Animated Solution for Physics - Electrostatics: The capacitance of a parallel plate capacitor with plate area $A$ and separation $d$, is $C$. The space between the plates is filled with two wedges of dielectric constants $K_1$ and $K_2$ respectively (figure). Find the capacitance of the resulting capacitor.

This is a classic and highly elegant problem in electrostatics that tests your ability to apply calculus to physical systems. When the dielectric medium inside a capacitor is not uniform, we cannot simply plug values into the standard formula $C=\frac{K{\epsilon }_{0}A}{d}$. Instead, we must break the system down into infinitesimally small, manageable pieces.

Imagine you are looking at the capacitor from the side. The space between the plates is filled with two wedge-shaped dielectrics, $K_1$ and $K_2$. The boundary between them is a diagonal line. Because the thickness of each dielectric changes continuously as you move from left to right, the electric field and the local capacitance also change continuously.

To tackle this, we need to slice the capacitor into tiny pieces where the properties are approximately constant.

Let's slice the capacitor vertically into infinitesimally thin strips of width $dx$, located at a distance $x$ from the left edge.

Why vertical strips? Because within a vertical strip, the electric field lines travel straight down from the top plate to the bottom plate, passing sequentially through the $K_2$ dielectric and then the $K_1$ dielectric. This sequential flow of electric flux means that each vertical strip acts as two tiny capacitors connected in series.

Furthermore, since all these vertical strips are connected across the same top and bottom conducting plates, they share the same potential difference. Therefore, all the vertical strips are connected in parallel with each other.

Let the length of the plates be $l$ and the breadth be $b$, so the total area is $A=bl$.

Using the geometry of similar triangles, we can find the thickness of each dielectric within our strip at position $x$: - The thickness of the top dielectric ($K_2$) is $y_2=x\frac{d}{l}$. - The thickness of the bottom dielectric ($K_1$) is $y_1=d-x\frac{d}{l}$.

The capacitance of the tiny series combination in this strip, $dC$, is given by the series formula:

Substituting our expressions for $y_1$ and $y_2$:

Rearranging the terms in the denominator to isolate $x$:

Since all these infinitesimal strips are in parallel, the total equivalent capacitance $C_R$ is simply the sum (integral) of all the individual $dC$ values from $x=0$ to $x=l$:

This is a standard integral of the form $\int \frac{1}{A+Bx}dx=\frac{1}{B}\ln (A+Bx)$. Applying this rule, we get:

Evaluating the limits from $0$ to $l$ and substituting the total area $A=bl$:

Notice that the term inside the first natural log simplifies beautifully:

So the expression becomes:

Using the property of logarithms, $\ln (A)-\ln (B)=\ln (A/B)$:

To match the standard convention (and the given answer), we can multiply the numerator and denominator by $-1$, which flips the sign of the denominator and inverts the fraction inside the logarithm:

where $C=\frac{{\epsilon }_{0}A}{d}$ is the capacitance of the empty capacitor.

Sanity Check: What if the two dielectrics were identical, i.e., $K_1=K_2=K$? If you plug this directly into our final formula, you get a $0/0$ indeterminate form. However, if you take the mathematical limit as $K_1\to K_2$ using L'Hôpital's rule, the expression elegantly reduces to $C_R=\frac{K{\epsilon }_{0}A}{d}$, which is exactly what we expect for a uniform dielectric! This confirms that our derivation is physically and mathematically robust.