An infinite non-conducting sheet has a surface charge density of $7 \times 10^{-7} \mathrm{C}…

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)
  1. $2.0 \mathrm{~mm}$
  2. $0.25 \mathrm{~mm}$
  3. $1.0 \mathrm{~mm}$
  4. $0.5 \mathrm{~mm}$

Solution

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} $

Asked in: AP EAMCET 2022 (06 Jul Shift 1)

Practice more Electrostatics questions on Aicharya