Question

The electrostatic pressure experienced by a small section of a charged conductor due to the remaining surface, in an electric field of intensity $6\text{V/m}$ is $x\times {10}^{-12}{\text{Nm}}^{-2}$. Then, the value of $x$ is

[Take ${ϵ}_0=8.85\times {10}^{-12}{\text{C}}^2{\text{/Nm}}^2$]

Answer 1

Given,
Electric field intensity, $E=6\text{V/m}$
Electrostatic pressure, $P=x\times {10}^{-12}{\text{N/m}}^{-2}$.

We know that electric field intensity outside a conductor is given by $E=\frac{\sigma }{{\epsilon }_0}...\left(1\right)$

Electrostatic pressure on any surface element of the conductor due to the remaining surface is
$P_e=\frac{{\sigma }^2}{2{\epsilon }_0}...\left(2\right)$

Using $\left(1\right)$ and $\left(2\right)$,
$P_e=\frac{1}{2}{\epsilon }_0{E}^2$

Substituting the values,
$P=\frac{1}{2}\times 8.85\times {10}^{-12}\times 6^2$
$\Rightarrow P=159.3\times {10}^{-12}{\text{N/m}}^2$
According to the problem,
$P=x\times {10}^{-12}{\text{N/m}}^{-2}$

$\therefore x=159.30$