Question

A parallel plate capacitor having a separation between the plates $d$, plate area $A$ and material with dielectric constant $K$ has capacitance $C_{0}$. 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 $\frac{1}{3} A,$ dielectric constant 2K and another with area $\frac{2}{3} A$ and dielectric constant $K$. If the capacitance of this new capacitor is $C$ then $\frac{C}{C_{0}}$ is

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

Answer 1

Answer: (B)

$C_{0} =\frac{k \in_{0} A}{d}$
$C=\frac{k \in_{0} 2 }{3 d}+\frac{2k \in_{0} A}{3d}=\frac{4k\in_{0} A}{3 d}$
$\therefore \frac{C}{C_{0}} = \frac{\frac{4}{3}\frac{k \in_{0} A}{d}}{\frac{k \in_{0} A}{d}} =\frac{4}{3}$