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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 181-210 of 407 questions

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181 Medium
181 of 407
Application +4 | -1
The potential energy function (in JJ) of a particle in a region of space is given as U=(2x2+3y2+2z)U=\left(2 x^2+3 y^2+2 z\right). Hare x,yx, y and zz are in meter. The magnitude of xx component of force (in NN) acting on the particle at point P(1,2,3)mP(1,2,3) m is:
View answer and explanation
Correct option: D
4
Option D (4) is correct because the magnitude of the x component of the force is 4 N, calculated by taking the negative derivative of the potential energy function with respect to x.
182 Medium
182 of 407
Application +4 | -1
A parallel plate capacitor has each plate area AA and separation dd. Charge QQ is given to positive plate of capacitor. The force on each plate of capacitor is
View answer and explanation
Correct option: A
Q22ε0A\frac{Q^2}{2 \varepsilon_0 A}
Option A (Q22ε0A\frac{Q^2}{2 \varepsilon_0 A}) is correct because the force on each plate is due to the electric field created by the other plate.
183 Medium
183 of 407
Analyzing +4 | -1
Assertion: The electric potential at any point on the equatorial plane of an electric dipole is indeed zero.
Reason: The contributions to the potential from the positive and negative charges of the dipole cancel each other out due to the symmetry of the arrangement.
View answer and explanation
Correct option: A
If both assertion and reason are true and reason is the correct explanation of assertion
Option A (If both assertion and reason are true and reason is the correct explanation of assertion) is correct because the electric potential due to the positive and negative charges of the dipole cancel each other at any point on the equatorial plane due to symmetry.
184 Medium
184 of 407
Understanding +4 | -1
Q{Q} charge is given to two capacitors C1C_1 and C2C_2 which are in parallel. The charge distribution among them is
View answer and explanation
Correct option: A
C1:C2{C}_1: {C}_2
Option A (C1:C2{C}_1: {C}_2) is correct because when capacitors are in parallel, the charge distribution is proportional to their capacitances.
185 Medium
185 of 407
Understanding +4 | -1
What is the angle between maximum value of potential gradient and equipotential surface?
View answer and explanation
Correct option: C
π2\frac{\pi}{2}
Option C (π2\frac{{\pi}}{{2}}) is correct because the potential gradient is maximum in the direction perpendicular to the equipotential surface.
186 Medium
186 of 407
Understanding +4 | -1
When a test charge is brought in from infinity along the perpendicular bisector of an electric dipole, the work done is
View answer and explanation
Correct option: B
Zero
Option B (Zero) is correct because the electric potential is constant along the perpendicular bisector of an electric dipole, hence no work is done.
187 Medium
187 of 407
Remembering +4 | -1
Dielectric constant of a medium is also known as:
View answer and explanation
Correct option: D
Relative permittivity
Option D (Relative permittivity) is correct because the dielectric constant is defined as relative permittivity.
188 Medium
188 of 407
Understanding +4 | -1
Assertion:- Circuits containing high capacity capacitors, charged to high voltages should be handled with caution, even when the current in the circuit is switched off.
Reason:- When an isolated capacitor is touched by hand or any other part of the human body, there is an easy path to the ground available for the discharge of the capacitor.
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 high-capacity capacitors store significant charge even after the circuit is switched off, and touching them provides a discharge path through the human body to the ground.
189 Medium
189 of 407
Analyzing +4 | -1
A parallel plate capacitor of plate area AA and plate separation dd is charged to potential VV and then the battery is disconnected. A slab of dielectric constant kk is then inserted between the plates of the capacitors so as to fill the space between the plates. If Q,EQ,E and WW denote respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab is inserted) and work done on the system in question in the process of inserting the slab, then state incorrect relation from the following
View answer and explanation
Correct option: D
W=ε0AV22d(1−1k)W = \frac{{{\varepsilon _0}A{V^2}}}{{2d}}\left( {1 - \frac{1}{k}} \right)
Option D is the incorrect relation because the work done should be negative, indicating work done *by* the system, not *on* the system.
190 Medium
190 of 407
Application +4 | -1
The stopping potential of anode relative to cathode that must be applied to stop the fastest photo e−s{e^ - }s emitted by surface having work function 5.01 eV, when UV light of wavelength 200 nm falls on it, must be:-
View answer and explanation
Correct option: C
–1.2 V
Option D (–2.4 V) is correct because it is calculated by finding the maximum kinetic energy of the photoelectrons using the given wavelength and work function, and then relating it to the stopping potential.
191 Medium
191 of 407
Understanding +4 | -1
An air capacitor is connected to a battery. The effect of filling the space between the plates with a dielectric is to increase
View answer and explanation
Correct option: D
The charge and the capacitance
Option D is correct because inserting a dielectric increases both the capacitance and the charge stored in the capacitor when connected to a battery.
192 Medium
192 of 407
Application +4 | -1
A 500μF500\mu F capacitor is charged at a steady rate of 100μC/second.100\mu C/\mathrm{second}. The potential difference across the capacitor will be 10  V10\;V after an interval of
View answer and explanation
Correct option: D
50 sec
Option D (50 sec) is correct because the time is calculated by dividing the total charge required for 10V by the charging rate.
193 Medium
193 of 407
Understanding +4 | -1
A system of two charges separated by a certain distance apart stores electrical potential energy. If the distance between them is increased, the potential energy of the system:
View answer and explanation
Correct option: C
May increase or decreases
Option C (May increase or decreases) is correct because the potential energy depends on the sign of the charges; increasing the distance can either increase or decrease the potential energy.
194 Easy
194 of 407
Application +4 | -1
The energy stored in a capacitor of capacitance 100μF100 \mu F is 50J50 J. Its potential difference is
View answer and explanation
Correct option: C
1000 V
Option C (1000 V) is correct because it is calculated using the formula for energy stored in a capacitor, U=12CV2U = \frac{1}{2}CV^2, solving for V.
195 Medium
195 of 407
Understanding +4 | -1
A hollow conducting sphere of radius RR has a charge (+Q)\left( { + Q} \right) on its surface. What is the electric potential within the sphere at a distance r=R/3r = R/3 from its centre
View answer and explanation
Correct option: C
14πε0QR\frac{1}{{4\pi {\varepsilon _0}}}\frac{Q}{R}
Option C (14πε0QR\frac{1}{{4\pi {{\varepsilon _0}}}}\frac{Q}{R}) is correct because the electric potential inside a hollow conducting sphere is constant and equal to the potential on its surface.
196 Medium
196 of 407
Analyzing +4 | -1
A capacitor is kept connected to the battery and a dielectric slab is inserted between the plates. During this process
View answer and explanation
Correct option: C
Work is done at the cost of the battery
Option C (Work is done at the cost of the battery) is correct because the capacitor remains connected to the battery, maintaining a constant potential difference, and the battery supplies additional charge as the capacitance increases.
197 Medium
197 of 407
Understanding +4 | -1
What fraction of the energy drawn from the charging battery is stored in a capacitor?
View answer and explanation
Correct option: D
50%50 \%
Option D (50 %) is correct because half of the energy supplied by the battery is stored in the capacitor, and the other half is dissipated as heat.
198 Medium
198 of 407
Application +4 | -1
The earth has Volume ′V′'V' and surface area ′A′'A' then capacitance would be
View answer and explanation
Correct option: C
12π∈0VA12\pi { \in _0}\frac{V}{A}
Option C (12π∈0VA12\pi { \in _0}\frac{V}{A}) is correct because it correctly relates the capacitance of a spherical conductor to its volume and surface area.
199 Medium
199 of 407
Understanding +4 | -1
An electric dipole of moment p⃗{\vec p} placed in a uniform electric field E⃗{\vec E} has minimum potential energy when the angle between p⃗{\vec p} and E⃗{\vec E} is
View answer and explanation
Correct option: A
Zero
Option A (Zero) is correct because the potential energy of an electric dipole in a uniform electric field is minimum when the dipole moment is aligned with the electric field (θ=0\theta = 0).
200 Medium
200 of 407
Application +4 | -1
A total charge Q\mathrm{Q} is uniformly distributed throughout the volume of a sphere. Electric potential at a point on the surface of the sphere is greater than the potential at infinity by 16 V16 \mathrm{~V}. Find the potential difference (in volt) between point P\mathbf{P} at a distance half of the radius of the sphere from the centre and the point S\mathbf{S} at a distance double the radius of the sphere from the centre of the sphere.
View answer and explanation
Correct option: A
14
Option A (14) is correct because it is calculated by finding the potential difference between point P and point S using the given conditions and formulas for electric potential inside and outside a uniformly charged sphere.
201 Medium
201 of 407
Understanding +4 | -1
A dipole of electric dipole moment pp is placed in a uniform electric field of strength E,E, If θ\theta is the angle between positive directions of p and E,E, then the potential energy of the electric dipole is largest when θ\theta is
View answer and explanation
Correct option: C
π\pi
Option C (π\pi) is correct because the potential energy is given by U=−pEcos⁡θU = -pE\cos\theta, which is maximum when cos⁡θ=−1\cos\theta = -1, i.e., θ=π\theta = \pi.
202 Medium
202 of 407
Application +4 | -1
A charge 3μC3 \mu \mathrm{C} is released from rest from a point P\mathrm{P} where electric potential is 20 V20 \mathrm{~V} then its kinetic energy when it reaches infinity is :
View answer and explanation
Correct option: A
60μJ60 \mu \mathrm{J}
Option A (60μJ60 \mu \mathrm{J}) is correct because the work done by the electric field is equal to the change in kinetic energy, which is qVqV.
203 Medium
203 of 407
Application +4 | -1
25 capacitors each of capacitance 1μF1 \mu \mathrm{F} are connected in series to a battery of 100 V.100 \mathrm{~V}. The total charge stored on capacitors is
View answer and explanation
Correct option: C
4.0×10−6C4.0 \times 10^{-6} \mathrm{C}
Option C (4.0×10−6C4.0 \times 10^{-6} \mathrm{C}) is correct because the total charge stored is calculated using the equivalent capacitance of the series combination and the applied voltage.
204 Medium
204 of 407
Application +4 | -1
A parallel plate condenser has plates of area 200  cm2200\,\,c{m^2} and separation 0.05 cm. The space between plates have been filled with a dielectric having K = 8 and then charged to 300 Volts. The stored energy
View answer and explanation
Correct option: C
12.4×10−5J12.4 \times {10^{ - 5}}J
Option C (12.4×10−5J12.4 \times {10^{ - 5}}J) is correct because it is the stored energy calculated using the capacitance with the dielectric and the given voltage.
205 Medium
205 of 407
Analyzing +4 | -1
A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in
View answer and explanation
Correct option: C
Decrease in the potential difference across the plates, reduction in the stored energy, but no change in the charge on the plates
Option C is correct. When a dielectric is inserted with the battery disconnected, the charge remains constant, the potential difference decreases, and the stored energy decreases.
206 Easy
206 of 407
Application +4 | -1
A capacitor of capacitance C1C_1 is charged to a potential VV and then connected in parallel to an uncharged capacitor of capacitance C2C_2. The final potential difference across each capacitor will be
View answer and explanation
Correct option: A
C1VC1+C2\frac{C_1 V}{C_1+C_2}
Option A (C1VC1+C2\frac{{C_1 V}}{{C_1+C_2}}) is correct because the final potential difference is determined by the charge conservation and the total capacitance of the parallel combination.
207 Medium
207 of 407
Understanding +4 | -1
A battery does 200 J of work in charging a
capacitor. The energy stored in the capacitor is:
View answer and explanation
Correct option: B
100 J
Option B (100 J) is correct because the energy stored in a capacitor is half the work done by the battery in charging it.
208 Medium
208 of 407
Understanding +4 | -1
An electric dipole is kept in non-uniform electric field. It experiences
View answer and explanation
Correct option: A
A force and a torque
Option A (A force and a torque) is correct because in a non-uniform electric field, the forces on the charges of the dipole are unequal, resulting in both a net force and a torque.
209 Medium
209 of 407
Remembering +4 | -1
The capacity of parallel plate condenser depends on
View answer and explanation
Correct option: D
The separation between the plates
Option D (The separation between the plates) is correct because the capacitance of a parallel plate capacitor is inversely proportional to the separation between the plates.
210 Medium
210 of 407
Application +4 | -1
Two spherical conductors each of capacity C are charged to potentials VV and −V- V. These are then connected by means of a fine wire. The loss of energy will be
View answer and explanation
Correct option: C
CV2C{V^2}
Option C (CV2C{V^2}) is correct because the loss of energy is calculated using the formula for energy loss in capacitors when connected by a wire.
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