Question

The capacitance of an air capacitor is $15\mu F$ the separation between the parallel plates is $6mm$. A copper plate of $3mm$ thickness is introduced symmetrically between the plates. The capacitance now becomes

• A. $5\mu F$
• B. $7.5\mu F$
• C. $22.5\mu F$
• D. $30\mu F$

Answer 1

Answer: (D)

By using $C_{\text{air }}=\frac{{\epsilon }_{0}A}{d},C_{\text{medium }}$
$=\frac{{\epsilon }_{0}A}{d-t+\frac{t}{K}}$
For $K=\infty ,C_{\text{medium }}=\frac{{\epsilon }_{0}A}{d-t}$
$\Rightarrow \frac{C_{\text{medium }}}{C_{\text{air }}}=\frac{d}{d-t}$
$\Rightarrow \frac{C_{\text{medium }}}{15}=\frac{6}{6-3}$
$\Rightarrow C_{\text{medium }}=30\mu F$