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NEET • PHYSICS • ELECTRIC POTENTIAL AND CAPACITANCE

Electric potential and capacitance practice · Hard

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1 Hard
1 of 23
Analyzing +4 | -1
Statement-1: A parallel plate capacitor is charged to a potential difference of 100 V100 \mathrm{~V}, and disconnected from the voltage source. A slab of dielectric is then slowly inserted between the plates. Compared to the energy before the slab was inserted, the energy stored in the capacitor with the dielectric is decreased.
Statement-2: When we insert a dielectric between the plates of a capacitor, the induced charges tend to draw in the dielectric into the field (just as neutral objects are attracted by charged objects due to induction). We resist this force while slowly inserting the dielectric, and thus do negative work on the system, removing electrostatic energy from the system.
View answer and explanation
Correct option: A
Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation for Statement-1
Option A (Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation for Statement-1) is correct because inserting a dielectric after disconnecting the voltage source decreases the energy stored, and this is due to the negative work done against the attractive force on the dielectric.
2 Hard
2 of 23
Application +4 | -1
A parallel plate capacitor consisting of two circular plates of radius 10 cm is being charged by a constant current of 0.15 A. If the rate of change of potential difference between the plates is 7×108V/s7 \times 10^8 \mathrm{V} / \mathrm{s} then the integer value of the distance between the parallel plates is ____ μm.\mu \mathrm{m}.

(Take ϵ0=9×10−12 F m,π=227)\left(\text{Take }\epsilon_0=9 \times 10^{-12} \frac{\mathrm{~F}}{\mathrm{~m}}, \pi=\frac{22}{7}\right)
View answer and explanation
Correct option: A
1320
Option A (1320) is correct because it is the calculated distance between the plates based on the given rate of change of potential difference and other parameters.
3 Hard
3 of 23
Application +4 | -1
A large insulated sphere of radius rr charged with QQ units of electricity is placed in contact with a small insulated uncharged sphere of radius r′r^{\prime} and in then separated. The charge on smaller sphere will now be
View answer and explanation
Correct option: B
Qr′r′+r\frac{Q r^{\prime}}{r^{\prime}+r}
Option B (Qr′r′+r\frac{Q r^{\prime}}{r^{\prime}+r}) is correct because the charge distributes proportionally to the radii of the spheres when they are connected.
4 Hard
4 of 23
Analyzing +4 | -1
A system consists of four equal charges - Q\mathrm{Q} placed at the four corners of a square and another charge q\mathrm{q} is placed at the centre of the square, then the value of q\mathrm{q} for which the system is equilibrium, is
View answer and explanation
Correct option: D
Q4(1+22)\frac{Q}{4}(1+2 \sqrt{2})
Option D (Q4(1+22)\frac{Q}{4}(1+2 \sqrt{2})) is correct because, at equilibrium, the net force on each charge at the corner of the square must be zero. This requires a specific value for q at the center.
5 Hard
5 of 23
Application +4 | -1
A charge QQ is uniformly distributed through out the volume of a dielectric sphere of radius RR. If potential at its surface is 24 volts. Then potential at r=R2r=\frac{R}{2} and r=3R2r=\frac{3 R}{2} respectively are. (where rr distance from the centre of sphere) :-
View answer and explanation
Correct option: B
33 V,16 V33 \mathrm{~V}, 16 \mathrm{~V}
Option B (33 V, 16 V) is correct. Inside the sphere (r < R), V = (kQ/2R3)(3R2 - r2). Outside the sphere (r > R), V = kQ/r. Using these, we find the potentials at r = R/2 and r = 3R/2.
6 Hard
6 of 23
Analyzing +4 | -1
Assertion (A): At a point electric potential is decreasing along xx-axis at a rate of 10 V/m10 \mathrm{~V} / \mathrm{m}. Therefore, xx-component of electric field at this point should be 10 V/m10 \mathrm{~V} / \mathrm{m} along xx-axis.

Reason (R) : Magnitude of Ex=∂V∂xE_x=\frac{\partial V}{\partial x}
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 assertion is false (it should be negative of the rate), but the reason is true (Ex=−∂V∂xE_x = -\frac{\partial V}{\partial x}).
7 Hard
7 of 23
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)}}
kA2ln⁡(2ε0+kd2ε0)\frac{{kA}}{{2\ln \left( {\frac{{2{\varepsilon _0} + kd}}{{2{\varepsilon _0}}}} \right)}} is correct because it is derived by integrating the capacitance over the varying permittivity with respect to distance.
8 Hard
8 of 23
Remembering +4 | -1
The energy stored in a parallel plate capacitor can be treated as the energy of the electric field. The energy per unit volume, due to the electric field is
View answer and explanation
Correct option: B
12ε0E2\frac{1}{2} \varepsilon_0 E^2
Option B (12ε0E2\frac{1}{2} \varepsilon_0 E^2) is correct because it represents the energy per unit volume (energy density) of the electric field in a capacitor.
9 Hard
9 of 23
Understanding +4 | -1
Given below are two statements.
Statement I : Electric potential is constant within
and at the surface of each conductor.
Statement II : Electric field just outside a charged
conductor is perpendicular to the surface of the
conductor at every point.
In the light of the above statements, choose the
most appropriate answer from the options given
below.
View answer and explanation
Correct option: A
Both statement I and statement II are correct
Option A (Both statement I and statement II are correct) is correct because electric potential is constant within and on the surface of a conductor, and the electric field is always perpendicular to the surface.
10 Hard
10 of 23
Understanding +4 | -1
While a capacitor remains connected to a battery and dielectric slab is applied between the plates, then
View answer and explanation
Correct option: B
Charge flows from the battery to the capacitor
Option B (Charge flows from the battery to the capacitor) is correct because the capacitance increases with the dielectric, and since the voltage is constant, the charge must increase.
11 Hard
11 of 23
Analyzing +4 | -1
Assertion (A): Two charges −q-q each are fixed at points AA and BB. When a third charge −q-q is moved from AA to BB, electrical potential energy first decreases than increases.

Reason (R) : Along the line joining AA and BB, the third charge is in stable equilibrium position at centre.
View answer and explanation
Correct option: A
If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.
Option A (If both Assertion and Reason are true and the Reason is correct explanation of the Assertion) is correct because the potential energy decreases as the third charge moves towards the midpoint (stable equilibrium) and then increases as it moves away.
12 Hard
12 of 23
Understanding +4 | -1
A point charge +Q+ Q is located at a distance of ′r′'r' from the centre of an uncharged conducting sphere of radius RR. The electric potential of that sphere is
View answer and explanation
Correct option: A
14π∈0QR\frac{1}{{4\pi { \in _0}}}\frac{Q}{R}
Option A is correct because the potential of a conducting sphere is constant throughout and equal to the potential at its surface, which is due to the induced charge distribution.
13 Hard
13 of 23
Analyzing +4 | -1
Statement-1: If temperature is increased, the dielectric constant of a polar dielectric decreases whereas that of a non-polar dielectric does not change significantly.
Statement-2: The magniude of dipole moment of individual polar molecule decreases significantly with increase in temperature.
View answer and explanation
Correct option: D
Statement-1 is false, Statement-2 is true
Option D (Statement-1 is false, Statement-2 is true) is correct. The dielectric constant of polar dielectrics decreases with increasing temperature, but Statement-2 is true because the dipole moment of individual polar molecules does not decrease significantly; rather, their alignment decreases due to thermal agitation.
14 Hard
14 of 23
Application +4 | -1
If the distance between the plates of a capacitor is d\mathrm{d} and potential difference is V\mathrm{V} then what is the energy density between the plates?
View answer and explanation
Correct option: C
12ε0V2 d2\frac{1}{2} \varepsilon_0 \frac{\mathrm{V}^2}{\mathrm{~d}^2}
Option C (12ε0V2 d2\frac{1}{2} \varepsilon_0 \frac{\mathrm{V}^2}{\mathrm{~d}^2}) is correct because energy density is given by 12ε0E2\frac{1}{2} \varepsilon_0 E^2 and E=VdE = \frac{V}{d}.
15 Hard
15 of 23
Analyzing +4 | -1
Assertion (A) : If we see along the axis of a charged ring, the magnitude of electric field is minimum at centre and magnitude of electric potential is maximum.

Reason (R): Electric field is a vector quantity while electric potential is scalar.
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 (If both Assertion and Reason are true but Reason is not the correct explanation of Assertion) is correct because while both statements are true, the scalar vs. vector nature doesn't directly explain the field/potential behavior on the ring's axis.
16 Hard
16 of 23
Application +4 | -1
A capacitor of capacitance CC has charge QQ and stored energy is W.W. If the charge is increased to 3Q,3Q, the stored energy will be
View answer and explanation
Correct option: C
9W9W
Option C (9W9W) is correct because the energy stored is proportional to the square of the charge, so tripling the charge increases the energy by a factor of 9.
17 Hard
17 of 23
Application +4 | -1
An electric field (−30i^+20j^)Vm−1(-30 \hat{\mathrm{i}}+20 \hat{\mathrm{j}}) \mathrm{Vm}^{-1} exists in the space. If potential at the origin is zero then potential at (5 m,3 m)(5 \mathrm{~m}, 3 \mathrm{~m}) is given by
View answer and explanation
Correct option: B
90 V90 \mathrm{~V}
Option B (90 V) is correct because the potential difference is calculated by integrating the electric field along the path from the origin to (5 m, 3 m).
18 Hard
18 of 23
Analyzing +4 | -1
Statement-1: The heat produced by a resistor in any time tt during the charging of a capacitor in a series circuit is half the energy stored in the capacitor by that time.
Statement-2: Current in the circuit is equal to the rate of increase in charge on the capacitor.
View answer and explanation
Correct option: D
Statement-1 is false, Statement-2 is true
Option D (Statement-1 is false, Statement-2 is true) is correct. The heat produced by the resistor is NOT half the energy stored in the capacitor; it's more than half. Statement-2 is true because current is indeed the rate of change of charge.
19 Hard
19 of 23
Analyzing +4 | -1
Assertion (A): Electric potential on the surface of a charged sphere of radius RR is VV. Then electric field at a distance r=R2r=\frac{R}{2} from centre is V2R\frac{V}{2 R}. Charge is distributed uniformly over the volume.

Reason (R) : From centre to surface, electric field varies linearly with rr. Here, rr is distance from centre.
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 (If both Assertion and Reason are true but Reason is not the correct explanation of Assertion) is correct because while both statements are true, the linear variation of the electric field doesn't directly explain the potential at the surface.
20 Hard
20 of 23
Analyzing +4 | -1
If the plates of a parallel plate capacitor connected to a battery are moved close to each other, then

A. the charge stored in it, increases.

B. the energy stored in it, decreases.

C. its capacitance increases.

D. the ratio of charge to its potential remains the same.

E. the product of charge and voltage increases.

Choose the most appropriate answer from the options given below:
View answer and explanation
Correct option: A
A, C and E only
Option A (A, C and E only) is correct because moving the plates closer increases capacitance and charge, and consequently increases the product of charge and voltage, while the energy stored increases, not decreases.
21 Hard
21 of 23
Application +4 | -1
Two particles of charges 4 nC and Q are kept in air with a separation of 10 cm between them. If the electrostatic potential energy of the system is 1.8 μJ,1.8 \, \mu \mathrm{J}, then Q =
View answer and explanation
Correct option: C
5 nC
Option C (5 nC) is correct because it is calculated using the formula for electrostatic potential energy between two point charges.
22 Hard
22 of 23
Analyzing +4 | -1
Statement-1: A dielectric is inserted between the plates of an isolated fully-charged capacitor. The dielectric completely fills the space between the plates. The magnitude of electrostatic force on either metal plate decreases, as it was before the insertion of dielectric medium.
Statement-2: Due to insertion of dielectric slab in an isolated parallel plate capacitor (the dielectric completely fills the space between the plates), the electrostatic potential energy of the capacitor decreases.
View answer and explanation
Correct option: D
Statement-1 is false, Statement-2 is true
Option D (Statement-1 is false, Statement-2 is true) is correct. The force increases, and the potential energy decreases.
23 Hard
23 of 23
Analyzing +4 | -1
Assertion (A): In the electric field E=(4i^+4j^)N/C\mathbf{E}=(4 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) \mathrm{N} / \mathrm{C}, electric potential at A(4 m,0)A(4 \mathrm{~m}, 0) is more than the electric potential at B(0,4 m)B(0,4 \mathrm{~m}).

Reason (R): Electric lines of forces always travel from higher potential to lower potential.
View answer and explanation
Correct option: D
If Assertion is false but the Reason is true.
Option D is correct. The assertion is false because electric potential decreases in the direction of the electric field, and the reason is true because electric field lines point from higher to lower potential.
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Key Takeaways

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