Question

An infinite non-conducting sheet has a surface charge density $2\times 10^{-7}C/m^2$ on one side. The distance between two equipotential surfaces whose potential differ by $90V$ is:
(Assume $\frac{1}{4\pi {\epsilon }_0}=9\times 10^9\frac{N{m}^2}{C^2})$

• A. $20\pi mm$
• B. $\frac{25}{\pi }mm$
• C. $\frac{12.5}{\pi }mm$
• D. $\frac{\pi }{20}mm$

Answer 1

Answer: (B)

Using $\left|\frac{dV}{dr}\right|=|B|$,
Distance between two equipotentials of a charged plate is
$r=\frac{\Delta V}{E}=\frac{\Delta V}{\sigma /2{\epsilon }_0}=\frac{\Delta V\cdot 2{\epsilon }_0}{\sigma }=\frac{\Delta V\cdot 4\pi {\epsilon }_0}{2\pi \cdot \sigma }$
Here, $\Delta V=90V,\frac{1}{4\pi {\epsilon }_0}=9\times 10^9\frac{N{m}^2}{C^2}$
$\sigma =2\times 10^{-7}C{m}^{-2}$
So, $r=\left(\frac{90}{2\times \pi \times 2\times 10^{-7}\times 9\times 10^9}\right)$
$=\frac{25}{\pi }\times 10^{-3}m=\frac{25}{\pi }mm$