A metal sphere of radius 'R' cm is charged with $4 \pi \mu \mathrm{C}$ situated in air. If ' $\sigma$ ' is…

A metal sphere of radius 'R' cm is charged with $4 \pi \mu \mathrm{C}$ situated in air. If ' $\sigma$ ' is surface density of charge , 'E' is electric intensity at a distance 'r' from the centre of sphere then 'r' is $\left(\epsilon_{0}=\right.$ permittivity of free space)
  1. $\mathrm{R} \sqrt{\frac{\epsilon_{0} \mathrm{E}}{\sigma}}$
  2. $\mathrm{R} \sqrt{\frac{\sigma}{\epsilon_{0} \mathrm{E}}}$
  3. $\sqrt{\frac{\epsilon_{0} \mathrm{E}}{\mathrm{R} \sigma}}$
  4. $\sqrt{\frac{\mathrm{R} \sigma}{\epsilon_{0} \mathrm{E}}}$

Solution

$\mathrm{E}=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{\mathrm{q}}{\mathrm{r}^{2}} ; \quad \sigma=\frac{\mathrm{q}}{4 \pi \mathrm{R}^{2}}$ $\therefore \mathrm{q}=4 \pi \mathrm{R}^{2} \sigma$ $\therefore \mathrm{E}=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{4 \pi \mathrm{R}^{2} \sigma}{\mathrm{r}^{2}}$ $\therefore r^{2}=\frac{R^{2} \sigma}{\epsilon_{0} E} \quad \therefore r=R \sqrt{\frac{\sigma}{\epsilon_{0} E}}$

Asked in: MHT CET 2020 (12 Oct Shift 2)

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