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

Answer 1

Answer: (A)

$C=\frac{{ϵ}_{0}A}{d}$
$C_1=\frac{{ϵ}_{0}KA}{d/2}=\frac{2{ϵ}_0\cdot KA}{d}$
Eq. $C_2=\frac{{ϵ}_{0}A}{d/2}=2\left(\frac{{ϵ}_{0}A}{d}\right)$
$C_1\&C_2$ are in series combination
Hence
$\frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{d}{2{ϵ}_{0}A}\left(\frac{1}{K}+\frac{1}{1}\right)$
$\Rightarrow \frac{1}{C_s}=\frac{d}{2{ϵ}_{0}A}\left(\frac{K+1}{K}\right)$
$C_s=\frac{2{ϵ}_{0}A}{d}\left(\frac{K}{K+1}\right)C_s=\frac{2CK}{K+1}$