Question

An infinite charged sheet has a surface charge density of ${10}^{-8}C/m^2$. In this situation the separation between two equipotential surfaces which are having a potential difference of $5volt$ is :

• A. $8.85nm$
• B. $8.85mm$
• C. $5mm$
• D. ${10}^{-8}$mm

Answer 1

Answer: (B)

$8.85mm$

given $\sigma ={10}^{-8}\frac{C}{m^2}$

we know $E=\frac{\sigma }{2{ϵ}_0}$
$E=\frac{{10}^{-8}}{2{ϵ}_0}$
we know that $E=\frac{V}{d}$

given the potential difference between two equipotential surfaces $=5V$
$\Rightarrow E\Delta d=\Delta V$
$\Rightarrow \Delta d=\frac{\Delta V}{E}=\frac{5\times 8.85\times {10}^{-12}\times 2}{{10}^{-8}}$

$\Rightarrow \Delta d=8.85\times {10}^{-3}m$