is a circle of radius 23a2+b2\frac{2}{3}\sqrt{a^{2} + b^{2}}32​a2+b2​ centered at (23a,23b)\left(\frac{2}{3}a, \frac{2}{3}b\right)(32​a,32​b)
  3. (C)For m=−2m = -2m=−2, the locus of these points is a circle of radius 2a2+b22\sqrt{a^{2} + b^{2}}2a2+b2​ centered at (2a,2b)(2a, 2b)(2a,2b)
  4. (D)For m=−3m = -3m=−3, locus of these points is 3bx+3ay=03bx + 3ay = 03bx+3ay=0.

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

Step-by-step solution →
Q2·PhysicsNumericalJEE Main 2026
A three coulomb charge moves from the point (0,−2,−5)(0, -2, -5)(0,−2,−5) to the point (5,1,2)(5, 1, 2)(5,1,2) in an electric field expressed as E⃗=2xi^+3y2j^+4k^\vec{E} = 2x\hat{i} + 3y^{2}\hat{j} + 4\hat{k}E=2xi^+3y2j^​+4k^ N/C. The work done in moving the charge is _______ J.

Correct answer: 186

Step-by-step solution →
Q3·PhysicsSingle correctJEE Main 2026
The electric potential as a function of x,yx, yx,y is given by V=5(x2−y2)V = 5(x^{2} - y^{2})V=5(x2−y2) V. The electric field at a point (2,3)(2, 3)(2,3) m is __________ V/m.
  1. (A)(−20i^+30j^)(-20\hat{i} + 30\hat{j})(−20i^+30j^​)
  2. (B)(20i^−30j^)(20\hat{i} - 30\hat{j})(20i^−30j^​)
  3. (C)(20i^+45j^)(20\hat{i} + 45\hat{j})(20i^+45j^​)
  4. (D)(−4i^+6j^)(-4\hat{i} + 6\hat{j})(−4i^+6j^​)

Correct answer: (A)

Step-by-step solution →
Q4·PhysicsSingle correctJEE Main 2026
Two metal plates (A,B)(A, B)(A,B) are kept horizontally with separation of (12π)\left(\frac{12}{\pi}\right)(π12​) cm, with plate A on the top. An atomizer jet sprays oil (density 1.5 g/cm3^{3}3) droplets of radius 1 mm horizontally. All oil droplets carry a charge 5 nC. The potentials VAV_AVA​ and VBV_BVB​ are required on plates A and B respectively in order to ensure the droplets do not descend. The values of VAV_AVA​ and VBV_BVB​ are ________. (Neglect the air resistance to the droplets and take g=10g = 10g=10 m/s2^{2}2)
  1. (A)100 V100\,V100V and 580 V580\,V580V
  2. (B)580 V580\,V580V and 100 V100\,V100V
  3. (C)60 V60\,V60V and 400 V400\,V400V
  4. (D)0 V0\,V0V and −200 V-200\,V−200V

Correct answer: (A)

Step-by-step solution →
Q5·PhysicsSingle correctJEE Main 2026
Two charged conducting spheres S1S_{1}S1​ and S2S_{2}S2​ of radii 8 cm and 18 cm are connected to each other by a wire. After equilibrium is established, the ratio of electric fields on S1S_{1}S1​ and S2S_{2}S2​ spheres are ES1E_{S_{1}}ES1​​ and ES2E_{S_{2}}ES2​​ respectively. The value of ES1ES2\frac{E_{S_{1}}}{E_{S_{2}}}ES2​​ES1​​​ is ______.
  1. (A)32\frac{3}{2}23​
  2. (B)23\frac{2}{3}32​
  3. (C)49\frac{4}{9}94​
  4. (D)94\frac{9}{4}49​

Correct answer: (D)

Step-by-step solution →
Q6·PhysicsSingle correctJEE Main 2026
Which one of the following is not a measurable quantity ?
  1. (A)Voltage difference
  2. (B)Resistance
  3. (C)Voltage
  4. (D)Displacement current

Correct answer: (C)

Step-by-step solution →
Q7·PhysicsSingle correctJEE Main 2026
Two point charges of 1 nC and 2 nC are placed at the two corners of equilateral triangle of side 3 cm. The work done in bringing a charge of 3 nC from infinity to the third corner of the triangle is _______ μJ. 14π∈0=9×109 N.m2/C2\frac{1}{4\pi \in_{0}} = 9 \times 10^{9}\,\text{N.m}^{2} / \text{C}^{2}4π∈0​1​=9×109N.m2/C2
  1. (A)2.7
  2. (B)5.4
  3. (C)3.3
  4. (D)27

Correct answer: (A)

Step-by-step solution →
Q8·PhysicsSingle correctJEE Main 2026
There are three co-centric conducting spherical shells A, B and C of radii a, b and c respectively. The potential of the spheres A, B and C respectively, are :
  1. (A)14π∈0(q1+q2+q3a),14π∈0(q1+q2+q3b),14π∈0(q1+q2+q3c)\frac{1}{4\pi \in_{0}}\left(\frac{q_{1}+q_{2}+q_{3}}{a}\right), \frac{1}{4\pi \in_{0}}\left(\frac{q_{1}+q_{2}+q_{3}}{b}\right), \frac{1}{4\pi \in_{0}}\left(\frac{q_{1}+q_{2}+q_{3}}{c}\right)4π∈0​1​(aq1​+q2​+q3​​),4π∈0​1​(bq1​+q2​+q3​​),4π∈0​1​(cq1​+q2​+q3​​)
  2. (B)14π∈0(q1+q2+q3a),14π∈0(q1+q2b+q3c),14π∈0(q1a+q2b+q3c)\frac{1}{4\pi \in_{0}}\left(\frac{q_{1}+q_{2}+q_{3}}{a}\right), \frac{1}{4\pi \in_{0}}\left(\frac{q_{1}+q_{2}}{b}+\frac{q_{3}}{c}\right), \frac{1}{4\pi \in_{0}}\left(\frac{q_{1}}{a}+\frac{q_{2}}{b}+\frac{q_{3}}{c}\right)4π∈0​1​(aq1​+q2​+q3​​),4π∈0​1​(bq1​+q2​​+cq3​​),4π∈0​1​(aq1​​+bq2​​+cq3​​)
  3. (C)14π∈0(q1a+q2b+q3c),14π∈0(q1+q2b+q3c),14π∈0(q1+q2+q3c)\frac{1}{4\pi \in_{0}}\left(\frac{q_{1}}{a}+\frac{q_{2}}{b}+\frac{q_{3}}{c}\right), \frac{1}{4\pi \in_{0}}\left(\frac{q_{1}+q_{2}}{b}+\frac{q_{3}}{c}\right), \frac{1}{4\pi \in_{0}}\left(\frac{q_{1}+q_{2}+q_{3}}{c}\right)4π∈0​1​(aq1​​+bq2​​+cq3​​),4π∈0​1​(bq1​+q2​​+cq3​​),4π∈0​1​(cq1​+q2​+q3​​)
  4. (D)14π∈0(q1a+q2b+q3c),14π∈0(q1+q2+q3b),14π∈0(q1+q2+q3c)\frac{1}{4\pi \in_{0}}\left(\frac{q_{1}}{a}+\frac{q_{2}}{b}+\frac{q_{3}}{c}\right), \frac{1}{4\pi \in_{0}}\left(\frac{q_{1}+q_{2}+q_{3}}{b}\right), \frac{1}{4\pi \in_{0}}\left(\frac{q_{1}+q_{2}+q_{3}}{c}\right)4π∈0​1​(aq1​​+bq2​​+cq3​​),4π∈0​1​(bq1​+q2​+q3​​),4π∈0​1​(cq1​+q2​+q3​​)

Correct answer: (C)

Step-by-step solution →
Q9·PhysicsSingle correctJEE Main 2026
Two charges 7μC and −2 μC are placed at (−9, 0, 0) cm and (9, 0, 0) cm respectively in an external field E=Ar2r^E = \frac{A}{r^{2}}\hat{r}E=r2A​r^ , where A=9×105A = 9 \times 10^{5}A=9×105 N/C.m2^{2}2. Considering the potential at infinity is 0, the electrostatic energy of the configuration is_____J.
  1. (A)1.4
  2. (B)−90.7
  3. (C)49.3
  4. (D)24.3

Correct answer: (C)

Step-by-step solution →
Q10·PhysicsSingle correctJEE Main 2026
Electric field in a region is given by E⃗=Axi^+Byj^\vec{E} = Ax\hat{i} + By\hat{j}E=Axi^+Byj^​, where AAA = 10 V/m2^22 and BBB = 5 V/m2^22. If the electric potential at a point (10, 20) is 500 V, then the electric potential at origin is ______ V.
  1. (A)1000
  2. (B)500
  3. (C)2000
  4. (D)0

Correct answer: (C)

Step-by-step solution →
Q11·PhysicsSingle correctJEE Main 2026
Three small identical bubbles of water having same charge on each coalesce to form a bigger bubble. Then the ratio of the potentials on one initial bubble and that on the resultant bigger bubble is :
  1. (A)1 : 31/3^{1/3}1/3
  2. (B)1 : 22/3^{2/3}2/3
  3. (C)32/3^{2/3}2/3 : 1
  4. (D)1 : 32/3^{2/3}2/3

Correct answer: (D)

Step-by-step solution →
Q12·PhysicsSingle correctJEE Main 2026
A point charge of 10−810^{-8}10−8 C is placed at origin. The work done in moving a point charge 2 μC from point A(4, 4, 2) m to B(2, 2, 1) m is ___________J. (14π∈0=9×109(\frac{1}{4\pi \in_0} = 9 \times 10^{9}(4π∈0​1​=9×109 in SI units)
  1. (A)45×10−645 \times 10^{-6}45×10−6
  2. (B)0
  3. (C)30×10−630 \times 10^{-6}30×10−6
  4. (D)15×10−615 \times 10^{-6}15×10−6

Correct answer: (C)

Step-by-step solution →
Q13·PhysicsSingle correctJEE Main 2026
Consider two identical metallic spheres of radius RRR each having charge QQQ and mass mmm. Their centers have an initial separation of 4R4R4R. Both the spheres are given an initial speed of uuu towards each other. The minimum value of u, so that they can just touch each other is : (Take k=14π∈0k = \frac{1}{4\pi \in_0}k=4π∈0​1​ and assume kQ2>Gm2kQ^2 > Gm^2kQ2>Gm2 where G is the Gravitational constant)
  1. (A)kQ24mR(1−Gm2kQ2)\sqrt{\frac{kQ^2}{4mR}\left(1 - \frac{Gm^2}{kQ^2}\right)}4mRkQ2​(1−kQ2Gm2​)​
  2. (B)kQ24mR(1+Gm2kQ2)\sqrt{\frac{kQ^2}{4mR}\left(1 + \frac{Gm^2}{kQ^2}\right)}4mRkQ2​(1+kQ2Gm2​)​
  3. (C)kQ22mR(1−Gm2kQ2)\sqrt{\frac{kQ^2}{2mR}\left(1 - \frac{Gm^2}{kQ^2}\right)}2mRkQ2​(1−kQ2Gm2​)​
  4. (D)kQ22mR(1−Gm22kQ2)\sqrt{\frac{kQ^2}{2mR}\left(1 - \frac{Gm^2}{2kQ^2}\right)}2mRkQ2​(1−2kQ2Gm2​)​

Correct answer: (A)

Step-by-step solution →
Q14·PhysicsMultiple correctJEE Advanced 2025
A positive point charge of 10−810^{-8}10−8 C is kept at a distance of 20 cm from the center of a neutral conducting sphere of radius 10 cm. The sphere is then grounded and the charge on the sphere is measured. The grounding is then removed and subsequently the point charge is moved by a distance of 10 cm further away from the center of the sphere along the radial direction. Taking 14πε0=9×109\frac{1}{4\pi\varepsilon_0} = 9\times 10^94πε0​1​=9×109 Nm2^22/C2^22 (where ε0\varepsilon_0ε0​ is the permittivity of free space), which of the following statements is/are correct:
  1. (A)Before the grounding, the electrostatic potential of the sphere is 450 V .
  2. (B)Charge flowing from the sphere to the ground because of grounding is 5×10−95\times 10^{-9}5×10−9 C.
  3. (C)After the grounding is removed, the charge on the sphere is −5×10−9-5\times 10^{-9}−5×10−9 C.
  4. (D)The final electrostatic potential of the sphere is 300 V.

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

Step-by-step solution →
Q15·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: Work done in moving a test charge between two points inside a uniformly charged spherical shell is zero, no matter which path is chosen. Reason R: Electrostatic potential inside a uniformly charged spherical shell is constant and is same as that on the surface of the shell. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)A is true but R is false
  2. (B)Both A and R are true and R is the correct explanation of A
  3. (C)A is false but R is true
  4. (D)Both A and R are true but R is NOT the correct explanation of A

Correct answer: (B)

Step-by-step solution →
Q16·PhysicsSingle correctJEE Main 2025
Two charges q1q_1q1​ and q2q_2q2​ are separated by a distance of 30 cm. A third charge q3q_3q3​ initially at C as shown in the figure, is moved along the circular path of radius 40 cm from C to D. If the difference in potential energy due to movement of q3q_3q3​ from C to D is given by q3K4πε0\dfrac{q_3K}{4\pi\varepsilon_0}4πε0​q3​K​, the value of K is:
  1. (A)8q28q_28q2​
  2. (B)6q26q_26q2​
  3. (C)8q18q_18q1​
  4. (D)6q16q_16q1​

Correct answer: (A)

Step-by-step solution →
Q17·PhysicsSingle correctJEE Main 2025
The electrostatic potential on the surface of a uniformly charged spherical shell of radius R=10R=10R=10 cm is 120 V. The potential at the centre of the shell, at a distance r=5r=5r=5 cm from the centre, and at a distance r=15r=15r=15 cm from the centre of the shell respectively, are:
  1. (A)120V, 120V, 80V
  2. (B)40V, 40V, 80V
  3. (C)0V, 0V, 80V
  4. (D)0V, 120V, 40V

Correct answer: (A)

Step-by-step solution →
Q18·PhysicsSingle correctJEE Main 2025
Two large plane parallel conducting plates are kept 10 cm apart as shown in the figure. The potential difference between them is V. The potential difference between the points A and B (shown in the figure) is:
  1. (A)14V\dfrac{1}{4}V41​V
  2. (B)25V\dfrac{2}{5}V52​V
  3. (C)34V\dfrac{3}{4}V43​V
  4. (D)1 V1\,V1V

Correct answer: (B)

Step-by-step solution →
Q19·PhysicsSingle correctJEE Main 2025
Three infinitely long wires with linear charge density λ\lambdaλ are placed along the x-axis, y-axis and z-axis respectively. Which of the following denotes an equipotential surface?
  1. (A)xy+yz+zx=xy+yz+zx=xy+yz+zx= constant
  2. (B)(x+y)(y+z)(z+x)=(x+y)(y+z)(z+x)=(x+y)(y+z)(z+x)= constant
  3. (C)(x2+y2)(y2+z2)(z2+x2)=(x^2+y^2)(y^2+z^2)(z^2+x^2)=(x2+y2)(y2+z2)(z2+x2)= constant
  4. (D)xyz=xyz=xyz= constant

Correct answer: (C)

Step-by-step solution →
Q20·PhysicsSingle correctJEE Main 2025
In the first configuration (1) as shown in the figure, four identical charges (q0)(q_0)(q0​) are kept at the corners A, B, C and D of a square of side length "a". In the second configuration (2), the same charges are shifted to mid points G, E, F and H, of the sides of the square. If K=14πε0K=\dfrac{1}{4\pi\varepsilon_0}K=4πε0​1​, the difference between the potential energy of configuration (2) and (1) is given by:
  1. (A)Kq02a(42−2)\dfrac{Kq_0^2}{a}\left(4\sqrt2-2\right)aKq02​​(42​−2)
  2. (B)Kq02a(3−2)\dfrac{Kq_0^2}{a}\left(3-\sqrt2\right)aKq02​​(3−2​)
  3. (C)Kq02a(4−22)\dfrac{Kq_0^2}{a}\left(4-2\sqrt2\right)aKq02​​(4−22​)
  4. (D)Kq02a(32−2)\dfrac{Kq_0^2}{a}\left(3\sqrt2-2\right)aKq02​​(32​−2)

Correct answer: (D)

Step-by-step solution →
Q21·PhysicsSingle correctJEE Main 2025
Two charges 7 μc7\,\mu c7μc and −4 μc-4\,\mu c−4μc are placed at (−7 cm,0,0)(-7\text{ cm}, 0, 0)(−7 cm,0,0) and (7 cm,0,0)(7\text{ cm}, 0, 0)(7 cm,0,0) respectively. Given, ε0=8.85×10−12 C2 N−1 m−2\varepsilon_0=8.85\times10^{-12}\ \mathrm{C^2\,N^{-1}\,m^{-2}}ε0​=8.85×10−12 C2N−1m−2, the electrostatic potential energy of the charge configuration is :
  1. (A)−1.5-1.5−1.5 J
  2. (B)−2.0-2.0−2.0 J
  3. (C)−1.2-1.2−1.2 J
  4. (D)−1.8-1.8−1.8 J

Correct answer: (D)

Step-by-step solution →
Q22·PhysicsNumericalJEE Advanced 2024
An infinitely long thin wire, having a uniform charge density per unit length of 5nC/m, is passing through a spherical shell of radius 1m, as shown in the figure. A 10nC charge is distributed uniformly over the spherical shell. If the configuration of the charges remains static, the magnitude of the potential difference between points P and R, in Volt, is [Given: In SI units 14πε0\frac{1}{4\pi\varepsilon_0}4πε0​1​ = 9 × 10910^9109, ln 2 = 0.7. Ignore the area pierced by the wire.]

Correct answer: 171

Step-by-step solution →
Q23·PhysicsSingle correctJEE Main 2024
Two charged conducting spheres of radii aaa and bbb are connected to each other by a conducting wire. The ratio of charges of the two spheres respectively is:
  1. (A)ab\sqrt{ab}ab​
  2. (B)ababab
  3. (C)ab\dfrac{a}{b}ba​
  4. (D)ba\dfrac{b}{a}ab​

Correct answer: (C)

Step-by-step solution →
Q24·PhysicsNumericalJEE Main 2024
The distance between charges +q+q+q and −q-q−q is 2ℓ2\ell2ℓ and between +2q+2q+2q and −2q-2q−2q is 4ℓ4\ell4ℓ. The electrostatic potential at point P at a distance rrr from centre O is −αqℓr2×109-\alpha\dfrac{q\ell}{r^2}\times10^9−αr2qℓ​×109 V, where the value of α\alphaα is ______. (Use 14πε0=9×109\dfrac{1}{4\pi\varepsilon_0}=9\times10^94πε0​1​=9×109 Nm2^22C−2^{-2}−2)

Correct answer: 27

Step-by-step solution →
Q25·PhysicsSingle correctJEE Main 2024
The electrostatic potential due to an electric dipole at a distance rrr varies as:
  1. (A)rrr
  2. (B)1r2\dfrac{1}{r^2}r21​
  3. (C)1r3\dfrac{1}{r^3}r31​
  4. (D)1r\dfrac{1}{r}r1​

Correct answer: (B)

Step-by-step solution →
Q26·PhysicsSingle correctJEE Main 2024
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Work done by electric field on moving a positive charge on an equipotential surface is always zero. Reason (R) : Electric lines of forces are always perpendicular to equipotential surfaces. 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 but (R) is not the correct explanation of (A)
  2. (B)(A) is correct but (R) is not correct
  3. (C)(A) is not correct but (R) is correct
  4. (D)Both (A) and (R) are correct and (R) is the correct explanation of (A)

Correct answer: (D)

Step-by-step solution →
Q27·PhysicsNumericalJEE Main 2024
The electric potential at the surface of an atomic nucleus (Z=50)(Z=50)(Z=50) of radius 9×10−139\times10^{-13}9×10−13 cm is ________ ×106\times10^6×106 V.

Correct answer: 8.00

Step-by-step solution →
Q28·PhysicsSingle correctJEE Main 2024
An electric charge 10−6 μ10^{-6}\,\mu10−6μC is placed at origin (0, 0) m of X–Y co-ordinate system. Two points P and Q are situated at (3,3)(\sqrt3,\sqrt3)(3​,3​) m and (6,0)(\sqrt6,0)(6​,0) m respectively. The potential difference between the points P and Q will be:
  1. (A)3\sqrt33​ V
  2. (B)6\sqrt66​ V
  3. (C)0 V
  4. (D)3 V

Correct answer: (C)

Step-by-step solution →
Q29·PhysicsSingle correctJEE Advanced 2023
An electric dipole is formed by two charges +q and - q located in xy-plane at (0, 2) mm and (0, -2) mm, respectively, as shown in the figure. The electric potential at point P (100, 100) mm due to the dipole is V0V_0V0​. The charges +q and -q are then moved to the points (-1, 2) mm and (1, -2) mm, respectively. What is the value of electric potential at P due to the new dipole?
  1. (A)V0/4V_0/4V0​/4
  2. (B)V0/2V_0/2V0​/2
  3. (C)V0/2V_0 / \sqrt{2}V0​/2​
  4. (D)3V0/43V_0/43V0​/4

Correct answer: (B)

Step-by-step solution →
Q30·PhysicsNumericalJEE Main 2023
64 identical drops each charged upto potential of 10 mV are combined to form a bigger drop. The potential of the bigger drop will be _________ mV.

Correct answer: 160

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Q31·PhysicsNumericalJEE Main 2023
Three concentric spherical metallic shells X, Y and Z of radius a, b and c respectively [a<b<c][a<b<c][a<b<c] have surface charge densities σ,−σ\sigma, -\sigmaσ,−σ and σ\sigmaσ respectively. The shells X and Z are at same potential. If the radii of X and Y are 2 cm and 3 cm respectively, the radius of shell Z is _________ cm.

Correct answer: 5

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Q32·PhysicsSingle correctJEE Main 2023
Electric potential at a point P due to a point charge of 5×10−95\times10^{-9}5×10−9 C is 50 V. The distance of P from the point charge is: (Assume 14πε0=9×109\dfrac{1}{4\pi\varepsilon_0}=9\times10^{9}4πε0​1​=9×109 Nm2^22C−2^{-2}−2)
  1. (A)3 cm
  2. (B)9 cm
  3. (C)90 cm
  4. (D)0.9 cm

Correct answer: (C)

Step-by-step solution →
Q33·PhysicsSingle correctJEE Main 2023
For a uniformly charged thin spherical shell, the electric potential (V)(V)(V) radially away from the centre (O)(O)(O) of shell can be graphically represented as:
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (A)

Step-by-step solution →
Q34·PhysicsSingle correctJEE Main 2023
Which of the following correctly represents the variation of electric potential (VVV) of a charged spherical conductor of radius (RRR) with radial distance (rrr) from the center?
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (A)

Step-by-step solution →
Q35·PhysicsSingle correctJEE Main 2023
Two isolated metallic solid spheres of radii RRR and 2R2R2R are charged such that both have same charge density σ\sigmaσ. The spheres are then connected by a thin conducting wire. If the new charge density of the bigger sphere is σ′\sigma'σ′, the ratio σ′σ\dfrac{\sigma'}{\sigma}σσ′​ is:
  1. (A)43\dfrac4334​
  2. (B)53\dfrac5335​
  3. (C)56\dfrac5665​
  4. (D)94\dfrac9449​

Correct answer: (C)

Step-by-step solution →
Q36·PhysicsSingle correctJEE Main 2023
A point charge 2×10−2 C2\times10^{-2}\,C2×10−2C is moved from PPP to SSS in a uniform electric field of 30 NC−130\,NC^{-1}30NC−1 directed along positive x-axis. If coordinates of PPP and SSS are (1,2,0)(1,2,0)(1,2,0) and (0,0,0)(0,0,0)(0,0,0) respectively, the work done by electric field will be:
  1. (A)1200 mJ1200\,mJ1200mJ
  2. (B)−1200 mJ-1200\,mJ−1200mJ
  3. (C)−600 mJ-600\,mJ−600mJ
  4. (D)600 mJ600\,mJ600mJ

Correct answer: (C)

Step-by-step solution →
Q37·PhysicsSingle correctJEE Main 2023
The electric potential at the centre of two concentric half rings of radii R1R_1R1​ and R2R_2R2​, having same linear charge density λ\lambdaλ is:
  1. (A)λ2ε0\dfrac{\lambda}{2\varepsilon_0}2ε0​λ​
  2. (B)λ4ε0\dfrac{\lambda}{4\varepsilon_0}4ε0​λ​
  3. (C)2λε0\dfrac{2\lambda}{\varepsilon_0}ε0​2λ​
  4. (D)λε0\dfrac{\lambda}{\varepsilon_0}ε0​λ​

Correct answer: (A)

Step-by-step solution →
Q38·PhysicsMultiple correctJEE Advanced 2022
A disk of radius R with uniform positive charge density σ\sigmaσ is placed on the xy plane with its centre at the origin. The Coulomb potential along the z-axis is V(z)=σ2 ϵ0(R2+z2−z).V(z) = \frac{\sigma}{2\,\epsilon_0}\left(\sqrt{R^2 + z^2} - z\right).V(z)=2ϵ0​σ​(R2+z2​−z). A particle of positive charge q is placed initially at rest at a point on the z axis with z=z0z = z_0z=z0​ and z0>0z_0 > 0z0​>0. In addition to the Coulomb force, the particle experiences a vertical force F⃗=−ck^\vec{F} = -c\hat{k}F=−ck^ with c>0c > 0c>0. Let β=2c ϵ0qσ\beta = \frac{2c\,\epsilon_0}{q\sigma}β=qσ2cϵ0​​ . Which of the following statements(s) is (are) correct?
  1. (A)For β=14\beta = \frac{1}{4}β=41​ and z0=257Rz_0 = \frac{25}{7}Rz0​=725​R , the particle reaches the origin.
  2. (B)For β=14\beta = \frac{1}{4}β=41​ and z0=37Rz_0 = \frac{3}{7}Rz0​=73​R , the particle reaches the origin.
  3. (C)For β=14\beta = \frac{1}{4}β=41​ and z0=R3z_0 = \frac{R}{\sqrt{3}}z0​=3​R​ , the particle returns back to z=z0z = z_0z=z0​ .
  4. (D)For β>1\beta > 1β>1 and z0>0z_0 > 0z0​>0, the particle always reaches the origin.

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

Step-by-step solution →
Q39·PhysicsSingle correctJEE Main 2022
Given below are two statements. Statement I : Electric potential is constant within and at the surface of each conductor. Statement II : Electric field just outside a charged conductor is perpendicular to the surface of the conductor at every point. In the light of the above statements, choose the most appropriate answer from the options give below.
  1. (A)Both statement I and statement II are correct
  2. (B)Both statement I and statement II are incorrect
  3. (C)Statement I is correct but statement II is incorrect
  4. (D)Statement I is incorrect but and statement II is correct

Correct answer: (A)

Step-by-step solution →
Q40·PhysicsSingle correctJEE Main 2022
Two uniformly charged spherical conductors A and B of radii 5 mm and 10 mm are separated by a distance of 2 cm. If the spheres are connected by a conducting wire, then in equilibrium condition, the ratio of the magnitudes of the electric fields at the surface of the sphere A and B will be :
  1. (A)1 : 2
  2. (B)2 : 1
  3. (C)1 : 1
  4. (D)1 : 4

Correct answer: (B)

Step-by-step solution →
Q41·PhysicsSingle correctJEE Main 2022
If the electric potential at any point (x, y, z)m in space is given by V=3x2V = 3x^2V=3x2 volt. The electric field at the point (1, 0, 3) m will be :
  1. (A)3 Vm−1^{-1}−1, directed along positive x-axis.
  2. (B)3 Vm−1^{-1}−1, directed along negative x-axis.
  3. (C)6 Vm−1^{-1}−1, directed along positive x-axis.
  4. (D)6 Vm−1^{-1}−1, directed along negative x-axis.

Correct answer: (D)

Step-by-step solution →
Q42·PhysicsNumericalJEE Main 2022
27 identical drops are charged at 22V each. They combine to form a bigger drop. The potential of the bigger drop will be _____ V.

Correct answer: 198

Step-by-step solution →
Q43·PhysicsNumericalJEE Advanced 2021
Two point charges −Q-Q−Q and +Q/3+Q/\sqrt{3}+Q/3​ are placed in the xy-plane at the origin (0, 0) and a point (2, 0), respectively, as shown in the figure. This results in an equipotential circle of radius RRR and potential V = 0 in the xy-plane with its center at (b,0)(b, 0)(b,0). All lengths are measured in meters. The value of bbb is ______ meter.

Correct answer: 3.00

Step-by-step solution →
Q44·PhysicsNumericalJEE Advanced 2021
Two point charges −Q-Q−Q and +Q/3+Q/\sqrt{3}+Q/3​ are placed in the xy-plane at the origin (0, 0) and a point (2, 0), respectively, as shown in the figure. This results in an equipotential circle of radius RRR and potential V = 0 in the xy-plane with its center at (b,0)(b, 0)(b,0). All lengths are measured in meters. The value of RRR is ______ meter.

Correct answer: 1.73

Step-by-step solution →
Q45·PhysicsSingle correctJEE Main 2021
The two thin coaxial rings, each of radius 'a' and having charges +Q and –Q respectively are separated by a distance of 's'. The potential difference between the centres of the two rings is :
  1. (A)Q2πε0[1a+1s2+a2]\frac{Q}{2\pi\varepsilon_{0}}\left[\frac{1}{a} + \frac{1}{\sqrt{s^{2} + a^{2}}}\right]2πε0​Q​[a1​+s2+a2​1​]
  2. (B)Q4πε0[1a+1s2+a2]\frac{Q}{4\pi\varepsilon_{0}}\left[\frac{1}{a} + \frac{1}{\sqrt{s^{2} + a^{2}}}\right]4πε0​Q​[a1​+s2+a2​1​]
  3. (C)Q4πε0[1a−1s2+a2]\frac{Q}{4\pi\varepsilon_{0}}\left[\frac{1}{a} - \frac{1}{\sqrt{s^{2} + a^{2}}}\right]4πε0​Q​[a1​−s2+a2​1​]
  4. (D)Q2πε0[1a−1s2+a2]\frac{Q}{2\pi\varepsilon_{0}}\left[\frac{1}{a} - \frac{1}{\sqrt{s^{2} + a^{2}}}\right]2πε0​Q​[a1​−s2+a2​1​]

Correct answer: (D)

Step-by-step solution →
Q46·PhysicsNumericalJEE Main 2021
27 similar drops of mercury are maintained at 10 V each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is ___________________ times that of a smaller drop.

Correct answer: 243

Step-by-step solution →
Q47·PhysicsNumericalJEE Main 2021
512 identical drops of mercury are charged to a potential of 2 V each. The drops are joined to form a single drop. The potential of this drop is _____ V.

Correct answer: 128

Step-by-step solution →
Q48·PhysicsIntegerJEE Advanced 2020
A point charge q of mass m is suspended vertically by a string of length lll. A point dipole of dipole moment p⃗\vec{p}p​ is now brought towards q from infinity so that the charge moves away. The final equilibrium position of the system including the direction of the dipole, the angles and distances is shown in the figure below. If the work done in bringing the dipole to this position is N×(mgh)N \times (mgh)N×(mgh), where g is the acceleration due to gravity, then the value of NNN is _________. (Note that for three coplanar forces keeping a point mass in equilibrium, Fsin⁡θ\frac{F}{\sin\theta}sinθF​ is the same for all forces, where F is any one of the forces and θ\thetaθ is the angle between the other two forces)

Correct answer: 2

Step-by-step solution →
Q49·PhysicsSingle correctJEE Main 2020
Two identical electric point dipoles have dipole moments p⃗1=pi^\vec{p}_{1} = p\hat{i}p​1​=pi^ and p⃗2=−pi^\vec{p}_{2} = -p\hat{i}p​2​=−pi^ and are held on the x axis at distance 'a' from each other. When released, they move along the x-axis with the direction of their dipole moments remaining unchanged. If the mass of each dipole is 'm', their speed when they are infinitely far apart is:
  1. (A)pa1π∈0ma\frac{p}{a}\sqrt{\frac{1}{\pi ∈_{0} ma}}ap​π∈0​ma1​​
  2. (B)pa12π∈0ma\frac{p}{a}\sqrt{\frac{1}{2\pi ∈_{0} ma}}ap​2π∈0​ma1​​
  3. (C)pa2π∈0ma\frac{p}{a}\sqrt{\frac{2}{\pi ∈_{0} ma}}ap​π∈0​ma2​​
  4. (D)pa32π∈0ma\frac{p}{a}\sqrt{\frac{3}{2\pi ∈_{0} ma}}ap​2π∈0​ma3​​

Correct answer: (B)

Step-by-step solution →
Q50·PhysicsSingle correctJEE Main 2020
Ten charges are placed on the circumference of a circle of radius R with constant angular separation between successive charges. Alternate charges 1, 3, 5, 7, 9 have charge (+ q) each, while 2, 4, 6, 8, 10 have charge (− q) each. The potential V and the electric field E at the centre of the circle are respectively: (Take V = 0 at infinity)
  1. (A)V=10q4πϵ0RV = \dfrac{10q}{4\pi \epsilon_0 R}V=4πϵ0​R10q​; E=10q4πϵ0R2E = \dfrac{10q}{4\pi \epsilon_0 R^{2}}E=4πϵ0​R210q​
  2. (B)V=0V = 0V=0; E=10q4πϵ0R2E = \dfrac{10q}{4\pi \epsilon_0 R^{2}}E=4πϵ0​R210q​
  3. (C)V=0V = 0V=0 ; E=0E = 0E=0
  4. (D)V=10q4πϵ0RV = \dfrac{10q}{4\pi \epsilon_0 R}V=4πϵ0​R10q​; E=0E = 0E=0

Correct answer: (C)

Step-by-step solution →
Q51·PhysicsSingle correctJEE Main 2020
A two point charges 4q and −q are fixed on the x-axis at x=−d2x = -\frac{d}{2}x=−2d​ and x=d2x = \frac{d}{2}x=2d​, respectively. If a third point charge 'q' is taken from the origin to x = d along the semicircle as shown in the figure, the energy of the charge will:
  1. (A)decrease by 4q23πϵ0d\frac{4q^{2}}{3\pi \epsilon_{0} d}3πϵ0​d4q2​
  2. (B)increase by 2q23πϵ0d\frac{2q^{2}}{3\pi \epsilon_{0} d}3πϵ0​d2q2​
  3. (C)increase by 3q24πϵ0d\frac{3q^{2}}{4\pi \epsilon_{0} d}4πϵ0​d3q2​
  4. (D)decrease by q24πϵ0d\frac{q^{2}}{4\pi \epsilon_{0} d}4πϵ0​dq2​

Correct answer: (A)

Step-by-step solution →
Q52·PhysicsSingle correctJEE Main 2020
Hydrogen ion and singly ionized helium atom are accelerated, from rest, through the same potential difference. The ratio of final speeds of hydrogen and helium ions is close to:
  1. (A)1 : 2
  2. (B)10 : 7
  3. (C)5 : 7
  4. (D)2 : 1

Correct answer: (D)

Step-by-step solution →
Q53·PhysicsSingle correctJEE Main 2020
Two isolated conducting spheres S1S_1S1​ and S2S_2S2​ of radius 23R\dfrac{2}{3}R32​R and 13R\dfrac{1}{3}R31​R have 12 μC and −3 μC charges, respectively, and are at a large distance from each other. They are now connected by a conducting wire. A long time after this is done the charges on S1S_1S1​ and S2S_2S2​ are respectively:
  1. (A)+4.5 μC and −4.5 μC
  2. (B)4.5 μC on both
  3. (C)6 μC and 3 μC
  4. (D)3 μC and 6 μC

Correct answer: (C)

Step-by-step solution →
Q54·PhysicsSingle correctJEE Main 2020
Concentric metallic hollow spheres of radii R and 4R hold charges Q1Q_1Q1​ and Q2Q_2Q2​ respectively. Given that surface charge densities of the concentric spheres are equal, the potential difference V (R) − V (4R) is:
  1. (A)3Q24πϵ0R\dfrac{3Q_2}{4\pi \epsilon_0 R}4πϵ0​R3Q2​​
  2. (B)Q24πϵ0R\dfrac{Q_2}{4\pi \epsilon_0 R}4πϵ0​RQ2​​
  3. (C)3Q116πϵ0R\dfrac{3Q_1}{16\pi \epsilon_0 R}16πϵ0​R3Q1​​
  4. (D)3Q14πϵ0R\dfrac{3Q_1}{4\pi \epsilon_0 R}4πϵ0​R3Q1​​

Correct answer: (C)

Step-by-step solution →
Q55·PhysicsSingle correctJEE Main 2020
Consider two charged metallic spheres S1_{1}1​ and S2_{2}2​ of radii R1_{1}1​ and R2_{2}2​, respectively. The electric fields E1_{1}1​ (on S1_{1}1​) and E2_{2}2​ (on S2_{2}2​) on their surfaces are such that E1_{1}1​/E2_{2}2​ = R1_{1}1​/R2_{2}2​. Then the ratio of V1_{1}1​ (on S1_{1}1​) / V2_{2}2​ (on S2_{2}2​) of the electrostatic potentials on each sphere is:
  1. (A)(R1/R2)2(R_{1}/R_{2})^{2}(R1​/R2​)2
  2. (B)(R1R2)3\left(\dfrac{R_{1}}{R_{2}}\right)^{3}(R2​R1​​)3
  3. (C)(R2/R1)(R_{2}/R_{1})(R2​/R1​)
  4. (D)R1/R2R_{1}/R_{2}R1​/R2​

Correct answer: (A)

Step-by-step solution →
Q56·PhysicsMultiple correctJEE Advanced 2019
An electric dipole with dipole moment p02(i^+j^)\frac{p_0}{\sqrt{2}}\left(\hat{i}+\hat{j}\right)2​p0​​(i^+j^​) is held fixed at the origin O in the presence of an uniform electric field of magnitude E0E_0E0​. If the potential is constant on a circle of radius R centered at the origin as shown in figure, then the correct statement(s) is/are : (ϵ0\epsilon_0ϵ0​ is permittivity of free space. R >> dipole size)
  1. (A)Total electric field at point A is E⃗A=2E0(i^+j^)\vec{E}_A = \sqrt{2}E_0\left(\hat{i}+\hat{j}\right)EA​=2​E0​(i^+j^​)
  2. (B)Total electric field at point B is E⃗B=0\vec{E}_B = 0EB​=0
  3. (C)R=(p04πϵ0E0)1/3R = \left(\frac{p_0}{4\pi\epsilon_0 E_0}\right)^{1/3}R=(4πϵ0​E0​p0​​)1/3
  4. (D)The magnitude of total electric field on any two points of the circle will be same.

Correct answer: (B), (C)

Step-by-step solution →
Q57·PhysicsSingle correctJEE Advanced 2019
A thin spherical insulating shell of radius R carries a uniformly distributed charge such that the potential at its surface is V0V_0V0​. A hole with a small area α4πR2(α<<1)\alpha 4\pi R^2 (\alpha << 1)α4πR2(α<<1) is made on the shell without affecting the rest of the shell. Which one of the following statements is correct?
  1. (A)The magnitude of electric field at a point, located on a line passing through the hole and shell's center, on a distance 2R from the center of the spherical shell will be reduced by αV02R\frac{\alpha V_0}{2R}2RαV0​​
  2. (B)The magnitude of electric field at the center of the shell is reduced by αV02R\frac{\alpha V_0}{2R}2RαV0​​
  3. (C)The ratio of the potential at the center of the shell to that of the point at 12\frac{1}{2}21​R from center towards the hole will be 1−α1−2α\frac{1-\alpha}{1-2\alpha}1−2α1−α​
  4. (D)The potential at the center of the shell is reduced by 2αV02\alpha V_02αV0​

Correct answer: (C)

Step-by-step solution →
Q58·PhysicsSingle correctJEE Main 2019
A point dipole p⃗=−p0x^\vec{p} = -p_0\hat{x}p​=−p0​x^ is kept at the origin. The potential and electric field due to this dipole on the y-axis at a distance d are, respectively : (Take V = 0 at infinity)
  1. (A)∣p⃗∣4πε0d2,−p⃗4πε0d3\dfrac{|\vec{p}|}{4\pi\varepsilon_0 d^{2}}, \dfrac{-\vec{p}}{4\pi\varepsilon_0 d^{3}}4πε0​d2∣p​∣​,4πε0​d3−p​​
  2. (B)0,p⃗4πε0d30, \dfrac{\vec{p}}{4\pi\varepsilon_0 d^{3}}0,4πε0​d3p​​
  3. (C)∣p⃗∣4πε0d2,p⃗4πε0d3\dfrac{|\vec{p}|}{4\pi\varepsilon_0 d^{2}}, \dfrac{\vec{p}}{4\pi\varepsilon_0 d^{3}}4πε0​d2∣p​∣​,4πε0​d3p​​
  4. (D)0,−p⃗4πε0d30, \dfrac{-\vec{p}}{4\pi\varepsilon_0 d^{3}}0,4πε0​d3−p​​

Correct answer: (D)

Step-by-step solution →
Q59·PhysicsSingle correctJEE Main 2019
A uniformly charged ring of radius 3a and total charge q is placed in xy-plane centered at origin. A point charge q is moving towards the ring along the z-axis and has speed v at z = 4a. The minimum value of v such that it crosses the origin is:
  1. (A)2m(15q24πϵ0a)1/2\sqrt{\dfrac{2}{m}\left(\dfrac{1}{5}\dfrac{q^{2}}{4\pi \epsilon_{0} a}\right)^{1/2}}m2​(51​4πϵ0​aq2​)1/2​
  2. (B)2m(115q24πϵ0a)1/2\sqrt{\dfrac{2}{m}\left(\dfrac{1}{15}\dfrac{q^{2}}{4\pi \epsilon_{0} a}\right)^{1/2}}m2​(151​4πϵ0​aq2​)1/2​
  3. (C)2m(415q24πϵ0a)1/2\sqrt{\dfrac{2}{m}\left(\dfrac{4}{15}\dfrac{q^{2}}{4\pi \epsilon_{0} a}\right)^{1/2}}m2​(154​4πϵ0​aq2​)1/2​
  4. (D)2m(215q24πϵ0a)1/2\sqrt{\dfrac{2}{m}\left(\dfrac{2}{15}\dfrac{q^{2}}{4\pi \epsilon_{0} a}\right)^{1/2}}m2​(152​4πϵ0​aq2​)1/2​

Correct answer: (D)

Step-by-step solution →
Q60·PhysicsSingle correctJEE Main 2019
A system of three charges are placed as shown in the figure: If D >> d, the potential energy of the system is best given by:
  1. (A)14πεo[−q2d−qQdD2]\frac{1}{4\pi\varepsilon_o}\left[-\frac{q^2}{d}-\frac{qQd}{D^2}\right]4πεo​1​[−dq2​−D2qQd​]
  2. (B)14πεo[−q2d−qQd2D2]\frac{1}{4\pi\varepsilon_o}\left[-\frac{q^2}{d}-\frac{qQd}{2D^2}\right]4πεo​1​[−dq2​−2D2qQd​]
  3. (C)14πεo[−q2d+2qQdD2]\frac{1}{4\pi\varepsilon_o}\left[-\frac{q^2}{d}+\frac{2qQd}{D^2}\right]4πεo​1​[−dq2​+D22qQd​]
  4. (D)14πεo[+q2d−qQdD2]\frac{1}{4\pi\varepsilon_o}\left[+\frac{q^2}{d}-\frac{qQd}{D^2}\right]4πεo​1​[+dq2​−D2qQd​]

Correct answer: (A)

Step-by-step solution →
Q61·PhysicsSingle correctJEE Main 2019
A solid conducting sphere, having a charge Q, is surrounded by an uncharged conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of −4 Q, the new potential difference between the same two surface is:
  1. (A)2 V
  2. (B)−2V
  3. (C)4 V
  4. (D)V

Correct answer: (D)

Step-by-step solution →
Q62·PhysicsSingle correctJEE Main 2019
The election field in a region is given by E⃗=(Ax+B)i^\vec{E} = (Ax + B)\hat{i}E=(Ax+B)i^ where E is in NC−1^{-1}−1 and x in meters. The values of constants are A = 20 SI unit and B = 10 SI unit. If the potential at x =1 is V1_11​ and that at x = −-−5 is V2_22​ then V1_11​ - V2_22​ is:
  1. (A)320 V
  2. (B)−-−48 V
  3. (C)−-−520 V
  4. (D)180 V

Correct answer: (D)

Step-by-step solution →
Q63·PhysicsSingle correctJEE Main 2019
A positive point charge is released from rest at a distance r0r_0r0​ from a positive line charge with uniform density. The speed (v) of the point charge, as a function of instantaneous distance r from line charge, is proportional to
  1. (A)v∝e+r/r0v \propto e^{+r/r_0}v∝e+r/r0​
  2. (B)v∝ln⁡(rr0)v \propto \ln\left(\dfrac{r}{r_0}\right)v∝ln(r0​r​)
  3. (C)v∝ln⁡(rr0)v \propto \sqrt{\ln\left(\dfrac{r}{r_0}\right)}v∝ln(r0​r​)​
  4. (D)v∝(rr0)v \propto \left(\dfrac{r}{r_0}\right)v∝(r0​r​)

Correct answer: (C)

Step-by-step solution →
Q64·PhysicsSingle correctJEE Main 2019
There is a uniform spherically symmetric surface charge density at a distance R0R_0R0​ from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that represents best the speed V(R(t)) of the distribution as a function of its instantaneous radius R(t) is:
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C)

Step-by-step solution →
Q65·PhysicsSingle correctJEE Main 2019
An electric field of 1000V/m is applied to an electric dipole at angle of 45°. The value of electric dipole moment is 10−29^{-29}−29 C.m. What is the potential energy of the electric dipole?
  1. (A)−20×1018-20\times10^{18}−20×1018 J
  2. (B)−7×10−27-7\times10^{-27}−7×10−27 J
  3. (C)−10×10−29-10\times10^{-29}−10×10−29 J
  4. (D)−9×10−20-9\times10^{-20}−9×10−20 J

Correct answer: (B)

Step-by-step solution →
Q66·PhysicsSingle correctJEE Main 2019
The give graph shown variation (with distance r from centre) of:
  1. (A)Electric field of a uniformly charged sphere
  2. (B)Potential of a uniformly charged spherical shell
  3. (C)Potential of a uniformly charged sphere
  4. (D)Electric field of a uniformly charged spherical shell

Correct answer: (B)

Step-by-step solution →
Q67·PhysicsSingle correctJEE Main 2019
A particle of mass m and charge q is in an electric and magnetic field given by: E⃗=2i^+3j^;B=4j^+6k^\vec{E}=2\hat{i}+3\hat{j}; B=4\hat{j}+6\hat{k}E=2i^+3j^​;B=4j^​+6k^ The charged particle is shifted from the origin to the point P(x = 1; y = 1) along a straight path. The magnitude of the total work done is:
  1. (A)(0.35)q
  2. (B)5q
  3. (C)(2.5)q
  4. (D)(0.15)q

Correct answer: (B)

Step-by-step solution →
Q68·PhysicsSingle correctJEE Main 2019
Three charges Q, +q and +q are placed at the vertices of a right – angle isosceles triangle as shown below. The net electrostatic energy of the configuration is zero, if the value of Q is:
  1. (A)+q
  2. (B)−2q2+1\frac{-\sqrt{2}q}{\sqrt{2}+1}2​+1−2​q​
  3. (C)−q1+2\frac{-q}{1+\sqrt{2}}1+2​−q​
  4. (D)–2q

Correct answer: (B)

Step-by-step solution →
Q69·PhysicsSingle correctJEE Main 2019
Two electric dipoles A, B with respective dipole moments dA⃗=−4qai^\vec{d_{A}}=-4qa\hat{i}dA​​=−4qai^ and dB⃗=2qai^\vec{d_{B}}=2qa\hat{i}dB​​=2qai^ are placed on the x-axis with a separation R, as shown in the figure. The distance from A at which both of them produce the same potential is:
  1. (A)R2+1\frac{R}{\sqrt{2}+1}2​+1R​
  2. (B)2R2+1\frac{\sqrt{2}R}{\sqrt{2}+1}2​+12​R​
  3. (C)R2−1\frac{R}{\sqrt{2}-1}2​−1R​
  4. (D)2R2−1\frac{\sqrt{2}R}{\sqrt{2}-1}2​−12​R​

Correct answer: (B)

Step-by-step solution →
Q70·PhysicsSingle correctJEE Main 2019
A charge Q is distributed over three concentric spherical shell of radii a, b, c (a < b < c) such that their surface charge densities are equal to one another. The total potential at a point at distance r from their common centre, where r < a, would be:
  1. (A)Q12πϵ0ab+bc+caabc\dfrac{Q}{12\pi\epsilon_{0}}\dfrac{ab+bc+ca}{abc}12πϵ0​Q​abcab+bc+ca​
  2. (B)Q(a2+b2+c2)4πϵ0(a3+b3+c3)\dfrac{Q\left(a^{2}+b^{2}+c^{2}\right)}{4\pi\epsilon_{0}\left(a^{3}+b^{3}+c^{3}\right)}4πϵ0​(a3+b3+c3)Q(a2+b2+c2)​
  3. (C)Q4πϵ0(a+b+c)\dfrac{Q}{4\pi\epsilon_{0}\left(a+b+c\right)}4πϵ0​(a+b+c)Q​
  4. (D)Q(a+b+C)4πϵ0(a2+b2+c2)\dfrac{Q\left(a+b+C\right)}{4\pi\epsilon_{0}\left(a^{2}+b^{2}+c^{2}\right)}4πϵ0​(a2+b2+c2)Q(a+b+C)​

Correct answer: (D)

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Electric Potential — frequently asked

How many questions from Electric Potential appear in JEE?

Electric Potential has appeared in 63 of the last 186 JEE Main and JEE Advanced papers — about 34% of them — contributing 70 questions in total across those papers.

Is Electric Potential an important chapter for JEE?

Judged by how often it is actually tested, it appears in roughly 34% 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 Potential 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.

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