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Question: A uniformly charged conducting sphere is having radius \[1\,m\] and surface charge density \(20\,C{m...

A uniformly charged conducting sphere is having radius 1m1\,m and surface charge density 20Cm220\,C{m^{ - 2}} . The total flux leaving the Gaussian surface enclosing the sphere is
A. 40πε0140\pi \varepsilon _0^{ - 1}
B. 80πε0180\pi \varepsilon _0^{ - 1}
C. 20πε0120\pi \varepsilon _0^{ - 1}
D. 60πε0160\pi \varepsilon _0^{ - 1}

Explanation

Solution

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 1ε0\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
ϕ=qε0\phi = \dfrac{q}{{{\varepsilon _0}}}
Here, ϕ\phi is the electric flux, qq is the charge in the sphere, and ε0{\varepsilon _0} is the permittivity.
Also, the formula of surface charge density is given by
σ=qA\sigma = \dfrac{q}{A}
Here, σ\sigma is the charge density of the sphere, qq is the charge, and AA is the area of the conducting sphere.

Complete step by step answer:
Consider a uniformly charged conducting sphere that is of the radius 1m1\,m.
The radius of the sphere, r=1mr = 1\,m
Also, surface charge density, σ=20Cm2\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 1ε0\dfrac{1}{{{\varepsilon _0}}} times the charge enclosed in that surface.
The formula of the electric flux is given by
ϕ=qε0\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
σ=qA\sigma = \dfrac{q}{A}
q=σA\Rightarrow \,q = \sigma A
Putting this value in the formula of electric flux as shown below
ϕ=σAε0\phi = \dfrac{{\sigma A}}{{{\varepsilon _0}}}
Now, the area of the sphere is given by
A=4πr2A = 4\pi {r^2}
Now, the formula of electric flux is given by
ϕ=σ(4πr2)ε0\phi = \dfrac{{\sigma \left( {4\pi {r^2}} \right)}}{{{\varepsilon _0}}}
ϕ=20(4π(1)2)ε0\Rightarrow \,\phi = \dfrac{{20\left( {4\pi {{(1)}^2}} \right)}}{{{\varepsilon _0}}}
ϕ=80πε0\Rightarrow \,\phi = \dfrac{{80\pi }}{{{\varepsilon _0}}}
ϕ=80πε01\therefore \,\phi = 80\pi \varepsilon _0^{ - 1}
Therefore, the total flux leaving the Gaussian surface enclosing the sphere is 80πε0180\pi \varepsilon _0^{ - 1} .

Hence, option B is the correct option.

Note: Here, in the above question ε0{\varepsilon _0} is the permittivity in free space. The value of permittivity in free space is ε0=8.8×1012m3kg1s4A2{\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 ε0{\varepsilon _0} .