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 Q1 and the rest of the area gets charge Q2. The electric field in the dielectric is E1 and that in the other portion is E2. Choose the correct option/options, ignoring edge effects.

Question

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 Q1 and the rest of the area gets charge Q2. The electric field in the dielectric is E1 and that in the other portion is E2. Choose the correct option/options, ignoring edge effects. (A)  (E1/E2)=1  (B)  (E1/E2)= (1/K)  (C)  (Q1/Q2)= (3/K)  (D)  (C/C1)= (2+K/K)

Tardigrade Admin · Accepted Answer

As E=V / d E 1 / E 2=1 (both parts have common potential difference) Assume C 0 be the capacitance without dielectric for whole capacitor. k ( C 0/3)+(2 C 0/3)= C ( C / C 1)=(2+ k / k ) ( Q 1/ Q 2)=( k /2)

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

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

$\frac{Q _{1}}{Q _{2}} = \frac{3}{K}$

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

As $E = V / d$
$E_{1} / E_{2} = 1$ (both parts have common potential difference)
Assume $C_{0}$ be the capacitance without dielectric for whole capacitor.
$k \frac{C _{0}}{3} + \frac{2 C _{0}}{3} = C$
$\frac{C}{C _{1}} = \frac{2 + k}{k}$
$\frac{Q _{1}}{Q _{2}} = \frac{k}{2}$

E2​E1​​=1

E2​E1​​=K1​

Q2​Q1​​=K3​

C1​C​=K2+K​

As E=V/d E1​/E2​=1 (both parts have common potential difference) Assume C0​ be the capacitance without dielectric for whole capacitor. k3C0​​+32C0​​=C C1​C​=k2+k​ Q2​Q1​​=2k​

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

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

$\frac{Q _{1}}{Q _{2}} = \frac{3}{K}$

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

As $E = V / d$
$E_{1} / E_{2} = 1$ (both parts have common potential difference)
Assume $C_{0}$ be the capacitance without dielectric for whole capacitor.
$k \frac{C _{0}}{3} + \frac{2 C _{0}}{3} = C$
$\frac{C}{C _{1}} = \frac{2 + k}{k}$
$\frac{Q _{1}}{Q _{2}} = \frac{k}{2}$