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JEE • PHYSICS • ELECTRIC POTENTIAL AND CAPACITANCE

Electric potential and capacitance practice · Medium

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Practice format and chapter quiz

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 372 questions

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181 Medium
181 of 372
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.
182 Medium
182 of 372
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.
183 Medium
183 of 372
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.
184 Medium
184 of 372
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).
185 Medium
185 of 372
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.
186 Medium
186 of 372
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.
187 Medium
187 of 372
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.
188 Medium
188 of 372
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.
189 Medium
189 of 372
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.
190 Medium
190 of 372
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.
191 Medium
191 of 372
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.
192 Medium
192 of 372
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.
193 Medium
193 of 372
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.
194 Medium
194 of 372
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.
195 Medium
195 of 372
Remembering +4 | -1
The energy stored in the condenser is
View answer and explanation
Correct option: B
12QV\frac{1}{2}QV
Option B (12QV\frac{1}{2}QV) is correct because the energy stored in a capacitor is given by the formula U=12QVU = \frac{1}{2}QV, where Q is the charge and V is the voltage.
196 Medium
196 of 372
Understanding +4 | -1
The electric field between the plates of a parallel plate capacitor when connected to a certain battery is E0.{E_0}. If the space between the plates of the capacitor is filled by introducing a material of dielectric constant KK without disturbing the battery connections, the field between the plates shall be
View answer and explanation
Correct option: B
E0{E_0}
Option B (E0{E_0}) is correct because the battery maintains a constant potential difference across the capacitor, and the electric field is directly proportional to the potential difference.
197 Medium
197 of 372
Understanding +4 | -1
When a capacitor is connected to a battery
View answer and explanation
Correct option: C
A current flow for sometime and finally it
decreases to zero
Option C (A current flow for sometime and finally it decreases to zero) is correct because the capacitor charges until its voltage equals the battery voltage, at which point the current stops.
198 Medium
198 of 372
Understanding +4 | -1
Two capacitors connected in parallel having the capacities C1{C_1} and C2{C_2} are given ′q′'q' charge, which is distributed among them. The ratio of the charge on C1{C_1}  and C2{C_2} will be
View answer and explanation
Correct option: A
C1C2\frac{{{C_1}}}{{{C_2}}}
Option A (C1C2\frac{{{C_1}}}{{{C_2}}}) is correct because in a parallel combination, the voltage across both capacitors is the same, leading to the charge ratio being equal to the capacitance ratio.
199 Medium
199 of 372
Analyzing +4 | -1
An uncharged capacitor is connected to a battery. Show that half the energy supplied by the battery is lost as heat while charging the capacitor.
View answer and explanation
Correct option: B
12QV\frac{1}{2} Q V
Option B (12QV\frac{1}{2} Q V) is correct because the energy stored in the capacitor is half the energy supplied by the battery, with the other half dissipated as heat.
200 Medium
200 of 372
Remembering +4 | -1
In an electric field E\mathbf{E}, the torque acting on a dipole moment p\mathbf{p} is
View answer and explanation
Correct option: B
p×E\mathbf{p} \times \mathbf{E}
Option B (p×E\mathbf{p} \times \mathbf{E}) is correct because the torque on a dipole in an electric field is given by the cross product of the dipole moment and the electric field.
201 Medium
201 of 372
Application +4 | -1
A half ring of radius RR has a charge of λ\lambda per unit length. The potential at the centre of the half ring is (k=14πε0)\left( {k = \frac{1}{{4\pi {\varepsilon _0}}}} \right)
View answer and explanation
Correct option: D
KπλK\pi \lambda
Option D (KπλK\pi \lambda) is correct because the potential at the center is calculated by integrating the potential due to each infinitesimal charge element.
202 Medium
202 of 372
Understanding +4 | -1
Equipotential surfaces associated with an electric field which is increasing in magnitude along the x−x - direction are
View answer and explanation
Correct option: A
Planes parallel to yz−yz - plane
Option A is correct because equipotential surfaces are always perpendicular to the electric field, and if the field increases along the x-direction, the equipotential surfaces must be planes parallel to the yz-plane.
203 Medium
203 of 372
Application +4 | -1
Dipole is placed parallel to the electric field. If Q is the work done in rotating the dipole by 60°, then work done in rotating it by 180° is
View answer and explanation
Correct option: C
4W
Option C (4W) is correct because the work done in rotating a dipole in an electric field depends on the change in potential energy, and rotating by 180° requires four times the work of rotating by 60° from the parallel position.
204 Medium
204 of 372
Understanding +4 | -1
A test charge is moved from lower potential point to a higher potential point. The potential energy of test charge will:
View answer and explanation
Correct option: B
Increase
Option B (Increase) is correct because moving a positive test charge to a higher potential increases its potential energy.
205 Medium
205 of 372
Application +4 | -1
Two points PP and QQ are maintained at the potential of 10  V10\;V and −4V- 4V, respectively. The work done in moving 100 electrons from PP to QQ is
View answer and explanation
Correct option: D
2.24×10−16J2.24 \times {10^{ - 16}}J
Option D (2.24×10−16J2.24 \times {10^{ - 16}}J) is correct because the work done is calculated by multiplying the number of electrons, the electron charge, and the potential difference.
206 Medium
206 of 372
Understanding +4 | -1
A parallel plate capacitor is charged. If the plates are pulled apart
View answer and explanation
Correct option: B
The potential difference increases
Option B (The potential difference increases) is correct because as the plates are pulled apart, the capacitance decreases, and since charge is constant, the potential difference must increase.
207 Medium
207 of 372
Application +4 | -1
If potential (in volts ) in a region is expressed as V(x,y,z)=6xy−y+2yz,V(x,y,z) = 6xy - y + 2yz, the electric field ( in N/C) at point (1,1,0) is
View answer and explanation
Correct option: D
−(6i^+5j^+2k^)- (6\hat i + 5\hat j + 2\hat k)
Option D (−(6i^+5j^+2k^)- (6\hat i + 5\hat j + 2\hat k)) is correct because it is found by taking the negative gradient of the potential function and evaluating it at the given point.
208 Medium
208 of 372
Understanding +4 | -1
Potentiometer measures potential more accurately because
View answer and explanation
Correct option: A
It measures potential in open circuit
Option A is correct because a potentiometer measures potential in an open circuit, preventing current draw and ensuring accurate measurement.
209 Medium
209 of 372
Understanding +4 | -1
The region surrounding a stationary electric dipole has:
View answer and explanation
Correct option: A
Electric field only
Option A (Electric field only) is correct because a stationary electric dipole produces an electric field in its surrounding region, but since the charges are not moving, there is no magnetic field.
210 Medium
210 of 372
Application +4 | -1
An electric field E=(20i^+25j^)E=(20 \hat{i}+25 \hat{j}) exist in space. If the potential at the origin is zero, then the potential at point (2 m,4 m)(2 \mathrm{~m}, 4 \mathrm{~m}) is:
View answer and explanation
Correct option: D
-140 volt
The correct answer is D (-140 volt). The potential at a point in an electric field is calculated by integrating the electric field along the path from the origin to that point.
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