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
($∈_0 =$ permittivity of free space)

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

Answer 1

Answer: (C)

According to Gauss's law,
Total flux through Gaussian surface
$\phi=\oint E \cdot d s=\oint_{s} E d s=E \cdot 4 \pi r^{2}$
If the charge enclosed by Gaussian surface is $q$, according to Gauss's theorem
$E \cdot 4 \pi r^{2}=\frac{q}{\varepsilon_{0}} \Rightarrow E=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}} \,\,\,\,\,\,\,...(i)$
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
$E=\frac{\sigma}{\varepsilon_{0}} \frac{R^{2}}{r^{2}} \,\,\,\, \therefore \sigma=\frac{\varepsilon_{0} E r^{2}}{R^{2}}$