Physics - Capacitor Question with Solution | TestHub

The given arrangement can be considered as two capacitors connected in series:

1. A capacitor with dielectric constant $K$, plate area $A$, and thickness $d_1=\frac{d}{2}$. Its capacitance is:
$C_1=\frac{K{\epsilon }_{0}A}{d/2}=\frac{2K{\epsilon }_{0}A}{d}$

2. A capacitor with air (dielectric constant $1$), plate area $A$, and thickness $d_2=\frac{d}{2}$. Its capacitance is:
$C_2=\frac{{\epsilon }_{0}A}{d/2}=\frac{2{\epsilon }_{0}A}{d}$

The equivalent capacitance $C_{eq}$ for the series combination is given by:
$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}$

Substituting the values of $C_1$ and $C_2$:
$\frac{1}{C_{eq}}=\frac{d}{2K{\epsilon }_{0}A}+\frac{d}{2{\epsilon }_{0}A}$

$\frac{1}{C_{eq}}=\frac{d}{2{\epsilon }_{0}A}\left(\frac{1}{K}+1\right)$

$\frac{1}{C_{eq}}=\frac{d(K+1)}{2K{\epsilon }_{0}A}$

Therefore, the equivalent capacitance is:
$C_{eq}=\frac{2K{\epsilon }_{0}A}{(K+1)d}$

Answer: $\frac{2KA{\epsilon }_0}{(K+1)d}$