Question

Two parallel plates of area $2\times {10}^4{\text{cm}}^2$ are each given equal and opposite charges of $500\times {10}^{-7}\text{C}$. The electric field within the dielectric material filling the space between the plates is $2.4\times {10}^6\text{V/m}$. Find the dielectric constant of the dielectric material and charge on the surface of the dielectric.

• A. $1.7,8.3\times {10}^{-6}\text{C}$
• B. $1.2,8.3\times {10}^{-6}\text{C}$
• C. $2.1,8.3\times {10}^{-6}\text{C}$
• D. $2.1,4.3\times {10}^{-6}\text{C}$

Answer 1

The correct option is B $1.2,8.3\times {10}^{-6}\text{C}$

Given:

$A=2\times {10}^4{\text{cm}}^2;Q=500\times {10}^{-7}\text{C}$
$E=2.4\times {10}^6\text{V/m}$

Electric field inside a capacitor filled by a dielectric is given by

$E=\frac{\sigma }{K{ϵ}_0}$

$\Rightarrow K=\frac{\sigma }{E{ϵ}_0}=\frac{Q}{A{ϵ}_{0}E}(\because \sigma =Q/A)$

$\Rightarrow K=\frac{500\times {10}^{-7}}{(2\times {10}^4\times {10}^{-4})\times 8.85\times {10}^{-12}\times 2.4\times {10}^6}$

$\therefore K=1.2$

Induced charge on the surface of the dielectric will be

$Q_i=Q(1-\frac{1}{K})$

$\Rightarrow Q_i=500\times {10}^{-7}(1-\frac{1}{1.2})$

$\Rightarrow Q_i\approx 8.3\times {10}^{-6}\text{C}$

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