Question

The energy density in a parallel plate capacitor is given by $3.8\times {10}^{-9}\frac{J}{m^3}$. Find the magnitude of electric field in the region between the plate.

• A. $21.6\frac{\text{N}}{\text{C}}$
• B. $2.1\frac{\text{N}}{\text{C}}$
• C. $2.9\frac{\text{N}}{\text{C}}$
• D. $29.29\frac{\text{N}}{\text{C}}$

Answer 1

The correct option is D $29.29\frac{\text{N}}{\text{C}}$

Given, $u=3.8\times {10}^{-9}\frac{J}{m^3}$
As we know, energy density is given by,

$u=\frac{\text{U}}{\text{V}}=\frac{1}{2}{\epsilon }_0{E}^2$

$\Rightarrow 3.8\times {10}^{-9}=\frac{1}{2}\times 8.85\times {10}^{-12}\times E^2$

$\frac{7.6}{8.85}\times {10}^3=E^2$

$E=29.29\frac{\text{N}}{\text{C}}$

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