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NEET • PHYSICS • ELECTRIC POTENTIAL AND CAPACITANCE

Electric potential and capacitance practice · Medium

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Showing 1-30 of 175 questions

1 Medium

1 of 175

Analyzing +4 | -1

A parallel plate capacitor is filled with a dielectric whose dimensions are the same as that of the capacitor. The capacitor is connected to a battery.
Statement 1: If the dielectric is pulled out, the electric field inside the capacitor increases.
Statement 2: The electric field inside the dielectric is lesser than the electric field in the region outside the dielectric.

View answer and explanation

Correct option: D

Statement–1 is False, Statement–2 is True

Statement–1 is False, Statement–2 is True because when a dielectric is pulled out with a battery connected, the electric field remains constant, and the electric field inside the dielectric is reduced due to polarization.

2 Medium

2 of 175

Understanding +4 | -1

An electric dipole of moment

{\vec p}

placed in a uniform electric field

{\vec E}

has minimum potential energy when the angle between

{\vec p}

and

{\vec E}

View answer and explanation

Correct option: A

Zero

Option A (Zero) is correct because the potential energy of a dipole in an electric field is minimum when the dipole moment is aligned with the electric field (angle is zero).

3 Medium

3 of 175

Analyzing +4 | -1

Assertion:- Two adjacent conductors carrying the same charge can be at different potentials.
Reason:- The conductors may have different sizes and hence different capacitances.

View answer and explanation

Correct option: A

If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion

Option A (If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion) is correct because the potential depends on capacitance, which is size-dependent.

4 Medium

4 of 175

Understanding +4 | -1

The electric field between the plates of a parallel plate capacitor when connected to a certain battery is

{E_0}.

If the space between the plates of the capacitor is filled by introducing a material of dielectric constant

K

without disturbing the battery connections, the field between the plates shall be

View answer and explanation

Correct option: B

{E_0}

Option B (

{E_0}

) is correct because when the battery remains connected, the potential difference across the capacitor remains constant, and therefore the electric field remains the same.

5 Medium

5 of 175

Analyzing +4 | -1

An air capacitor is charged with an amount of charge

q

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

Increase because when the oil is pumped out, the dielectric constant decreases, and since the charge remains constant, the electric field increases.

6 Medium

6 of 175

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 battery supplies the additional energy required to increase the charge on the capacitor when the dielectric is inserted.

7 Medium

7 of 175

Analyzing +4 | -1

A parallel plate capacitor of plate area

A

and plate separation

d

is charged to potential

V

and then the battery is disconnected. A slab of dielectric constant

k

is then inserted between the plates of the capacitors so as to fill the space between the plates. If

Q,E

and

W

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 = \frac{{{\varepsilon _0}A{V^2}}}{{2d}}\left( {1 - \frac{1}{k}} \right)

Option D (

W = \frac{{{\varepsilon _0}A{V^2}}}{{2d}}\left( {1 - \frac{1}{k}} \right)

) is the incorrect relation because it represents the work done in inserting the dielectric, which should be negative as energy is reduced.

8 Medium

8 of 175

Remembering +4 | -1

E

is the electric field intensity of an electrostatic field, then the electrostatic energy density is proportional to

View answer and explanation

Correct option: B

{E^2}

Option B (

{E^2}

) is correct because the electrostatic energy density is proportional to the square of the electric field intensity.

9 Medium

9 of 175

Remembering +4 | -1

One volt is equivalent to:

View answer and explanation

Correct option: C

Joule/coulomb

Option C (Joule/coulomb) is correct because one volt is defined as one joule of energy per coulomb of charge.

10 Medium

10 of 175

Application +4 | -1

N

identical spherical drops charged to the same potential

V

are combined to form a big drop. The potential of the new drop will be

View answer and explanation

Correct option: D

V \times {N^{2/3}}

Option D (V x N^(2/3)) is correct because when N identical drops combine, the volume increases by a factor of N, and the radius increases by N^(1/3). Since potential is proportional to Q/R and Q increases by N, the potential becomes V x N^(2/3).

11 Medium

11 of 175

Analyzing +4 | -1

Statement-I: Due to two point charges electric field and electric potential can't be zero at some point simultaneously.
Statement-II: Field is a vector quantity and potential a scalar quantity.

View answer and explanation

Correct option: A

Statement-I is True, statement-II is True

Option A (Statement-I is True, statement-II is True) is correct because it's possible to have zero potential but non-zero field, or vice versa, but not both zero simultaneously with just two point charges.

12 Medium

12 of 175

Application +4 | -1

A parallel plate capacitor having a plate separation of

2mm

is charged by connecting it to a

300V

supply. The energy density is

View answer and explanation

Correct option: B

0.1J/{m^3}

Option B (

0.1J/{m^3}

) is correct because it is the energy density calculated using the electric field and permittivity of free space.

13 Medium

13 of 175

Understanding +4 | -1

When a positive

q

charge is taken from lower potential to a higher potential point, then its potential energy will

View answer and explanation

Correct option: B

Increase

Option B (Increase) is correct because moving a positive charge to a higher potential increases its potential energy.

14 Medium

14 of 175

Understanding +4 | -1

An electric dipole in a uniform electric field experiences (When it is placed at an angle

\theta

with the field)

View answer and explanation

Correct option: C

Torque but no force

Option C (Torque but no force) is correct because a uniform electric field exerts a torque on a dipole when it's at an angle, but the net force is zero.

15 Medium

15 of 175

Understanding +4 | -1

Two charges are at a distance

'd'

apart. If a copper plate (conducting medium) of thickness

\frac{d}{2}

is placed between them, the effective force will be

View answer and explanation

Correct option: C

0

Option C (0) is correct because the electric field inside a conductor is always zero, effectively short-circuiting the region between the charges.

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16 Medium

16 of 175

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 assertion is false, while the reason is true.

17 Medium

17 of 175

Understanding +4 | -1

The work done in bringing a unit positive charge from infinite distance to a point at distance

x

from a positive charge

Q

W.

Then the potential

\phi

at that point is

View answer and explanation

Correct option: B

W

Option B is correct because the electric potential at a point is defined as the work done in bringing a unit positive charge from infinity to that point, which is given as

W

18 Medium

18 of 175

Application +4 | -1

A body of capacity

4\;\mu \;F

is charged to

80\;V

and another body of capacity

6\mu \;F

is charged to

30V.

When they are connected the energy lost by

4\mu \;F

capacitor is

View answer and explanation

Correct option: A

7.8\;mJ

Option A (7.8 mJ) is correct because it represents the energy lost when the capacitors are connected, calculated using the difference in initial and final energies.

19 Medium

19 of 175

Remembering +4 | -1

The capacity of a parallel plate capacitor increases with the

View answer and explanation

Correct option: C

Increase of its area

Option C (Increase of its area) is correct because the capacitance of a parallel plate capacitor is directly proportional to the area of the plates.

20 Medium

20 of 175

Application +4 | -1

A capacitor of capacity

C

has charge

Q

and stored energy is

W

. If the charge is increased to

2Q

, the stored energy will be

View answer and explanation

Correct option: C

4W

Option C (

4W

) is correct because the stored energy is proportional to the square of the charge, so doubling the charge quadruples the energy.

21 Medium

21 of 175

Understanding +4 | -1

Two capacitors connected in parallel having the capacities

{C_1}

and

{C_2}

are given

'q'

charge, which is distributed among them. The ratio of the charge on

{C_1}

  and

{C_2}

 will be

View answer and explanation

Correct option: A

\frac{{{C_1}}}{{{C_2}}}

Option A (

\frac{{{C_1}}}{{{C_2}}}

) is correct because in parallel, the voltage is the same across both capacitors, and the charge is proportional to the capacitance.

22 Medium

22 of 175

Understanding +4 | -1

The outer sphere of a spherical air capacitor is earthed. For increasing its capacitance

View answer and explanation

Correct option: B

Dielectric material is filled between the two spheres

Option B (Dielectric material is filled between the two spheres) is correct because introducing a dielectric increases the capacitance of a capacitor.

23 Medium

23 of 175

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. Filling the space with a dielectric increases the capacitance and, since the voltage is constant, it also increases the charge stored.

24 Medium

24 of 175

Analyzing +4 | -1

A parallel plate capacitor with dielectric is charged completely and then the battery is disconnected. Now the dielectric is pulled out. Due to such pulling the

View answer and explanation

Correct option: D

Energy stored in the capacitor will increase

Option D (Energy stored in the capacitor will increase) is correct because when the dielectric is pulled out after disconnecting the battery, the voltage increases, leading to an increase in stored energy.

25 Medium

25 of 175

Application +4 | -1

Three capacitors of capacitance

3\;\mu \;F

are connected in a circuit. Then their maximum and minimum capacitance will be

View answer and explanation

Correct option: A

9\;\mu \;F,\;1\;\mu \;F

Option A (

9 \;\mu \;F,\;1 \;\mu \;F

) is correct because maximum capacitance is achieved in series and minimum in parallel.

26 Medium

26 of 175

Application +4 | -1

Two spherical conductors A and B of radii

1\;mm

and 2

mm

are separated by a distance of

5\;cm

and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres A and B is

View answer and explanation

Correct option: B

2:1

Option B (2:1) is correct because when the spheres are connected, they have the same potential, and the electric field at the surface is E = V/R, so the ratio of electric fields is inversely proportional to the ratio of radii.

27 Medium

27 of 175

Understanding +4 | -1

Statement I : The circuits containing capacitor be handed cautiously, even when there is no current.

Statement II : A dielectric differs from an insulator.

View answer and explanation

Correct option: B

Statement I is true, Statement II is true; Statement II is not the correct explanation for Statement I

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 the dielectric property doesn't explain the cautious handling of capacitors.

28 Medium

28 of 175

Remembering +4 | -1

‘Debye’ is the unit of

View answer and explanation

Correct option: B

Electric dipolement

Option B (Electric dipolement) is correct because Debye is the unit of electric dipole moment.

29 Medium

29 of 175

Analyzing +4 | -1

Statement-1: Two concentric conducting spherical shells are charged. The charge on the outer shell is varied keeping the charge on inner shell constant, as a result the electric potential difference between the two shells does not change.
Statement-2: If charge is changed on an isolated thin conducting spherical shell, the potential at all points inside the shell changes by same amount.

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 changing the outer shell's charge changes the potential by the same amount everywhere inside, including the inner shell, thus not affecting the potential difference.

30 Medium

30 of 175

Application +4 | -1

If 3 charges are placed at the vertices of equilateral triangle of charge

'q'

each. What is the net potential energy, if the side of equilateral

\Delta

l\;cm

View answer and explanation

Correct option: C

\frac{1}{{4\pi {\varepsilon _0}}}\frac{{3{q^2}}}{l}

Option C (

\frac{1}{{4\pi {\varepsilon _0}}}\frac{{3{q^2}}}{l}

) is correct because the total potential energy is the sum of the potential energies of each pair of charges, and there are three such pairs.

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Key Takeaways

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