A parallel plate capacitor with plate area 'A' and distance of separation 'd' is filled with a dielectric. What is the capacity of the capacitor when permittivity of the dielectric varies as:
$$\varepsilon_x = \varepsilon_0 + kx$$, for $$0 < x \le \frac{d}{2}$$
$$\varepsilon_x = \varepsilon_0 + k(d-x)$$, for $$\frac{d}{2} \le x \le d$$

For a parallel-plate capacitor we assume the plates have free charge $$+Q$$ and $$-Q$$. Throughout the dielectric the displacement field is uniform, so by Gauss’s law we have

$$D=\frac{Q}{A}\;.$$

The local permittivity is a function of the distance $$x$$ measured from one plate. Therefore the electric field at that point is obtained from the relation $$D=\varepsilon_x\,E(x)$$, giving

$$E(x)=\frac{D}{\varepsilon_x}= \frac{Q}{A\,\varepsilon_x}\;.$$

The total potential difference between the plates equals the integral of the electric field across the whole separation,

$$V=\int_0^{d}E(x)\,dx=\frac{Q}{A}\int_0^{d}\frac{dx}{\varepsilon_x}\;.$$

The capacitance is defined by $$C=\dfrac{Q}{V}$$, hence

$$C=\frac{A}{\displaystyle\int_{0}^{d}\frac{dx}{\varepsilon_x}}\;.$$

Now we substitute the given piece-wise form of the permittivity.

For $$0<x\le\dfrac{d}{2}$$ we have $$\varepsilon_x=\varepsilon_0+kx$$, and for $$\dfrac{d}{2}\le x\le d$$ we have $$\varepsilon_x=\varepsilon_0+k(d-x)$$. Thus the integral in the denominator splits into two parts:

$$\int_{0}^{d}\frac{dx}{\varepsilon_x}= \int_{0}^{d/2}\frac{dx}{\varepsilon_0+kx}+ \int_{d/2}^{d}\frac{dx}{\varepsilon_0+k(d-x)}\;.$$

In the second integral we set $$u=d-x\; (du=-dx)$$. When $$x=d/2$$, $$u=d/2$$ and when $$x=d$$, $$u=0$$. Therefore

$$\int_{d/2}^{d}\frac{dx}{\varepsilon_0+k(d-x)}=\int_{d/2}^{0}\frac{-du}{\varepsilon_0+ku}= \int_{0}^{d/2}\frac{du}{\varepsilon_0+ku}\;.$$

Both integrals are now identical, so the total becomes

$$\int_{0}^{d}\frac{dx}{\varepsilon_x}=2\int_{0}^{d/2}\frac{dx}{\varepsilon_0+kx}\;.$$

We evaluate the remaining logarithmic integral. Using the standard result

$$\int\frac{dx}{a+bx}=\frac{1}{b}\ln(a+bx)\;,$$

we get

$$\int_{0}^{d/2}\frac{dx}{\varepsilon_0+kx}= \left[\frac{1}{k}\ln\bigl(\varepsilon_0+kx\bigr)\right]_{0}^{d/2}= \frac{1}{k}\Bigl[\ln\!\left(\varepsilon_0+\frac{kd}{2}\right)-\ln(\varepsilon_0)\Bigr] =\frac{1}{k}\ln\!\left(\frac{\varepsilon_0+\dfrac{kd}{2}}{\varepsilon_0}\right).$$

Multiplying by 2 gives

$$\int_{0}^{d}\frac{dx}{\varepsilon_x}= \frac{2}{k}\ln\!\left(\frac{\varepsilon_0+\dfrac{kd}{2}}{\varepsilon_0}\right) =\frac{2}{k}\ln\!\left(\frac{2\varepsilon_0+kd}{2\varepsilon_0}\right).$$

Substituting this result into the capacitance formula we finally obtain

$$C=\frac{A}{\dfrac{2}{k}\ln\!\left(\dfrac{2\varepsilon_0+kd}{2\varepsilon_0}\right)} =\frac{kA}{2\,\ln\!\left(\dfrac{2\varepsilon_0+kd}{2\varepsilon_0}\right)}\;.$$

Hence, the correct answer is Option B.