Question

A uniformly charged conducting sphere of $2.4m$ diameter has a surface charge density of $80\mu C/{{m}^{2}}$.
(a) Find the charge on the sphere.
(b) What is the total electric flux leaving the surface of the sphere?

Answer 1

Hint: For a conductor, all the charge of the body resides on the surface. We can get the charge on the surface by multiplying the surface charge density with the surface area of the body. We can find the electric flux by using Gauss’ law for the outer surface of the sphere.

Formula used:
$\sigma =\dfrac{Q}{A}$
$A=4\pi {{R}^{2}}$
$\phi =\dfrac{Q}{{{\varepsilon }_{0}}}$ (Gauss’ Law)

Complete step-by-step answer:
We will find the charge of the conducting sphere by multiplying the surface charge density with the surface area of the sphere.
The surface charge density $\sigma $ of a surface is given by
$\sigma =\dfrac{Q}{A}$ --(1)
where $Q$ is the total charge on the surface and $A$ is the surface area.
The surface area $A$ of a sphere of radius $R$ is given by
$A=4\pi {{R}^{2}}$ --(2)
Now, let us analyze the question.
The surface charge density of the uniformly conducting sphere is given to be
$\sigma =80\mu C/{{m}^{2}}=80\times {{10}^{-6}}C/{{m}^{2}}$ $\left( \because 1\mu C={{10}^{-6}}C \right)$
Let the charge on the body be $Q$.
The diameter of the sphere is given to be $d=2.4m$.
Hence, the radius $R$ of the sphere will be $R=\dfrac{d}{2}=\dfrac{2.4}{2}=1.2m$ $\left( \because \text{Radius = }\dfrac{\text{Diameter}}{2} \right)$

Therefore, using (2), we get the surface area $A$ of the sphere as
$A=4\pi {{R}^{2}}$ --(3)
Using (1), we get,
$\sigma =\dfrac{Q}{A}$
$\therefore Q=\sigma \times A$
Using (3), we get,
$Q=\sigma \times 4\pi {{R}^{2}}=80\times {{10}^{-6}}\times 4\pi \times {{\left( 1.2 \right)}^{2}}=80\times {{10}^{-6}}\times 4\pi \times 1.44\approx 1.45\times {{10}^{-3}}C$ --(4)
Therefore, the charge on the sphere is $1.45\times {{10}^{-3}}C$
We can find the total electric flux leaving the sphere by the application of Gauss' Law.
According to Gauss’ Law, the total electric flux $\phi $ through a closed surface is equal to the ratio of the charge enclosed ${{Q}_{enclosed}}$ within the closed surface to the permittivity $\varepsilon $ of the medium.
$\therefore \phi =\dfrac{{{Q}_{enclosed}}}{\varepsilon }$ --(5)
Let us analyze the question.
Let the electric flux leaving the closed outer surface of the sphere be $\phi $.
Since, the medium is vacuum, the permittivity is $\varepsilon ={{\varepsilon }_{0}}=8.85\times {{10}^{-12}}{{m}^{-3}}k{{g}^{-1}}{{s}^{4}}{{A}^{2}}$.
The charge enclosed in the surface is nothing but the total charge on the spherical surface.
$\therefore {{Q}_{enclosed}}=Q$
Using (5), we get,
$\phi =\dfrac{Q}{{{\varepsilon }_{0}}}$
$\therefore \phi =\dfrac{1.45\times {{10}^{-3}}}{8.85\times {{10}^{-12}}}=1.63\times {{10}^{8}}N.{{C}^{-1}}{{m}^{2}}$
Hence, the flux leaving the surface of the sphere is $1.63\times {{10}^{8}}N.{{C}^{-1}}{{m}^{2}}$.

Note: Students must remember that in Gauss’ law, the permittivity depends upon the medium under which the question is considered. Students must properly analyze a given question to determine whether the medium in which Gauss’ Law is being applied is free space (vacuum) or any other material medium. If some other medium is mentioned in the question and the student considers it to be free space, then the student will arrive at an erroneous result.