Animated Solution for Physics - Electrostatics: 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 $\epsilon (x)={\epsilon }_0+kx$, for $\left(0<x\le \frac{d}{2}\right)$ $\epsilon (x)={\epsilon }_0+k(d-x)$, for $\left(\frac{d}{2}\le x\le d\right)$

Imagine a standard parallel plate capacitor, but instead of a uniform vacuum or a single block of material, the space between the plates is filled with a highly specialized dielectric. The permittivity of this dielectric, $\epsilon (x)$, is not constant. It changes as you move from the left plate ($x=0$) to the right plate ($x=d$).

Specifically, the permittivity increases linearly from the left plate up to the exact midpoint ($x=d/2$), and then it decreases symmetrically until it reaches the right plate. This creates a beautiful, symmetric "tent-like" profile for the permittivity function. Our goal is to find the total equivalent capacitance of this entire system.

When dealing with a continuously varying medium, the standard formula $C=\frac{{\epsilon }_{0}A}{d}$ breaks down. We must resort to calculus. We can visualize the entire capacitor as being constructed from an infinite number of infinitesimally thin, parallel slices.

Each slice has a thickness $dx$ and the same cross-sectional area $A$. Because the electric field lines must pass through each of these slices sequentially from one plate to the other, these elemental capacitors are effectively connected in series. The capacitance of one such infinitesimally thin slice is given by:

For capacitors connected in series, we know that the reciprocal of the equivalent capacitance is the sum of the reciprocals of the individual capacitances. Extending this to an infinite sum (an integral), we get our master equation:

Here is where we must be careful. The function $\epsilon (x)$ does not have a single mathematical definition across the entire distance $d$. It changes its behavior exactly at the midpoint. Therefore, we cannot evaluate this as a single integral. We must split it into two distinct parts:

Let's tackle the first integral. It is a standard logarithmic form. Remember that $\int \frac{1}{a+bx}dx=\frac{1}{b}\ln (a+bx)$. Applying this to our first half:

Now for the second integral. We must watch out for the minus sign! The term inside the denominator is ${\epsilon }_0+kd-kx$. The coefficient of $x$ is $-k$, which means a negative sign will emerge during integration:

Evaluating this at the upper limit $d$ gives $\ln ({\epsilon }_0)$, and at the lower limit $d/2$ gives $\ln ({\epsilon }_0+kd/2)$. Distributing the negative sign, we get:

Notice how the second integral evaluates to the exact same expression as the first! This is a direct mathematical consequence of the physical symmetry of the dielectric.

Now, we simply add the two identical parts together:

Using the fundamental property of logarithms, $\ln (a)-\ln (b)=\ln (a/b)$, we can condense the expression:

Finally, to find the equivalent capacitance $C_{eq}$, we take the reciprocal of both sides:

This perfectly matches option (b). The problem beautifully demonstrates how calculus and physical symmetry work hand-in-hand to solve complex electrostatic setups.