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JEE • PHYSICS • ELECTRIC POTENTIAL AND CAPACITANCE

Electric potential and capacitance practice · Medium

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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 241-270 of 372 questions

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241 Medium
241 of 372
Application +4 | -1
A capacitor of unknown capacitance C\mathrm{C} is connected across a battery of V\text{V} volt. The charge stored in it becomes Q\text{Q} coulomb. When potential across the capacitor is reduced by V′\text{V}^{\prime} volt, the charge stored in it becomes Q′\text{Q}^{\prime} coulomb. The capacitance C\text{C} is
View answer and explanation
Correct option: A
Q−Q′V′\frac{\mathrm{Q}-\mathrm{Q}^{\prime}}{\mathrm{V}^{\prime}}
Option A (Q−Q′V′\frac{\mathrm{Q}-\mathrm{Q}^{\prime}}{\mathrm{V}^{\prime}}) is correct because the capacitance is calculated using the change in charge and the change in voltage across the capacitor.
242 Medium
242 of 372
Understanding +4 | -1
If combining two charged bodies, the current does not flow then
View answer and explanation
Correct option: C
Potential is equal on both
Option B (Statement I is true, Statement II is true; Statement II is not the correct explanation for Statement I) is correct because both statements are true, but they are independent of each other.
243 Medium
243 of 372
Application +4 | -1
The plates in a parallel plate capacitor are separated by a distance dd with air as the medium in between the plates. In order to increase the capacity by 66 %, a dielectric slab of dielectric constant 5 is introduced in between the plates. The thick ness of the dielectric slab required is
View answer and explanation
Correct option: B
d2\frac{d}{2}
Option B (d2\frac{d}{2}) is correct because the thickness of the dielectric slab is calculated by equating the new capacitance to 1.66 times the original capacitance.
244 Medium
244 of 372
Application +4 | -1
Consider the fusion in helium plasma. Find the temperature at which the average thermal energy 1.5 kT1.5 \mathrm{~kT} equals the Coulomb potential energy at 2 fm2 \mathrm{~fm}.
View answer and explanation
Correct option: A
2.23×1010 K2.23 \times 10^{10} \mathrm{~K}
Option A (2.23×1010 K2.23 \times 10^{10} \mathrm{~K}) is correct because it is the temperature at which the average thermal energy equals the Coulomb potential energy at 2 fm, calculated using the given formula.
245 Medium
245 of 372
Application +4 | -1
The kinetic energy of a charged particle decreases by 10 J10 \mathrm{~J} as it moves from a point at potential 100 V100 \mathrm{~V} to a point at potential 200 V200 \mathrm{~V}. Find the charge on the particle.
View answer and explanation
Correct option: C
0.1 C
Option C (0.1 C) is correct because the charge is calculated by equating the change in kinetic energy to the work done by the electric potential difference.
246 Medium
246 of 372
Remembering +4 | -1
If EE is the electric field intensity of an electrostatic field, then the electrostatic energy density is proportional to
View answer and explanation
Correct option: B
E2{E^2}
Option B (E2{E^2}) is correct because the electrostatic energy density is directly proportional to the square of the electric field intensity.
247 Medium
247 of 372
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.
248 Medium
248 of 372
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.
249 Medium
249 of 372
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.
250 Medium
250 of 372
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.
251 Medium
251 of 372
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.
252 Medium
252 of 372
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.
253 Medium
253 of 372
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.
254 Medium
254 of 372
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.
255 Medium
255 of 372
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.
256 Medium
256 of 372
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.
257 Medium
257 of 372
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.
258 Medium
258 of 372
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.
259 Medium
259 of 372
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.
260 Medium
260 of 372
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.
261 Medium
261 of 372
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.
262 Medium
262 of 372
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.
263 Medium
263 of 372
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.
264 Medium
264 of 372
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.
265 Medium
265 of 372
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.
266 Medium
266 of 372
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.
267 Medium
267 of 372
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.
268 Medium
268 of 372
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.
269 Medium
269 of 372
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).
270 Medium
270 of 372
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].
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