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Question: The inward and outward electric flux for a closed surface is \(N{m^2}/C\)are respectively \(8 \times...

The inward and outward electric flux for a closed surface is Nm2/CN{m^2}/Care respectively 8×1038 \times {10^3}and 4×1034 \times {10^3}. Then the total charge inside the surface is ________ C.
(A) 4×103ε0\dfrac{{ - 4 \times {{10}^3}}}{{{\varepsilon _0}}}
(B) 4×103- 4 \times {10^3}
(C) 4×1034 \times {10^3}
(D) 4×103ε04 \times {10^3}{\varepsilon _0}

Explanation

Solution

Hint We should know the amount of something, which can be electric field or any other physical quantity, that passes through the surface. The total amount of flux is dependent on the strength of the field, the size of the surface through which the flux is passing through and also the orientation.

Complete step by step answer
We know that according to the convention, the inward flux is always taken as negative and the outward flux is always taken as positive.
Hence we can say that the flux, which is denoted by ϕ\phiis given as : 8×103+4×103=4×103Nm2/C- 8 \times {10^3} + 4 \times {10^3} = - 4 \times {10^3}N{m^2}/C
Now we know that, ϕ\phior the flux is denoted as qε0\dfrac{q}{{{\varepsilon _0}}}
So, we can evaluate to find that:
q=ϕε0 \Rightarrow q = \phi {\varepsilon _0}{\text{ }}
q=4×103ε0\Rightarrow q = - 4 \times {10^3}{\varepsilon _0}
Hence, we can say that the total charge inside the surface is 4×103ε04 \times {10^3}{\varepsilon _0}.

Hence the correct answer is option D.

q=ϕε0\Rightarrow q = \phi {\varepsilon _0}
q=4×103ε0\Rightarrow q = - 4 \times {10^3}{\varepsilon _0}

Note We should know that if we consider a closed surface then the orientation of the surface is usually defined as the flux which is flowing from the inside to the outside as positive in nature. This is also known as the inward flux. And the flux which is flowing from the outside to the inside, is considered as negative and is termed as the outward flux.