Q146 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Physics
An infinite non-conducting sheet has a surface charge density $2 \times 10^{-7} \mathrm{C} / \mathrm{m}^2$ on one side. The distance between two equipotential surfaces whose potential differ by $90 \mathrm{~V}$ is (assume, $\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \frac{\mathrm{Nm}^2}{\mathrm{C}^2}$ )
- A$20 \pi \mathrm{mm}$
- Correct answerB$\frac{25}{\pi} \mathrm{mm}$
- C$\frac{12.5}{\pi} \mathrm{mm}$
- D$\frac{\pi}{20} \mathrm{~mm}$
Solution
Using $\left|\frac{d \mathbf{V}}{d r}\right|=|\mathbf{E}|$,
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, $\Delta V=90 \mathrm{~V}, \frac{\mathrm{l}}{4 \pi \varepsilon_0}=9 \times 10^9 \frac{\mathrm{Nm}^2}{\mathrm{C}^2}$
$\sigma=2 \times 10^{-7} \mathrm{Cm}^{-2}$
So, $r=\left(\frac{90}{2 \times \pi \times 2 \times 10^{-7} \times 9 \times 10^9}\right)$
$=\frac{25}{\pi} \times 10^{-3} \mathrm{~m}=\frac{25}{\pi} \mathrm{mm}$
Hence, the correct answer is option: 'B'.