A parallel plate condenser is filled with two dielectrics as shown. Area of each plate is A metre 2 and the separation is t metre. The dielectric constants are K1 and K2, respectively. Its capacitance in farad will be

Question

A parallel plate condenser is filled with two dielectrics as shown. Area of each plate is A metre 2 and the separation is t metre. The dielectric constants are K1 and K2, respectively. Its capacitance in farad will be (A) (ε0 A/t)(K1+K2) (B) (2 ε0 A/t)(K1+K2) (C) (ε0 A/t) ⋅ (K1+K2/2) (D) (ε0 A/t) ⋅ (K1-K2/2)

Tardigrade Admin · Accepted Answer

The given arrangement is equivalent to two capacitors in parallel. In each capacitor, the area of the plate will be

A /2

.

∴

Equivalent capacitance,

C^{'} = C_{1} + C_{2}

= \frac{K _{1} ε _{0} A /2}{t} + \frac{K _{2} ε _{0} A /2}{t}

= \frac{( K _{1} + K _{2} ) ε _{0} A}{2 t}

Q. A parallel plate condenser is filled with two dielectrics as shown. Area of each plate is $A$ metre $^{2}$ and the separation is $t$ metre. The dielectric constants are $K_{1}$ and $K_{2}$ , respectively. Its capacitance in farad will be

$\frac{ε _{0} A}{t} (K_{1} + K_{2})$

$\frac{2 ε _{0} A}{t} (K_{1} + K_{2})$

$\frac{ε _{0} A}{t} \cdot \frac{K _{1} + K _{2}}{2}$

$\frac{ε _{0} A}{t} \cdot \frac{K _{1} - K _{2}}{2}$

The given arrangement is equivalent to two capacitors in parallel. In each capacitor, the area of the plate will be $A /2$ .
$∴$ Equivalent capacitance, $C^{'} = C_{1} + C_{2}$
$= \frac{K _{1} ε _{0} A /2}{t} + \frac{K _{2} ε _{0} A /2}{t}$
$= \frac{( K _{1} + K _{2} ) ε _{0} A}{2 t}$

Q. A parallel plate condenser is filled with two dielectrics as shown. Area of each plate is A metre 2 and the separation is t metre. The dielectric constants are K1​ and K2​, respectively. Its capacitance in farad will be

tε0​A​(K1​+K2​)

t2ε0​A​(K1​+K2​)

tε0​A​⋅2K1​+K2​​

tε0​A​⋅2K1​−K2​​