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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. enter image description here

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 ?

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  • $\begingroup$ 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 $\endgroup$
    – lineage
    Jun 16, 2021 at 0:26

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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 $$ C= \begin{cases} C_0 \frac{k_1k_2}{k_1-k_2}\ln\frac {k_1}{k_2} & k_1\ne k_2\\ C_0 k & k_1=k_2=k \end{cases} $$

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