Question

A parallel plate capacitor having a separation between the plates $d$ , plate area $A$ and material with dielectric constant $k$ has capacitance $C_o$ . Now one third of the material is replaced by another material with dielectric constant $2k$, so that effectively there are two capacitors one with area $\frac{A}{3}$, dielectric constant $2k$ and another with area $\frac{2A}{3}$ and dielectric constant $k$. If the capacitance of this new capacitor is $C$ then the value of $\frac{C}{C_0}$ is:

• A. $1$
• B. $\frac{4}{3}$
• C. $\frac{1}{3}$
• D. $\frac{2}{3}$

Answer 1

The correct option is B $\frac{4}{3}$

Initial arrangement of the capacitor with capacitance,
$$C_0=\frac{kA{\epsilon }_0}{d}....\left(1\right)$$


After the one third material is replaced by another material of dielectric constant $2k$:


Thus equivalent capacitance for the parallel combination of two capacitors $C_1$ and $C_2$ will be,$$C_{eq}=C_1+C_2$$ On substituting the values,

$\Rightarrow C_{eq}=\left(2k\right)\left[\frac{\left(\frac{A}{3}\right){\epsilon }_0}{d}\right]+\left(k\right)\left[\frac{\left(\frac{2A}{3}\right){\epsilon }_0}{d}\right]$

$\Rightarrow C_{eq}=\frac{2kA{\epsilon }_0}{3d}+\frac{2kA{\epsilon }_0}{3d}$

$\therefore C_{eq}=\frac{4kA{\epsilon }_0}{3d}$

As, $C_{eq}=C=\frac{4kA{\epsilon }_0}{3d}....\left(2\right)$

From equations (1) and (2)

$\frac{C}{C_0}=\frac{\left(\frac{4kA{\epsilon }_0}{3d}\right)}{\left(\frac{kA{\epsilon }_0}{d}\right)}$

$\therefore \frac{C}{C_0}=\frac{4}{3}$

Hence, option (b) is the correct answer.