Question

For changing the capacitance of a given paralle. plate capacitor, a dielectric material of dielectric constant $K$ is used, which has the same area as the plates of the capacitor. The thickness of thev dielectric slab is $\frac{3}{4}d$, where ${}^{\prime }{d}^{\prime }$ is the separation between the plates of parallel plate capacitor The new capacitance $(C^{\prime })$ in terms of original capacitance $\left(C_0\right)$ is given by the following relation:

• A. $C^{\prime }=\frac{3+K}{4K}{C}_0$
• B. $C^{\prime }=\frac{4+K}{3}{C}_0$
• C. $C^{\prime }=\frac{4K}{K+3}{C}_0$
• D. $C^{\prime }=\frac{4}{3+K}{C}_0$

Answer 1

Answer: (C)

$C_0=\frac{{\in }_{0}A}{d}$
$C^{\prime }=C_1$ and $C_2$ in series.
i.e. $\frac{1}{C^{\prime }}=\frac{1}{C_1}+\frac{1}{C_2}$
$\frac{1}{C^{\prime }}=\frac{(3d/4)}{{ϵ}_{0}KA}+\frac{d/4}{{ϵ}_{0}A}$
$\frac{1}{C^{\prime }}=\frac{d}{4{\in }_{0}A}\left(\frac{3+K}{K}\right)$
$C^{\prime }=\frac{4K{C}_0}{(3+K)}$