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An infinite non-conducting sheet has a surface charge density of $7 \times 10^{-7} \mathrm{C} \mathrm{m}^{-2}$ on one side. The distance between equipotential surfaces whose potentials differ by $19.8 \mathrm{~V}$, will be
(Take $\frac{1}{4 \pi \varepsilon_{\mathrm{o}}}=9 \times 10^9$ SI units)
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(Take $\frac{1}{4 \pi \varepsilon_{\mathrm{o}}}=9 \times 10^9$ SI units)
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Verified Answer
The correct answer is:
$0.5 \mathrm{~mm}$
We have
$$
\begin{aligned}
& |\mathrm{E}|=\frac{\Delta \mathrm{V}}{\Delta \mathrm{r}}=\frac{19.8}{\Delta \mathrm{r}} \\
\Rightarrow & \frac{\sigma}{2 \varepsilon_0}=\frac{19.8}{\Delta \mathrm{r}} \\
\Rightarrow & \Delta \mathrm{r}=\frac{19.8 \times 2 \varepsilon_0}{\sigma}=\frac{19.8 \times 2 \times 8.85 \times 10^{-12}}{7 \times 10^{-7}} \\
& =5 \times 10^{-4} \mathrm{~m} \\
& =0.5 \mathrm{~mm}
\end{aligned}
$$
$$
\begin{aligned}
& |\mathrm{E}|=\frac{\Delta \mathrm{V}}{\Delta \mathrm{r}}=\frac{19.8}{\Delta \mathrm{r}} \\
\Rightarrow & \frac{\sigma}{2 \varepsilon_0}=\frac{19.8}{\Delta \mathrm{r}} \\
\Rightarrow & \Delta \mathrm{r}=\frac{19.8 \times 2 \varepsilon_0}{\sigma}=\frac{19.8 \times 2 \times 8.85 \times 10^{-12}}{7 \times 10^{-7}} \\
& =5 \times 10^{-4} \mathrm{~m} \\
& =0.5 \mathrm{~mm}
\end{aligned}
$$
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