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JEE • PHYSICS • ELECTRIC POTENTIAL AND CAPACITANCE

Electric potential and capacitance practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 271-300 of 407 questions

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271 Medium
271 of 407
Understanding +4 | -1
When a dielectric slab is introduced between the plates of a capacitor connected to a battery, then
View answer and explanation
Correct option: A
Charge on capacitor increases
Option A is correct because introducing a dielectric increases the capacitance, and since the voltage is constant (connected to a battery), the charge on the capacitor increases.
272 Medium
272 of 407
Application +4 | -1
The electric potential is +100 V+100 \mathrm{~V} at a distance of 10 cm10 \mathrm{~cm} from a point charge q.\text{q}. Then, q\text{q} is equal to
View answer and explanation
Correct option: A
+1.1×10−9C+1.1 \times 10^{-9} \mathrm{C}
Option A (+1.1×10−9 C+1.1 \times 10^{-9} \text{ C}) is correct because it is the charge calculated using the formula for electric potential due to a point charge.
273 Medium
273 of 407
Application +4 | -1
Two spheres of radii 12 cm and 16 cm have equal charge. The ratio of their energies is
View answer and explanation
Correct option: A
4:34: 3
Option A (4:34:3) is correct because the energy of a charged sphere is inversely proportional to its radius when the charge is constant.
274 Medium
274 of 407
Analyzing +4 | -1
Assertion (A) : A small electric dipole is moved translationally from higher potential to lower potential in uniform electric field. Work done by electric field is positive.

Reason (R) : When a positive charge is moved from higher potential to lower potential, work done by electric field is positive.
View answer and explanation
Correct option: D
If Assertion is false but the Reason is true.
Option D (If Assertion is false but the Reason is true) is correct because the work done by the electric field on a dipole depends on the change in potential energy, and the reason correctly states the work done on a positive charge.
275 Medium
275 of 407
Application +4 | -1
A sphere of radius 0.03 mis suspended within a hollow sphere of radius 0.05 m0.05 \mathrm{~m}. If the inner sphere is charged to a potential of 1500 volt and outer sphere is earthed, find the capacitance and the charge on the inner sphere.
View answer and explanation
Correct option: A
1.25×10−8C1.25 \times 10^{-8} \mathrm{C}
The correct answer is A (1.25×10−8C1.25 \times 10^{-8} \mathrm{C}) because it is the charge calculated using the capacitance of the spherical capacitor and the given potential.
276 Medium
276 of 407
Application +4 | -1
An electric dipole consists of two opposite charges each 0.05μC\mathrm{\mu C} separated by 30 mm. The dipole is placed in an uniform external electric field of 106NC−1.{10^6}\mathrm{N}{\mathrm{C}^{ - 1}}. The maximum torque exerted by the field on the dipole is
View answer and explanation
Correct option: D
1.5×10−3N−m1.5 \times {10^{ - 3}}\mathrm{N} - \mathrm{m}
Option D (1.5×10−3N−m1.5 \times {10^{ - 3}}\mathrm{N} - \mathrm{m}) is correct because it is calculated using the formula for maximum torque on a dipole in an electric field, τ=pE\tau = pE, where pp is the dipole moment and EE is the electric field.
277 Medium
277 of 407
Application +4 | -1
A 6 μF6\,\mu F capacitor is charged to 200 V. If its plates are joined through a resistance of 4 KΩ,4\,K\Omega , then heat produced in the resistance is
View answer and explanation
Correct option: A
0.12 J
Option A (0.12 J) is correct because the heat produced is equal to the energy stored in the capacitor, which is calculated using U=12CV2U = \frac{1}{2}CV^2.
278 Medium
278 of 407
Application +4 | -1
Taking the Earth to be a spherical conductor of diameter 12.8×103 km.12.8 \times 10^3 \mathrm{~ k m}. Its capacity will be
View answer and explanation
Correct option: A
711μF711 \mu \mathrm{F}
Option A (711μF711 \mu \mathrm{F}) is correct because it is the capacitance calculated using the formula for the capacitance of a spherical conductor.
279 Medium
279 of 407
Application +4 | -1
If a slab of insulating material 4×10−3m4 \times {10^{ - 3}}m thick is introduced between the plates of a parallel plate capacitor, the separation between plates has to be increased by 3.5×10−3m3.5 \times {10^{ - 3}}m to restore the capacity to original value. The dielectric constant of the material will be
View answer and explanation
Correct option: B
8
Option B (8) is correct because the dielectric constant is calculated using the formula relating the thickness of the slab and the increase in separation.
280 Medium
280 of 407
Application +4 | -1
Electric field is given by E=100x2E=\frac{100}{x^2} potential difference between x=10x=10 and x=20 mx=20 \mathrm{~m}.
View answer and explanation
Correct option: A
55 volt
Option A (5 volt) is correct because the potential difference is calculated by integrating the electric field over the given distance.
281 Medium
281 of 407
Application +4 | -1
When charge of magnitude 10μC10 \mu \mathrm{C} is shifted from infinity to a point in an electric field, then it was found that work done by electrostatic forces is −10μJ-10 \mu \mathrm{J}. If this charge is doubled and moved again from infinity to the same point without changing its velocity, then find out the work done by electric field and against this electric field.
View answer and explanation
Correct option: A
20μJ20 \mu \mathrm{J} ,−20μJ-20 \mu \mathrm{J}
Option A (20μJ20 \mu \mathrm{J} ,−20μJ-20 \mu \mathrm{J}) is correct because the work done is directly proportional to the charge moved in the electric field.
282 Medium
282 of 407
Understanding +4 | -1
The electric potential at the centre of a charged
conductor is
View answer and explanation
Correct option: D
Same as on the surface
Option D (Same as on the surface) is correct because the electric field inside a charged conductor is zero, implying a constant potential.
283 Medium
283 of 407
Application +4 | -1
Electric potential at any point is V=−5x+3y+15z,V = - 5x + 3y + \sqrt {15z} , then the magnitude of the electric field is
View answer and explanation
Correct option: D
7
Option D (7) is correct because the electric field components are the negative gradients of the potential, and the magnitude is the square root of the sum of the squares of the components.
284 Medium
284 of 407
Application +4 | -1
A condenser has a capacity 2μF2 \mu \mathrm{F} and is charged to a voltage of 50 V.50 \mathrm{~V}. The energy stored is
View answer and explanation
Correct option: D
25×103 erg25 \times 10^3 \mathrm{~erg}
Option D (25×103 erg25 \times 10^3 \mathrm{~erg}) is correct because the energy stored in a capacitor is calculated using the formula E = (1/2)CV², and then converted from Joules to ergs.
285 Medium
285 of 407
Understanding +4 | -1
A metallic hollow sphere of radius 3 cm3 \mathrm{~cm} is charged so that potential at its surface is 5 V.5 \mathrm{~V}. The potential at the centre of sphere will be
View answer and explanation
Correct option: B
5 V5 \mathrm{~V}
Option B (5 V5 \mathrm{~V}) is correct because the electric potential inside a hollow metallic sphere is constant and equal to the potential at the surface.
286 Medium
286 of 407
Understanding +4 | -1
A ring with uniform distribution of charge qq and radius aa is placed in Y−ZY-Z plane with its centre at origin.
View answer and explanation
Correct option: B
Potential at origin is q4πε0a\frac{q}{4 \pi \varepsilon_0 a}
Option B (Potential at origin is q4πε0a\frac{q}{4 \pi \varepsilon_0 a}) is correct because the electric potential at the center of a uniformly charged ring is given by this formula.
287 Medium
287 of 407
Understanding +4 | -1
The separation between the plates of a charged parallel- plate capacitor is increased. The force between the plates
View answer and explanation
Correct option: C
remains same
Option C (remains same) is correct because the force between the plates depends on the charge density and electric field, which remain constant.
288 Medium
288 of 407
Application +4 | -1
Capacitance of a parallel plate capacitor becomes 43\frac{4}{3} times its original value, if a dielectric slab of thickness t=d2t = \frac{d}{2} is inserted between the plates [dd is the separation between the plates]. The dielectric constant of the slab is
View answer and explanation
Correct option: C
2
Option C (2) is correct because the dielectric constant is calculated using the formula for capacitance with a dielectric slab inserted.
289 Medium
289 of 407
Application +4 | -1
Two equal charges ‘q’ of opposite sign are separated by small distance ‘2a’. The electric intensity ‘E’ at a point on the perpendicular bisector of the line joining the two charges at a very distance ‘r’ from the line is
View answer and explanation
Correct option: B
14πε0 2qar3\frac{1}{{4\pi {\varepsilon _0}}}\,\frac{{2qa}}{{{r^3}}}
Option B is correct because the electric field due to a dipole at a point on the perpendicular bisector is inversely proportional to the cube of the distance.
290 Medium
290 of 407
Analyzing +4 | -1
Assertion:- A parallel plate capacitor is connected across a battery through a key. A dielectric slab of dielectric constant KK is introduced between the plates. The energy which is stored becomes KK times.
Reason:- The potential difference between the plates remains constant or unchanged.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion
Option A is correct because inserting a dielectric increases the stored energy by a factor of K when the capacitor is connected to a battery, and the potential difference remains constant.
291 Hard
291 of 407
Analyzing +4 | -1
A parallel plate capacitor with plate area AA and distance of separation dd is filled with a dielectric. What is the capacity of the capacitor when permittivity of the dielectric varies as
ε (X)=ε0+  kx,\varepsilon \,(X) = {\varepsilon _0} + \,\,kx, for (0<x≤d2)\left( {0 < x \le \frac{d}{2}} \right)
ε (X)  =  ε0+k (d−x)\varepsilon \,(X)\,\, = \,\,{\varepsilon _0} + k\,(d - x) for (d2≤x≤d)\left( {\frac{d}{2} \le x \le d} \right)
View answer and explanation
Correct option: B
kA2ln⁡(2ε0+kd2ε0)\frac{{kA}}{{2\ln \left( {\frac{{2{\varepsilon _0} + kd}}{{2{\varepsilon _0}}}} \right)}}
Option D is the correct answer. The capacitance is calculated by integrating the inverse of the permittivity function over the distance between the plates, considering the varying permittivity.
292 Medium
292 of 407
Application +4 | -1
A sample of HCl\mathrm{HCl} gas is placed in an electric field of 2.5×104 NC−12.5 \times 10^4 \mathrm{~N} \mathrm{C}^{-1}. The dipole moment of each HCl\mathrm{HCl} molecule is 3.4×10−30Cm3.4 \times 10^{-30} \mathrm{Cm}. Find the maximum torque that can act on a molecule.
View answer and explanation
Correct option: A
8.5×10−268.5 \times 10^{-26}
Option A (8.5×10−268.5 \times 10^{-26}) is correct because the maximum torque is calculated by multiplying the dipole moment and the electric field strength.
293 Easy
293 of 407
Application +4 | -1
A point AA is at potential +500 Volt+500 \mathrm{~Volt} and BB is a point of potential −500 Volt.-500 \mathrm{~Volt}. The work done by an external agent to take a unit positive charge from AA to BB is
View answer and explanation
Correct option: B
−1000J-1000J
Option B (-1000J) is correct because the work done by an external agent is the negative of the potential difference multiplied by the charge.
294 Medium
294 of 407
Application +4 | -1
You are given an arrangement of three point charges q,2qq, 2 q and xqx q separated by equal finite distances that electric potential energy of the system is zero. Then the value of xx is:
View answer and explanation
Correct option: A
−2/3-2 / 3
Option A (-2/3) is correct because it is the value of x that makes the total electric potential energy of the system zero.
295 Medium
295 of 407
Application +4 | -1
A uniform electric field of 100 N/C100 \mathrm{~N} / \mathrm{C} exists in XY\mathrm{XY} plane making an angle 60∘60^{\circ} with +x+\mathrm{x} direction. Consider points A(1,2,3)mA(1,2,3) \mathrm{m} and B(3,2,1)mB(3,2,1) \mathrm{m} then VA−VB\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}} is:
View answer and explanation
Correct option: A
100 volt
Option A (100 volt) is correct because the potential difference is calculated using the formula VA−VB=E‾⋅(r‾B−r‾A)V_A - V_B = \overline{E} \cdot (\overline{r}_B - \overline{r}_A).
296 Medium
296 of 407
Application +4 | -1
The electric intensity due to a dipole of length 10 cm and having a charge of 500  μC500\,\,\mu C at a point on the axis at a distance 20 cm from one of the charges in air, is
View answer and explanation
Correct option: A
6.25×107N/C6.25 \times {10^7}N/C
The correct answer is A (6.25×107N/C6.25 \times {10^7}N/C) because the electric intensity due to a dipole is calculated using the formula E=9×109[2pr(r2−l2)2]E = 9 \times {10^9}\left[ {\frac{{2pr}}{{{{\left( {{r^2} - {l^2}} \right)}^2}}}} \right].
297 Medium
297 of 407
Analyzing +4 | -1
Assertion:- Electric potential energy of any positive charge is always positive.
Reason:-Potential energy is a vector quantity.
View answer and explanation
Correct option: D
If both Assertion & Reason are False
Option D (If both Assertion & Reason are False) is correct because the electric potential energy of a positive charge can be negative, and potential energy is a scalar quantity.
298 Medium
298 of 407
Understanding +4 | -1
Assertion: Electrings always move from a region of lower potential to a region of high potential.

Reason: Electrings carryy a negative charge.
View answer and explanation
Correct option: A
Both Assertion & Reason are true. Reason is the correct explanation for assertion
Option A (Both Assertion & Reason are true. Reason is the correct explanation for assertion) is correct because electrons are negatively charged and move from lower to higher potential.
299 Medium
299 of 407
Understanding +4 | -1
Assertion:- The capacitance of a capacitor depends on the shape, size and geometrical placing of the conductors and the medium between them.
Reason:- When a charge qq passes through a battery of emf E{E} from the negative terminal to the positive terminal, an amount of work qE{qE} is done by the battery.
View answer and explanation
Correct option: B
If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion
Option B (If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion) is correct because both statements are true, but the reason doesn't explain why capacitance depends on geometry and the medium.
300 Medium
300 of 407
Analyzing +4 | -1
A point charge is situated at an axial point of a small electric dipole at a large distance from it. The charge
experiences a force F. If the distance of the charge is doubled, the force acting on the charge will become
View answer and explanation
Correct option: B
F8\frac{F}{8}
Option B (F8\frac{F}{8}) is correct because the electric field due to a dipole varies inversely with the cube of the distance.
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