Question

A parallel plate capacitor is made of two circular plates separated by a distance of $5\text{mm}$ and with a dielectric of dielectric constant $2.2$ between them. When the electric field in the dielectric is $3\times {10}^4\text{V/m}$, the charge density of the positive plate will be close to:

• A. $6\times {10}^4{\text{C/m}}^2$
• B. $6\times {10}^{-7}{\text{C/m}}^2$
• C. $3\times {10}^{-7}{\text{C/m}}^2$
• D. $3\times {10}^4{\text{C/m}}^2$

Answer 1

The correct option is B $6\times {10}^{-7}{\text{C/m}}^2$

Given: $d=5\text{mm},K=2.2,E=3\times {10}^4\text{V/m},\sigma =?$

Electric field in presence of dielectric between the two plates of a parallel plate capacitor is:

$\frac{\sigma }{K{\epsilon }_0}=E$

$\Rightarrow \sigma =K{ϵ}_{0}E$

$\Rightarrow \sigma =2.2\times 8.85\times {10}^{-12}\times 3\times {10}^4$

$\therefore \sigma \approx 6\times {10}^{-7}{\text{C/m}}^2$

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