Question

In the figure, a capacitor is filled with dielectrics. The resultant capacitance is

• A. $\frac{2{\epsilon }_{0}A}{d}\left[\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}\right]$
• B. $\frac{{\epsilon }_{0}A}{d}\left[\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}\right]$
• C. $\frac{2{\epsilon }_{0}A}{d}\left[k_1+k_2+k_3\right]$
• D. None of these

Answer 1

Answer: (D)

$C_1=\frac{k_1{\epsilon }_0\frac{A}{2}}{\left(\frac{d}{2}\right)}=\frac{k_1{\epsilon }_{0}A}{d}$
$C_2=\frac{k_2{\epsilon }_0\left(\frac{A}{2}\right)}{\left(\frac{d}{2}\right)}=\frac{k_2{\epsilon }_{0}A}{d}$
and $C_3=\frac{k_3{\epsilon }_{0}A}{2d}=\frac{k_3{\epsilon }_{0}A}{2d}$
Now, $C_{eq}=C_3+\frac{C_1{C}_2}{C_1+C_2}$
$=\left(\frac{k_3}{2}+\frac{k_1{k}_2}{k_1+k_2}\right)\cdot \frac{{\epsilon }_{0}A}{d}$