Question

Surface density of charge on a sphere of radius ‘$R$’ in terms of electric intensity ‘$E$’ at a distance ‘$r$’ in free space is
(${\in }_0=$ permittivity of free space)

• A. ${\epsilon }_{0}E{\left(\frac{R}{r}\right)}^2$
• B. $\frac{{\epsilon }_{0}ER}{r^2}$
• C. ${\epsilon }_{0}E{\left(\frac{r}{R}\right)}^2$
• D. $\frac{{\epsilon }_{0}Er}{R^2}$

Answer 1

Answer: (C)

According to Gauss's law,
Total flux through Gaussian surface
$\varphi =\oint E\cdot ds={\oint }_{s}Eds=E\cdot 4\pi r^2$
If the charge enclosed by Gaussian surface is $q$, according to Gauss's theorem
$E\cdot 4\pi r^2=\frac{q}{{\epsilon }_0}\Rightarrow E=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}...(i)$
If $\sigma$ is uniform surface charge density of spherical shell, then
$q=4\pi R^2\sigma ...(ii)$
Substituting Eq. (ii) in Eq. (i), we get
$E=\frac{\sigma }{{\epsilon }_0}\frac{R^2}{r^2}\therefore \sigma =\frac{{\epsilon }_{0}E{r}^2}{R^2}$