Two parallel plates of area A are separated by two different dielectric as shown in the figure. The net capacitance is

Question

Two parallel plates of area A are separated by two different dielectric as shown in the figure. The net capacitance is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-s0xpestqynjrks0t.jpg" /> (A) (15 ϵ 0 A/4 d) (B) (10 ϵ 0 A/7 d) (C) (11 ϵ 0 A/5 d) (D) (12 ϵ 0 A/3 d)

Tardigrade Admin · Accepted Answer

Since, we know,
Capacitance of a parallel plate capacitance,

C = \frac{K ϵ _{0} A}{d}

Where, symbols have their usual meaning.
In the given figure,

\frac{1}{C _{n e t}} = \frac{1}{C _{1}} + \frac{1}{C _{2}} = \frac{1}{\frac{K _{1} ϵ _{0} A}{d /2}} + \frac{1}{\frac{K _{2} ϵ _{0} A}{d /2}}

= \frac{1}{\frac{3 ϵ _{0} A}{d /2}} + \frac{1}{\frac{5 ϵ _{0} A}{d /2}}

= \frac{d}{6 ϵ _{0} A} + \frac{d}{10 ϵ _{0} A} = \frac{d}{2 ϵ _{0} A} [\frac{1}{3} + \frac{1}{5}]

= \frac{8 d}{15 \times 2 \times ϵ _{0} A} = \frac{4 d}{15 ϵ _{0} A}

Thus,

C_{n e t} = \frac{15 ϵ _{0} A}{4 d}

Q. Two parallel plates of area $A$ are separated by two different dielectric as shown in the figure. The net capacitance is

$\frac{15 ϵ _{0} A}{4 d}$

$\frac{10 ϵ _{0} A}{7 d}$

$\frac{11 ϵ _{0} A}{5 d}$

$\frac{12 ϵ _{0} A}{3 d}$

Q. Two parallel plates of area A are separated by two different dielectric as shown in the figure. The net capacitance is

4d15ϵ0​A​

7d10ϵ0​A​

5d11ϵ0​A​

3d12ϵ0​A​

Q. Two parallel plates of area $A$ are separated by two different dielectric as shown in the figure. The net capacitance is

$\frac{15 ϵ _{0} A}{4 d}$

$\frac{10 ϵ _{0} A}{7 d}$

$\frac{11 ϵ _{0} A}{5 d}$

$\frac{12 ϵ _{0} A}{3 d}$