Question

The surface density of charge on a sphere of a radius $R$ in terms of electric intensity $E$ at a distance $r$ in free space is:
(A). ${\varepsilon _0}E{\left( {\dfrac{R}{r}} \right)^2}$
(B). $\dfrac{{{\varepsilon _0}ER}}{{{r^2}}}$
(C). ${\varepsilon _0}E{\left( {\dfrac{r}{R}} \right)^2}$
(D). $\dfrac{{{\varepsilon _0}Er}}{{{R^2}}}$

Answer 1

In this problem, we have to calculate the surface charge density ( $\sigma$ ) of the sphere by using the equation $\sigma = \dfrac{Q}{A}$ . So, first calculate the value of $Q$ using the equation for the electric field, i.e. $E = \dfrac{Q}{{4\pi {\varepsilon _0}{r^2}}}$ . Then calculate the value of $A$ using the equation for the surface area of a sphere, i.e. $A = 4\pi {R^2}$ . Then put these values in the equation $\sigma = \dfrac{Q}{A}$ to reach the solution.

Complete step-by-step answer :
In this problem, we are given a charged sphere that has a radius $R$ and the electric field intensity at a distance $r$ is $E$ and we have to calculate the surface charge density (the amount of charge distributed per unit surface area).
Let the surface charge density of the sphere be $\sigma$ , the total charge on the sphere be $Q$ and the surface area of the sphere be $A$ .
So, $\sigma = \dfrac{Q}{A}$
After seeing the above equation it becomes clear that to calculate the value of surface charge density ( $\sigma$ ) we have to first calculate the total charge on the sphere ( $Q$ ) and the surface area of the entire sphere ( $A$ ).
We know that the electric field $E$ due to a charger sphere having a total charge $Q$ at any distance $r$ is given by the equation
$E = \dfrac{Q}{{4\pi {\varepsilon _0}{r^2}}}$
This equation is calculated assuming that the total charge in the sphere will behave as a point charge.
$Q = 4\pi {\varepsilon _0}{r^2}E$
We also know that the surface area of the sphere of a radius $R$ is given by the equation
$A = 4\pi {R^2}$
So, the surface charge density $\sigma$ is
$\sigma = \dfrac{Q}{A}$
$\sigma = \dfrac{{4\pi {\varepsilon _0}{r^2}E}}{{4\pi {R^2}}}$
$\sigma = \dfrac{{{\varepsilon _0}E{r^2}}}{{{R^2}}}$
$\sigma = {\varepsilon _0}E{\left( {\dfrac{r}{R}} \right)^2}$
Hence, option C is the correct choice.

Note : In the solution above, during the process of calculating the electric field due to the charged sphere at a distance $r$ , we assumed the sphere to be a point charge. This was done to simplify the solution because, in reality, all the charges spread on the surface of the sphere have different distances from the point for which we have to calculate the electric field.