A medium having dielectric constant K\gt 1 fills the space between the plates of a parallel plate capacitor. The plates have large area, and the distance between them is d . The capacitor is connected to a battery of voltage V , as shown in Figure (a). Now, both the plates are moved by a distance of \fracd2 from their original positions, as shown in Figure (b). In the process of going from the configuration depicted in Figure (a) to that in Figure (b), which of the following statement(s) is(are) correct?

The capacitance is decreased by a factor of \frac1K+1 .

The electric field inside the dielectric material is reduced by a factor of 2 K .

The voltage between the capacitor plates is increased by a factor of (K+1) .

The work done in the process DOES NOT depend on the presence of the dielectric material.

Capacitor PYQ — JEE Advanced 2022 · Shift 1 · Q45 with Solution

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Correct answer: B

Initial configuration (Figure a)

A parallel plate capacitor of plate separation $d$ is completely filled with dielectric of dielectric constant $K$ . Since it is connected to a battery, the potential difference remains constant at $V$ .

So the initial capacitance is

C_i = \frac{K\varepsilon_0 A}{d}

The initial electric field inside the dielectric is

E_i = \frac{V}{d}

because the entire separation is filled with dielectric and the battery maintains voltage $V$ across distance $d$ .

Final configuration (Figure b)

Both plates are moved by $\frac d2$ from their original positions as shown. This means the dielectric slab remains in the middle region, while vacuum gaps of thickness $\frac d2$ are created on the two sides together. Equivalently, the space between plates now consists of:

dielectric region of thickness $d$
air/vacuum region of total thickness $d$

Hence the system behaves like two capacitors in series:

dielectric-filled capacitor of thickness $d$ , capacitance $C_1 = \frac{K\varepsilon_0 A}{d}$
air capacitor of thickness $d$ , capacitance $C_2 = \frac{\varepsilon_0 A}{d}$

Thus,

\frac{1}{C_f} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{d}{K\varepsilon_0 A} + \frac{d}{\varepsilon_0 A} = \frac{d}{\varepsilon_0 A}\left(\frac{1}{K}+1\right)

So,

C_f = \frac{\varepsilon_0 A}{d\left(1+\frac{1}{K}\right)} = \frac{K\varepsilon_0 A}{d(K+1)}

Now compare with initial capacitance:

C_i = \frac{K\varepsilon_0 A}{d}

Therefore,

\frac{C_f}{C_i} = \frac{1}{K+1}

Hence the capacitance becomes

C_f = \frac{C_i}{K+1}

So it is decreased by a factor of $\frac{1}{K+1}$ .

Thus, Option B is correct.

Check Option A: electric field inside dielectric

Since the battery remains connected, total voltage across the combination remains $V$ .

Let electric field in dielectric be $E_d$ and in air be $E_a$ . For series arrangement, electric displacement is same:

\varepsilon_0 E_a = K\varepsilon_0 E_d

E_a = K E_d

Total voltage is

V = E_d(d) + E_a(d) = E_d d + K E_d d = (K+1)E_d d

Thus

E_d = \frac{V}{(K+1)d}

Initially,

E_i = \frac{V}{d}

So the field inside dielectric is reduced by factor

\frac{E_d}{E_i} = \frac{1}{K+1}

not by $\frac{1}{2K}$ .

Therefore, Option A is incorrect.

Check Option C: voltage between capacitor plates

The capacitor remains connected to the battery throughout, so the voltage across the plates remains constant:

V_f = V_i = V

It does not increase by factor $(K+1)$ .

Therefore, Option C is incorrect.

Check Option D: work done dependence on dielectric

Since battery is connected, the mechanical work involved depends on the change in electrostatic energy and battery work, both of which depend on capacitance.

Initial energy:

U_i = \frac{1}{2}C_iV^2 = \frac{1}{2}\cdot \frac{K\varepsilon_0 A}{d}V^2

Final energy:

U_f = \frac{1}{2}C_fV^2 = \frac{1}{2}\cdot \frac{K\varepsilon_0 A}{d(K+1)}V^2

These clearly depend on $K$ , so the work done in the process also depends on the dielectric constant.

Therefore, Option D is incorrect.

Final answer

Only Option B is correct.

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View written solutionFree

Correct answer: B

Initial configuration (Figure a)

So the initial capacitance is

C_i = \frac{K\varepsilon_0 A}{d}

The initial electric field inside the dielectric is

E_i = \frac{V}{d}

because the entire separation is filled with dielectric and the battery maintains voltage $V$ across distance $d$ .

Final configuration (Figure b)

dielectric region of thickness $d$
air/vacuum region of total thickness $d$

Hence the system behaves like two capacitors in series:

dielectric-filled capacitor of thickness $d$ , capacitance $C_1 = \frac{K\varepsilon_0 A}{d}$
air capacitor of thickness $d$ , capacitance $C_2 = \frac{\varepsilon_0 A}{d}$

Thus,

\frac{1}{C_f} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{d}{K\varepsilon_0 A} + \frac{d}{\varepsilon_0 A} = \frac{d}{\varepsilon_0 A}\left(\frac{1}{K}+1\right)

So,

C_f = \frac{\varepsilon_0 A}{d\left(1+\frac{1}{K}\right)} = \frac{K\varepsilon_0 A}{d(K+1)}

Now compare with initial capacitance:

C_i = \frac{K\varepsilon_0 A}{d}

Therefore,

\frac{C_f}{C_i} = \frac{1}{K+1}

Hence the capacitance becomes

C_f = \frac{C_i}{K+1}

So it is decreased by a factor of $\frac{1}{K+1}$ .

Thus, Option B is correct.

Check Option A: electric field inside dielectric

Since the battery remains connected, total voltage across the combination remains $V$ .

Let electric field in dielectric be $E_d$ and in air be $E_a$ . For series arrangement, electric displacement is same:

\varepsilon_0 E_a = K\varepsilon_0 E_d

E_a = K E_d

Total voltage is

V = E_d(d) + E_a(d) = E_d d + K E_d d = (K+1)E_d d

Thus

E_d = \frac{V}{(K+1)d}

Initially,

E_i = \frac{V}{d}

So the field inside dielectric is reduced by factor

\frac{E_d}{E_i} = \frac{1}{K+1}

not by $\frac{1}{2K}$ .

Therefore, Option A is incorrect.

Check Option C: voltage between capacitor plates

The capacitor remains connected to the battery throughout, so the voltage across the plates remains constant:

V_f = V_i = V

It does not increase by factor $(K+1)$ .

Therefore, Option C is incorrect.

Check Option D: work done dependence on dielectric

Since battery is connected, the mechanical work involved depends on the change in electrostatic energy and battery work, both of which depend on capacitance.

Initial energy:

U_i = \frac{1}{2}C_iV^2 = \frac{1}{2}\cdot \frac{K\varepsilon_0 A}{d}V^2

Final energy:

U_f = \frac{1}{2}C_fV^2 = \frac{1}{2}\cdot \frac{K\varepsilon_0 A}{d(K+1)}V^2

These clearly depend on $K$ , so the work done in the process also depends on the dielectric constant.

Therefore, Option D is incorrect.

Final answer

Only Option B is correct.

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