Question

For changing the capacitance of a given parallel 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 the dielectric slab is $\frac{3}{4}d$, where $\text{'}d\text{'}$ is the separation between the plates of parallel plate capacitors. The new capacitance $(C\text{'})$ in terms of original capacitance $\left(C_0\right)$ is given by the following relation:

• A. $\begin{array}{l}C’=\frac{4K}{K+3}{C}_0\end{array}$
• B. $\begin{array}{l}C’=\frac{4}{3+K}{C}_0\end{array}$
• C. $\begin{array}{l}C’=\frac{3+K}{4K}{C}_0\end{array}$
• D. $\begin{array}{l}C’=\frac{4+K}{3}{C}_0\end{array}$

Answer 1

The correct option is A

$\begin{array}{l}C’=\frac{4K}{K+3}{C}_0\end{array}$

Step 1: Apply the Formula

We know that the capacitance of a parallel plate capacitor is given as

$\begin{array}{l}{C}_0=\frac{{ϵ}_{0}A}{d}\end{array}$

where ${\epsilon }_0$ is absolute permittivity, $A$ is the area of the plate, and $\text{'}d\text{'}$ is the separation between the plates.

Given that the two capacitors,

$C_1=\frac{{\epsilon }_{0}KA}{3d/4}$ and $C_2=\frac{{\epsilon }_{0}A}{d/4}$

$C_1$ and $C_2$ are in series and $C’$ is the new capacitance

Step 2: Calculate the new Capacitance

Therefore,

$\begin{array}{l}\frac{1}{C’}=\frac{1}{C_1}+\frac{1}{C_2}\end{array}$

$\begin{array}{l}\frac{1}{C’}=\frac{(3d/4)}{{ϵ}_{0}KA}+\frac{d/4}{{ϵ}_{0}A}\end{array}$

$\begin{array}{l}\frac{1}{C’}=\frac{d}{4{ϵ}_{0}A}\left(\frac{3+K}{K}\right)\end{array}$

$\begin{array}{l}C’=\frac{4K}{(K+3)}{C}_0\end{array}$

Hence option $\left(A\right)$ is the correct answer.