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 :
(${\epsilon }_0=$ 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

The correct option is C ${\epsilon }_{0}E{\left(\frac{r}{R}\right)}^2$

According to Gauss's law,

Total flux through Gaussian surface

$\varphi =\oint E\cdot ds={\oint }_{s}E\cdot ds=E\cdot 4\pi r^2$

If the charge enclosed by Gaussian surface is $q$, according to Gauss's law

$E\cdot 4\pi r^2=\frac{q}{{\epsilon }_0}\Rightarrow E=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{q}{r^2}$ ......(i)

If $\sigma$ is uniform surface charge density of spherical shell, then

$q=4\pi R^2\sigma$ .......(ii)

Substituting equation (ii) in equation (i), we get

$E=\frac{\sigma }{{\epsilon }_0}\cdot \frac{R^2}{r^2}\therefore \sigma =\frac{{\epsilon }_{0}E{r}^2}{R^2}$