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 $90\, V$ is:
(Assume $\left.\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \frac{ Nm ^{2}}{C^{2}}\right)$

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

Answer 1

Answer: (B)

Using $\left|\frac{d V}{d r}\right|=|B|$,
Distance between two equipotentials of a charged plate is
$r=\frac{\Delta V}{E}=\frac{\Delta V}{\sigma / 2 \varepsilon_{0}}=\frac{\Delta V \cdot 2 \varepsilon_{0}}{\sigma}=\frac{\Delta V \cdot 4 \pi \varepsilon_{0}}{2 \pi \cdot \sigma}$
Here, $\Delta V=90 V , \frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \frac{ Nm ^{2}}{ C ^{2}} $
$\sigma=2 \times 10^{-7} Cm ^{-2}$
So, $r=\left(\frac{90}{2 \times \pi \times 2 \times 10^{-7} \times 9 \times 10^{9}}\right) $
$=\frac{25}{\pi} \times 10^{-3} m =\frac{25}{\pi} mm$