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Question

A parallel plate capacitor is made of two square plates of side '$$a$$', separated by a distance $$d(d<<a)$$. The lower triangular portion is filled with a dielectric of dielectric constant $$K$$, as shown in the figure. 
The capacitance of this capacitor is :
1330036_3d90ffd7d8e24202873a4d397df3e4d1.png


A
12k0a2d
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B
k0a2dlnK
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C
k0a2d(K1)lnK
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D
k0a22d(K+1)
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Solution

The correct option is C $$\dfrac{k\in_0a^2}{d(K-1)}ln\, K$$
Lets consider a strip of thickness $$dx$$ at a distance of x from the left end a shown in the figure. 
$$\dfrac{y}{x} = \dfrac{d}{a} \implies (\dfrac{d}{a})x$$

$$C_1 = \dfrac{\epsilon_o a \ dx }{d-y}$$    and  $$C_2 = \dfrac{k \epsilon_o \ a \ dx}{y}  $$

$$C_{eq} = \dfrac{C_1 . C_2}{C_1 + C_2} = \dfrac{k \epsilon_o a dx}{kd + (1-k)y}$$

On integrating it from 0 to a, we will get $$\dfrac{k\in_0a^2}{d(K-1)}ln\, K$$

1142154_1330036_ans_725ca63a39e04a07b91bac66a983f791.PNG

Physics

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