Question

A parallel plate capacitor has plates of area 'A' separated by distance 'd' between them. It is filled with a dielectric which has a dielectric constant that varies as $$k(x)=K(1+\alpha x)$$ where 'x' is the distance measured from one of the plates. If $$(\alpha d)<<1$$, the total capacitance of the system is best by the expression:

Answer 1

Correct option is C. $$\dfrac{A\epsilon_0 K}{d}(1+\dfrac{\alpha d}{2})$$

Refer image 1.

Let $$k(x)=k(1+\alpha x)$$

be the dielectric constant

The capacitor can be considered to be a series combination of a large number of capacitors of thickness $$\Delta x$$.

Refer image 2.

$$\displaystyle \dfrac{1}{C_{eq}}=\sum\dfrac{\Delta x}{k(x)A\epsilon_0}=\int^d_0\dfrac{dx}{k(x)A\epsilon_0}$$

where $$\Delta x\rightarrow 0$$

$$\displaystyle \dfrac{1}{C_{eq}}=\int^d_0\dfrac{dx}{k(1+\alpha x)A\epsilon_0}=\dfrac{1}{KA\epsilon_0}\int^d_0\dfrac{dx}{1+\alpha x}$$

$$=\dfrac{1}{KA\epsilon_0}\log_e(\dfrac{1+\alpha d}{1})\times \dfrac{1}{\alpha}$$

$$=\dfrac{1}{K\alpha A\epsilon_0}\log_e(1+\alpha d)$$

Now, $$\log (1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-------$$

$$\therefore \log_e(1+\alpha d)\simeq \alpha d-\dfrac{\alpha^2d^2}{2}$$

$$\therefore \dfrac{1}{C_{eq}}=\dfrac{1}{k\alpha A\epsilon_e}\log_e(1+\alpha d)$$

$$=\dfrac{1}{k\alpha A\epsilon_0}[\alpha d-\dfrac{\alpha^2d^2}{2}+-------]$$

$$\simeq \dfrac{\alpha d}{k\alpha A\epsilon_0}[1-\dfrac{\alpha d}{2}]\simeq \dfrac{d}{KA\epsilon_0}[1-\dfrac{\alpha d}{2}]$$

$$C_{eq}=\dfrac{KA\epsilon_0}{d[1-\dfrac{\alpha d}{2}]}=\dfrac{KA\epsilon_0}{d}[1-\dfrac{\alpha d}{2}]^{-1}$$

Now, $$(1-x)^{-1}\simeq 1+x$$ $$[\because x<<1]$$

$$\therefore C_{eq}=\dfrac{KA\epsilon_0}{d}[1+\dfrac{\alpha d}{2}]$$

option (C) is correct.