Question

A uniformly charged conducting sphere is having radius $$1\,m$$ and surface charge density $20\,C{m^{ - 2}}$ . The total flux leaving the Gaussian surface enclosing the sphere is
A. $40\pi \varepsilon _0^{ - 1}$
B. $80\pi \varepsilon _0^{ - 1}$
C. $20\pi \varepsilon _0^{ - 1}$
D. $60\pi \varepsilon _0^{ - 1}$

Answer 1

Hint: Here, we will use the concept of Gauss law in the case of the conducting sphere. Gauss law states that the electric flux linked with the surface will be equal to the $\dfrac{1}{{{\varepsilon _0}}}$ times the charge enclosed in that surface. The charge in this case can be calculated by using the formula of surface charge density.

Formula used:
The formula of electric flux using Gauss law is given by
$\phi = \dfrac{q}{{{\varepsilon _0}}}$
Here, $\phi $ is the electric flux, $q$ is the charge in the sphere, and ${\varepsilon _0}$ is the permittivity.
Also, the formula of surface charge density is given by
$\sigma = \dfrac{q}{A}$
Here, $\sigma $ is the charge density of the sphere, $q$ is the charge, and $A$ is the area of the conducting sphere.

Complete step by step answer:
Consider a uniformly charged conducting sphere that is of the radius $1\,m$.
The radius of the sphere, $r = 1\,m$
Also, surface charge density, $\sigma = 20\,C{m^{ - 2}}$
The electric flux in this case can be calculated by using Gauss law. Gauss law states that the total flux inside a charged sphere is equal to the $\dfrac{1}{{{\varepsilon _0}}}$ times the charge enclosed in that surface.
The formula of the electric flux is given by
$\phi = \dfrac{q}{{{\varepsilon _0}}}$
Surface charge density is defined as the charge in the sphere per unit area of the sphere. The formula of the surface charge density is given by
$\sigma = \dfrac{q}{A}$
$ \Rightarrow \,q = \sigma A$
Putting this value in the formula of electric flux as shown below
$\phi = \dfrac{{\sigma A}}{{{\varepsilon _0}}}$
Now, the area of the sphere is given by
$A = 4\pi {r^2}$
Now, the formula of electric flux is given by
$\phi = \dfrac{{\sigma \left( {4\pi {r^2}} \right)}}{{{\varepsilon _0}}}$
$ \Rightarrow \,\phi = \dfrac{{20\left( {4\pi {{(1)}^2}} \right)}}{{{\varepsilon _0}}}$
$ \Rightarrow \,\phi = \dfrac{{80\pi }}{{{\varepsilon _0}}}$
$ \therefore \,\phi = 80\pi \varepsilon _0^{ - 1}$
Therefore, the total flux leaving the Gaussian surface enclosing the sphere is $80\pi \varepsilon _0^{ - 1}$ .

Hence, option B is the correct option.

Note:Here, in the above question ${\varepsilon _0}$ is the permittivity in free space. The value of permittivity in free space is ${\varepsilon _0} = 8.8 \times {10^{ - 12}}\,{m^{ - 3}}k{g^{ - 1}}{s^4}{A^2}$ . Here, we do not have this value in the above example because we want the answer in terms of ${\varepsilon _0}$ .