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MHT CET · Physics · Electrostatics

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. A \(\mathrm{R} \sqrt{\frac{\epsilon_{0} \mathrm{E}}{\sigma}}\)
  2. B \(\mathrm{R} \sqrt{\frac{\sigma}{\epsilon_{0} \mathrm{E}}}\)
  3. C \(\sqrt{\frac{\epsilon_{0} \mathrm{E}}{\mathrm{R} \sigma}}\)
  4. D \(\sqrt{\frac{\mathrm{R} \sigma}{\epsilon_{0} \mathrm{E}}}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\mathrm{R} \sqrt{\frac{\sigma}{\epsilon_{0} \mathrm{E}}}\)

Step-by-step Solution

Detailed explanation

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