Question

A parallel plate capacitor has a plate area $100{\text{cm}}^2$ and separation $1\text{mm}$. After filling the region between the plates with a dielectric, the capacitor is able to hold charge of $0.12\mu \text{C}$ when connected to a $120\text{V}$ battery. The dielectric constant of the dielectric is

• A. $12.7$
• B. $11.3$
• C. $10.1$
• D. $6.8$

Answer 1

The correct option is B $11.3$

Given:

$A=100{\text{cm}}^2={10}^{-2}{\text{m}}^2$

$d=1\text{mm}={10}^{-3}\text{m}$

Let the dielectric constant be $K$.

The capacitance of a capacitor increase by a factor of $K$ when a dielectric is fully introduced between its plates.

$C=\frac{KA{ϵ}_0}{d}$

$\Rightarrow C=\frac{K\times {10}^{-2}\times 8.85\times {10}^{-12}}{{10}^{-3}}$

$\Rightarrow C=K(8.85\times {10}^{-11})\text{F}$

Now let's apply,

$Q=CV$

$\Rightarrow 0.12\times {10}^{-6}=K\times 8.85\times {10}^{-11}\times 120$

$\Rightarrow K=11.3$

Hence, option $\left(b\right)$ is correct.