Animated Solution for Physics - Electrostatics: A parallel plate capacitor is made of two square plates of side '$a$' separated by a distance $d$ ($d\ll a$). The lower triangular portions is filled with a dielectric of dielectric constant $K$, as shown in the figure. Capacitance of this capacitor is

Imagine looking at the capacitor from the side. We have two square plates of side $a$, separated by a distance $d$. The lower triangular region is filled with a dielectric of constant $K$. Because the thickness of the dielectric varies continuously as we move along the plate, we cannot use the standard formula $C=\frac{{\epsilon }_{0}A}{d}$ directly. The electric field lines experience different amounts of dielectric material depending on their horizontal position.

The Differential Element

To find the total capacitance, we must slice the capacitor into infinitesimally thin vertical strips. Each strip acts as a tiny, independent capacitor. Since these strips are placed side-by-side between the same two equipotential plates, they are connected in parallel. The total capacitance will simply be the integral (sum) of the capacitance of these tiny strips.

Let's look at one such strip of width $dx$ at a distance $x$ from the left edge. Inside this strip, the electric field passes through the dielectric and then through the air. This means we have two capacitors in series: the lower part filled with dielectric of height $y$, and the upper part filled with air of height $d-y$.

First, we need to express the height $y$ in terms of $x$. By observing the similar triangles formed by the dielectric boundary, the ratio of $y$ to $x$ is equal to the ratio of the total height $d$ to the total length $a$.

Now, let's write the capacitance for the two parts of our strip. The area of the strip is $a\cdot dx$ (since the plates are square, the strip extends a distance $a$ into the page).

The air part has capacitance $C_1=\frac{{\epsilon }_0(a\cdot dx)}{d-y}$.

The dielectric part has capacitance $C_2=\frac{K{\epsilon }_0(a\cdot dx)}{y}$.

Since they are in series, their equivalent capacitance $dC$ is given by $\frac{C_1{C}_2}{C_1+C_2}$. Substituting $C_1$ and $C_2$ into the series formula and simplifying the algebra, we get:

To find the total capacitance $C$, we integrate $dC$ from $x=0$ to $x=a$. We substitute $y$ with $\frac{d}{a}x$ in the denominator to express everything in terms of $x$.

This is an integral of the form $\int \frac{1}{A+Bx}dx$. Integrating it gives a natural logarithm. Let's use substitution. Let $u=Kd+\frac{(1-K)d}{a}x$. This implies $dx=\frac{a}{(1-K)d}du$.

When $x=0$, $u=Kd$. When $x=a$, $u=Kd+(1-K)d=d$.

Substituting these into our integral, we get:

Evaluating the integral yields $\ln u$ from $Kd$ to $d$:

Using the property of logarithms, $\ln d-\ln Kd=\ln \left(\frac{d}{Kd}\right)=\ln \left(\frac{1}{K}\right)=-\ln K$.

Finally, absorbing the negative sign into the denominator gives us our elegant final expression: