Question

Half of the space between parallel plate capacitor is filled with a medium of dielectric constant $K$ parallel to the plates. If initially the capacity is $C ,$ then the new capacity will be :

• A. $2 KC /(1+ K )$
• B. $C ( K +1) / 2$
• C. $CK /(1+ K )$
• D. $KC$

Answer 1

Answer: (A)

$C =\frac{\epsilon_{0} A}{d}$
$C _{1}=\frac{\epsilon_{0} K A}{d / 2}=\frac{2 \epsilon_{0} \cdot K A}{d}$
Eq. $C _{2}=\frac{\epsilon_{0} A}{d / 2}=2\left(\frac{\epsilon_{0} A}{d}\right)$
$C_{1} \& C_{2}$ are in series combination
Hence
$\frac{1}{C_{s}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{d}{2 \epsilon_{0} A}\left(\frac{1}{K}+\frac{1}{1}\right)$
$\Rightarrow \frac{1}{C_{s}}=\frac{d}{2 \epsilon_{0} A}\left(\frac{K+1}{K}\right)$
$C _{ s }=\frac{2 \epsilon_{0} A}{d}\left(\frac{K}{K+1}\right) C _{ s }=\frac{2 C K}{K+1}$