# Capacitance with two different dielectrics [closed]

A parallel-plate capacitor with plates of area $$LW$$ and plate separation $$t$$ has the region between its plates filled with wedges of two dielectric materials as shown in figure down below. Assume $$t$$ is much less than both $$L$$ and $$W$$. Determine its capacitance.

I tried to study the capacitance on a small portion $$dx$$ on $$L$$ but I am not able to use the formula of the capacitance for a parallel-plate $$C=\epsilon_0 A/d$$. Can someone help me ?

• u are on the right track. The formula for cap. with different dieelctrics inside is different....btw does the wedge extend in $\hat{W}$ direction Jun 16, 2021 at 0:26

First of all, the capacitance of a cap. with stacked dielectrics of thickness $$d_i$$ and dielectric constant $$k_i$$ is given by

$$\frac 1 {C_{stack}}=\sum_i\frac 1 {C_i}\tag{1}$$

where $$C_i=k_i\epsilon_0 A/d_i$$. This is obtained by treating the inner space of the multi-dielectric cap. as a series connection of multiple single dielectric caps. filled with just those dielectrics.

For your question imagine a $$dC_{stack,x}$$ in every $$(x,x+dx)$$. The full cap. $$C$$, is then just the parallel connection of each of these caps. $$C=\int_0^L dC_{stack,x}\tag{2}$$

Sol.

The dividing line starts on the bottom left corner at $$(0,0)$$ and stops at the top right corner $$(L,H)$$, where I have taken $$H=t$$. The line's eqn. is $$y=H x/L\tag{3}$$ The dielectric thicknesses are $$d_1=H-y\hspace{1cm} d_2=y \tag{4}$$ The caps are

$$C_1=\frac{k_1\epsilon_0 dx\,W}{H(1-x/L)} \hspace{1cm} C_2=\frac{k_2\epsilon_0 dx\,W}{Hx/L} \tag{5}$$ The cap at $$x$$ using eqns. $$1,5$$ is

\begin{align} dC_{stack,x}&=\frac{C_1 C_2}{C_1+C_2}\tag{6}\\ &=C_0\frac{k_1 k_2 dx}{(k_1-k_2)x+k_2L}\tag{7} \end{align} where $$C_0=\frac{\epsilon_0 LW}{H}$$ is the capacitance of the whole system without any dielectric.

The total capacitance using eqns. $$2,7$$ is

$$C=C_0\int_0^L\frac{k_1 k_2 dx}{(k_1-k_2)x+k_2L}\tag{7}$$ giving $$C= \begin{cases} C_0 \frac{k_1k_2}{k_1-k_2}\ln\frac {k_1}{k_2} & k_1\ne k_2\\ C_0 k & k_1=k_2=k \end{cases}$$