Question

A parallel plate capacitor of capacitance C is equally filled with parallel layers of materials of dielectric constants $K_1$ and $K_2$. Then the ratio of new capacitance to the previous capacitance is

• A. $\frac{2K_1{K}_2}{K_1+K_2}$
• B. $K_1+K_2$
• C. $\frac{K_1{K}_2}{K_1+K_2}$
• D. None of the above

Answer 1

The correct option is A $\frac{2K_1{K}_2}{K_1+K_2}$

We can redraw circuit as

$C_1=K_1\frac{{\epsilon }_{0}A}{d/2}=\frac{2K_1{\epsilon }_{0}A}{d}$
$C_2=K_2\frac{{\epsilon }_{0}A}{d/2}=\frac{2K_2{\epsilon }_{0}A}{d}$
Now,
$C_{eq}=\frac{C_1\times C_2}{C_1+C_2}$
$C_{eq}=\frac{\frac{2K_1{\epsilon }_{0}A}{d}\times \frac{2K_2{\epsilon }_{0}A}{d}}{\frac{2K_1{\epsilon }_{0}A}{d}+\frac{2K_2{\epsilon }_{0}A}{d}}$

On Solving we get
$C_{eq}=\frac{2K_1{K}_2\left[\frac{{\epsilon }_{0}A}{d}\right]}{[K_1+K_2]}$
Now,
$\frac{C_{final}}{C_{initial}}=\frac{\frac{2K_1{K}_2}{(K_1+K_2)}.\left[\frac{{\epsilon }_{0}A}{d}\right]}{\frac{{\epsilon }_{0}A}{d}}=\frac{2K_1{K}_2}{(K_1+K_2)}$