Question

A parallel plate capacitor with plate area A and plate separation d, is filled with a dielectric slab as shown. What is the capacitance of the arrangement?​

Answer 1

Answer:

The capacitance of the arrangement shown in the figure is given as

$C = \frac{k(k + 3)\epsilon_0A}{2(k + 1) d}$

Explanation:

For capacitance of left half of the capacitor we will have

$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}$

$\frac{1}{C} = \frac{d/2}{\epsilon_0 A/2} + \frac{d/2}{k\epsilon_0 A/2}$

$\frac{1}{C} = \frac{d}{\epsilon_0A}(1 + \frac{1}{k})$

$C_1 = \frac{k\epsilon_0A}{(k + 1)d}$

now for right half we have

$\frac{1}{C} = \frac{d/2}{k\epsilon_0 A/2} + \frac{d/2}{k\epsilon_0 A/2}$

$C_2 = \frac{k\epsilon_0A}{2d}$

now we have

$C = C_1 + C_2$

$C = \frac{k\epsilon_0A}{(k + 1)d} + \frac{k\epsilon_0A}{2d}$

$C = \frac{k\epsilon_0A}{d}(\frac{1}{k + 1} + \frac{1}{2})$

$C = \frac{k(k + 3)\epsilon_0A}{2(k + 1) d}$

#Learn

Topic : Capacitor

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