Animated Solution for Physics - Electrostatics Potential and Capacitance: A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants $K_1,K_2,K_3,K_4$ arranged as shown in the figure. The effective dielectric constant $K$ will be:

We have a square plate capacitor, separated by a distance $d$, and the area of each plate is $A=L^2$. The space between the plates is filled with four different dielectrics, $K_1,K_2,K_3,$ and $K_4$.

Because the plates are on the left and right, the electric field goes horizontally. We can think of this system as two parallel branches. The top branch has dielectrics $K_1$ and $K_2$ in series, and the bottom branch has $K_3$ and $K_4$ in series.

Each section has an area of $A/2$ and a thickness of $d/2$.

For the first section, $C_1$ is given by:

Similarly, for the other sections, we have:

Since $C_1$ and $C_2$ are in series, their equivalent capacitance $C_{12}$ is:

By the exact same logic, the bottom branch with $K_3$ and $K_4$ in series will have an equivalent capacitance $C_{34}$:

Now, these two branches are in parallel. So, the total capacitance $C_{net}$ is simply the sum of $C_{12}$ and $C_{34}$:

Canceling $\frac{{\epsilon }_{0}A}{d}$, we find that $K$ equals:

Notice that none of the given options match this exact expression, which is why this was a bonus question in the exam!

There is a catch here. What if we inserted a thin conducting foil between the left and right halves? That would force the boundary to be an equipotential surface, putting $K_1$ and $K_3$ in parallel, and $K_2$ and $K_4$ in parallel. If you calculate the effective $K$ for that case, you'll find it matches option (c)! But without that foil, our derived answer is the physically correct one.