Question

What is the electrostatic pressure on the plate of a PPC of capacitance $2\mu \text{F}$ separated by $2\text{mm}$ and $20\text{V}$ potential difference is applied across the capacitor?

• A. $2.425\times {10}^{-5}\text{Pa}$
• B. $1.425\times {10}^{-5}\text{Pa}$
• C. $4.425\times {10}^{-4}\text{Pa}$
• D. $7.425\times {10}^{-4}\text{Pa}$

Answer 1

The correct option is C $4.425\times {10}^{-4}\text{Pa}$

Given:
$C=2\mu \text{F}$
$d=2\text{mm}$
$V=20\text{volt}$

$P=\frac{F}{A}=\frac{{\sigma }^2}{2{ϵ}_0}=\frac{1}{2}{ϵ}_0{E}^2$

[where $E=\frac{V}{d}$]
$P=\frac{1}{2}{ϵ}_0\frac{V^2}{d^2}$
$P=\frac{1}{2}\times 8.85\times {10}^{-12}\times \frac{400}{4\times {10}^{-6}}$
$P=4.425\times {10}^{-4}\text{Pa}$
Hence, option $\left(B\right)$ is correct.