Question

A parallel-plate capacitor of a plate area A and plate separation d. The left half

of the gap is filled with a material with dielectric constant κ1; the top of the right half is filled with

material of dielectric constant κ2; the bottom of the right half is filled material of dielectric constant

κ3. What is the capacitance C?

Answer 1

Gives

Dielectric material=$k_1,k_2,k_3$

Let us assume d'=2d

$C_1=\frac{\epsilon_0( \frac{A}{2} )k_1}{2d'}=\frac{A\epsilon_0 k_1}{4d'}$

$C_2=\frac{\epsilon_0( \frac{A}{2} )k_2}{d'}=\frac{A\epsilon_0 k_2}{2d'}$

$C_3=\frac{\epsilon_0( {A}{} )k_3}{2d'}$

$C=C_1+\frac{C_2C_3}{C_1+C_3}$

$C=\frac{A\epsilon_0}{4d'}[k_1+\frac{2k_2k_3}{k_2+k_3}]$

$C=\frac{A\epsilon_0}{2d'}[\frac{k_1}{2}+\frac{k_2k_3}{k_2+k_3}]$

$C=\frac{A\epsilon_0}{d}[\frac{k_1}{2}+\frac{k_2k_3}{k_2+k_3}]$