JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Capacitance

A parallel plate capacitor has a dielectric slab of dielectric constant

K

between its plates that covers 1/3 of the area of its plates, as shown in the figure. The total capacitance of the capacitor is

C

while that of the portion with dielectric in between is

C_{1}

. When the capacitor is charged, the plate area covered by the dielectric gets charge

Q_{1}

and the rest of the area gets charge

Q_{2}

. The electric field in the dielectric is

E_{1}

and that in the other portion is

E_{2}

. Choose the correct option/options, ignoring edge effects.

Answer: (A), (D)

Solution

The capacitor is effectively two capacitors in parallel: one with dielectric (area A/3, capacitance

C_{1} = K \frac{ε _{0} A /3}{d}

) and one without (area 2A/3, capacitance

C_{2} = \frac{ε _{0} ( 2 A /3 )}{d}

).

Let

c = \frac{ε _{0} A}{3 d}

. Then

C_{1} = K c

and

C_{2} = 2 c

. Total:

C = C_{1} + C_{2} = (K + 2) c

, so

c = \frac{C}{K + 2}

.

Thus

C_{1} = \frac{K C}{K + 2}

, giving

\frac{C}{C _{1}} = \frac{K + 2}{K} = \frac{2 + K}{K}

. So (D) is correct.

Both portions are at the same potential difference V (parallel connection). Electric field

E = \frac{V}{d}

is the same in both regions, so

\frac{E _{1}}{E _{2}} = 1

. So (A) is correct.

Charges:

Q_{1} = C_{1} V

and

Q_{2} = C_{2} V

, so

\frac{Q _{1}}{Q _{2}} = \frac{C _{1}}{C _{2}} = \frac{K c}{2 c} = \frac{K}{2}

, not

\frac{3}{K}

. So (C) is wrong.

Option (B) is wrong since

E_{1} = E_{2}

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · One or more correct options