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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 151-180 of 407 questions

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151 Medium
151 of 407
Application +4 | -1
A series combination of N1\text{N}_1 capacitors (each of capacity C1\text{C}_1) is charged to potential difference 3 V.3 \mathrm{~V}. Another parallel combination of N2\mathrm{N}_2 capacitors (each of capacity C2\text{C}_2) is charged to potential difference V.\text{V}. The total energy stored in both the combinations is same. The value of C1\text{C}_1 in terms of C2\text{C}_2 is
View answer and explanation
Correct option: A
C2N1N29\frac{\mathrm{C}_2 \mathrm{N}_1 \mathrm{N}_2}{9}
Option A (C2N1N29\frac{\mathrm{C}_2 \mathrm{N}_1 \mathrm{N}_2}{9}) is correct because equating the energies stored in the series and parallel combinations allows us to solve for C1 in terms of C2, N1, and N2.
152 Medium
152 of 407
Understanding +4 | -1
The potential to which a conductor is raised, depends on
View answer and explanation
Correct option: C
Both (1) and (2)
Option C (Both (1) and (2)) is correct because the potential of a conductor depends on both the amount of charge and its geometry and size.
153 Medium
153 of 407
Application +4 | -1
A parallel plate condenser has a capacitance 50μF50\mu F in air and 110μF110\mu F when immersed in an oil. The dielectric constant ′k′'k' of the oil is
View answer and explanation
Correct option: D
2.2
Option D (2.2) is correct because the dielectric constant is the ratio of capacitance in the medium to the capacitance in air.
154 Medium
154 of 407
Application +4 | -1
The equation of an equi-potential line in an electric field is y=2xy=2 x, then the electric field strength vector at (1,2)(1,2) may be:
View answer and explanation
Correct option: D
−8i^+4j^-8 \hat{i}+4 \hat{j}
Option D (-8i^\hat{i}+4j^\hat{j}) is correct because the electric field must be perpendicular to the equipotential line, and this option satisfies that condition.
155 Medium
155 of 407
Understanding +4 | -1
Assertion: For a point charge, concentric spheres centered at a location of the charge are equipotential surfaces.
Reason: An equipotential surface is a surface over which potential has constant value.
View answer and explanation
Correct option: B
If both assertion and reason are true but reason is not the correct explanation of assertion
Option B is correct because both the assertion and reason are true, but the reason doesn't explain why concentric spheres are equipotential surfaces for a point charge.
156 Medium
156 of 407
Understanding +4 | -1
When two conductors of charges and potentials C1,V1C_1, V_1 and C2,V2C_2, V_2 respectively are joined, the common potential will be:
View answer and explanation
Correct option: D
C1V1+C2V2C1+C2\frac{C_1 V_1+C_2 V_2}{C_1+C_2}
Option D (C1V1+C2V2C1+C2\frac{C_1 V_1+C_2 V_2}{C_1+C_2}) is correct because the common potential is found by equating the total charge before and after joining the conductors and solving for the final potential.
157 Medium
157 of 407
Analyzing +4 | -1
The electrostatic force between the metal plates of an isolated parallel plate capacitor CC having a charge QQ and area A{A}, is
View answer and explanation
Correct option: A
Independent of the distance between the plates
Option A is correct. The electrostatic force between the plates of an isolated parallel plate capacitor is independent of the distance between the plates because the charge Q is constant.
158 Medium
158 of 407
Understanding +4 | -1
Potential at any point inside a charged hollow
sphere:
View answer and explanation
Correct option: B
Is a constant
Option B (Is a constant) is correct because the electric field inside a charged hollow sphere is zero, implying that the potential is constant.
159 Medium
159 of 407
Application +4 | -1
12 J12 \mathrm{~J} of work has to be done against an existing electric field to take a charge of 0.01C0.01 \mathrm{C} from AA to BB. How much is the potential difference VB−VAV_B-V_A ?
View answer and explanation
Correct option: C
1200 V
Option C (1200 V) is correct because the potential difference is the work done per unit charge.
160 Medium
160 of 407
Understanding +4 | -1
The plates of a charged condenser is connected to a voltmeter. If the plates are moved apart, the reading of voltmeter will
View answer and explanation
Correct option: A
Increase
Option A (Increase) is correct because as the plates move apart, capacitance decreases, and since charge is constant, voltage increases.
161 Medium
161 of 407
Application +4 | -1
The difference between equivalent capacitances of two identical capacitors connected in parallel to that in series is 6μF.6 \mu \mathrm{F}. The value of capacitance of each capacitor is:
View answer and explanation
Correct option: C
4μF4 \mu \mathrm{F}
Option C (4μF4 \mu \mathrm{F}) is correct because the capacitance of each capacitor is found by setting up and solving an equation based on the difference between parallel and series combinations.
162 Medium
162 of 407
Application +4 | -1
The capacitance of two concentrie spherical shells of radii R1R_1 and R2(R2>R1)R_2\left(R_2>R_1\right) is
View answer and explanation
Correct option: D
4πε0R2R1R2−R14 \pi \varepsilon_{0} \frac{{R}_2 {R}_1}{{R}_2-{R}_1}
Option D (4πε0R2R1R2−R14 \pi \varepsilon_{0} \frac{{R}_2 {R}_1}{{R}_2-{R}_1}) is correct because it is the formula for the capacitance of two concentric spherical shells.
163 Medium
163 of 407
Application +4 | -1
An infinite sheet carrying a uniform surface charge density σ\sigma lies on the xyx y plane. The work done to carry a charge qq from the point A=a(i^+2j^+3k^)A=a(\hat{{i}}+2 \hat{{j}}+3 \hat{{k}}) to the point B=a(i^−2j^+6k^)B=a(\hat{{i}}-2 \hat{{j}}+6 \hat{{k}}) (where aa is a constant with the dimension of length and ε0\varepsilon_0 is the permittivity of free space) is
View answer and explanation
Correct option: A
3σaq2ε0\frac{3 \sigma a q}{2 \varepsilon_0}
Option A (3σaq2ε0\frac{3 \sigma a q}{2 \varepsilon_0}) is correct because the work done depends only on the change in the z-coordinate, given the electric field is only in the z-direction.
164 Medium
164 of 407
Application +4 | -1
A variable condenser is permanently connected to a 100  V100\;V battery. If the capacity is changed from 2μ  F2\mu \;F to 10  μ  F,10\;\mu \;F, then change in energy is equal to
View answer and explanation
Correct option: D
4×10−2  J4 \times {10^{ - 2}}\;J
Option D (4×10−2  J4 \times {10^{ - 2}}\;J) is correct because the change in energy is calculated using the formula ΔU=V22(C2−C1)\Delta U = \frac{{{V^2}}}{2}\left( {{C_2} - {C_1}} \right).
165 Medium
165 of 407
Analyzing +4 | -1
The potential energies associated with four orientations of an electric dipole in an electric field are:

(i)−5U0(ii)−7U0\text{(i)}-5 \mathrm{U}_0 \quad \text{(ii)}-7 \mathrm{U}_0 \quad (iii) 3U0(iv) 5U0\text{(iii) }3 \mathrm{U}_0 \quad \text{(iv) }5 \mathrm{U}_0

Where U0\text{U}_0 is positive. Rank the orientations according to the angle between the electric dipole moment P→\overrightarrow{\mathrm{P}} and the electric field E→\overrightarrow{\mathrm{E}}, greatest first
View answer and explanation
Correct option: C
(iv), (iii), (i), (ii)
Option D ((iv), (i), (iii), (ii)) is correct because the potential energy is inversely proportional to the cosine of the angle between the dipole moment and the electric field, so the most positive potential energy corresponds to the largest angle.
166 Easy
166 of 407
Application +4 | -1
The capacity of a parallel plate capacitor is 8.85×10−12  F8.85 \times {10^{ - 12}}\,\,F and area of each plate is 4  cm24\,\,c{m^2} then the distance between two plates is
View answer and explanation
Correct option: B
4 cm
Option B (4 cm) is correct because the distance between the plates is calculated using the formula for the capacitance of a parallel plate capacitor.
167 Medium
167 of 407
Understanding +4 | -1
Can two equipotential surfaces intersect each other?
View answer and explanation
Correct option: B
No
Option B (No) is correct because if equipotential surfaces intersected, there would be two different electric potentials at the point of intersection, which is impossible.
168 Medium
168 of 407
Understanding +4 | -1
A hollow metal sphere of radius R\text{R} is charged with a charge Q.\text{Q}. The electric potential and intensity inside the sphere are respectively
View answer and explanation
Correct option: B
Q4πε0R\frac{\mathrm{Q}}{4 \pi \varepsilon_0 \mathrm{R}} and zero
Option B (Q4πε0R\frac{\mathrm{Q}}{4 \pi \varepsilon_0 \mathrm{R}} and zero) is correct because the electric potential inside a charged hollow sphere is constant and equal to the potential at the surface, while the electric field inside is zero.
169 Easy
169 of 407
Analyzing +4 | -1
Assertion (A) : When an uncharged capacitor is charged by a battery, only 50%50 \% of the energy supplied by a battery is stored in the capacitor.

Reason (R): Rest 50%50 \% is lost.
View answer and explanation
Correct option: B
If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
Option B is correct because both the assertion and the reason are true, but the reason does not fully explain the assertion. The energy loss is due to resistance in the connecting wires.
170 Medium
170 of 407
Understanding +4 | -1
A uniform electric field pointing in positive x-direction exists in a region let AA be the origin BB be the point on x-axis at x=+1 cmx = + 1\,cm and CC be the point on y-axis at y=+1 cmy = + 1\,cm. Then the potentials at points A,B, and C satisfy.
View answer and explanation
Correct option: B
VA>VB{V_A} > {V_B}
Option B (VA>VB{V_A} > {V_B}) is correct because electric potential decreases along the direction of the electric field.
171 Medium
171 of 407
Application +4 | -1
Two capacitors of capacitances 20.0pF20.0 \mathrm{pF} and 50.0pF50.0 \mathrm{pF} are connected in series with a 6.00 V6.00 \mathrm{~V} battery. Find the potential difference across each capacitor.
View answer and explanation
Correct option: A
4.284 V; 1.71 V
Option A (4.284 V; 1.71 V) is correct because it is calculated by finding the equivalent capacitance and then determining the voltage across each capacitor in series.
172 Medium
172 of 407
Understanding +4 | -1
Can a metal be used as a medium for dielectric
View answer and explanation
Correct option: B
No
Option B (No) is correct because metals are conductors, not insulators, and dielectrics must be insulators to store electrical energy.
173 Medium
173 of 407
Analyzing +4 | -1
An air capacitor is charged with an amount of charge qq and dipped into an oil tank. If the oil is pumped out, the electric field between the plates of capacitor will
View answer and explanation
Correct option: A
Increase
Option A (Increase) is correct because when the oil is pumped out, the voltage increases, leading to an increase in the electric field.
174 Medium
174 of 407
Application +4 | -1
A capacitor stores 10μC10 \mu \mathrm{C} charge when connected across a battery. When the gap between the plates is filled with a dielectric, a charge of 20μC20 \mu \mathrm{C} flows through the battery. Find the dielectric constant of the dielectric.
View answer and explanation
Correct option: C
k=3\mathrm{k}=3
Option C (k=3) is correct because the dielectric constant is the ratio of the charge stored with the dielectric to the charge stored without it, given the same voltage.
175 Medium
175 of 407
Understanding +4 | -1
The potential energy of a system of two charges is negative when:
View answer and explanation
Correct option: C
One charge is positive and other is negative
Option C (One charge is positive and other is negative) is correct because the potential energy of a system of two charges is negative when the charges have opposite signs, indicating an attractive force.
176 Medium
176 of 407
Application +4 | -1
Two circular plates each of radius 'r\text{r}' and charge 'q\text{q}' form a parallel plate capacitor. Then the force of attraction between the plates is (Medium between the plate is air)
View answer and explanation
Correct option: D
q22πε0r2\frac{\mathrm{q}^2}{2 \pi \varepsilon_0 \mathrm{r}^2}
Option D (q22πε0r2\frac{\mathrm{q}^2}{2 \pi \varepsilon_0 \mathrm{r}^2}) is correct because it is the force of attraction calculated using the electric field due to one plate and the charge on the other plate.
177 Medium
177 of 407
Understanding +4 | -1
Torque acting on an electric dipole in a uniform electric field is maximum if the angle between p and E is
View answer and explanation
Correct option: C
90°
Option C (90°) is correct because the torque on an electric dipole in a uniform electric field is maximum when the angle between the dipole moment and the electric field is 90 degrees.
178 Easy
178 of 407
Understanding +4 | -1
If the plates of a charged capacitor are brought closer, then the pp.(dd) between the plates will
View answer and explanation
Correct option: A
Decrease.
Option A (Decrease) is correct because as the distance between the plates decreases, the capacitance increases, leading to a decrease in potential difference.
179 Hard
179 of 407
Application +4 | -1
Consider the combination of two capacitors C1{C_1} and C2,{C_2}, with C2 >C1, {C_2}\, > {C_1},\, when connected in parallel, the equivalent capacitance is 15/415/4 time the equivalent capacitance of the same connected in series. Calculate the ratio of capacitors C2C1\frac{{{C_2}}}{{{C_1}}}.
View answer and explanation
Correct option: D
None of the above
The correct answer is D (None of the above) because the calculated ratio results in a complex number, which is not among the provided options.
180 Medium
180 of 407
Remembering +4 | -1
Stored electrostatic energy in a capacitor of capacity C,C, which is charged upto QQ charge, is:
View answer and explanation
Correct option: B
Q22C\frac{Q^2}{2 {C}}
Option B (Q22C\frac{Q^2}{2 {C}}) is correct because the electrostatic energy stored in a capacitor is given by the formula U=Q22CU = \frac{Q^2}{2C}.
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