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Law of Conservation of Energy

Fig. shows fo...

Question

Fig. shows four charges $q_{1}, q_{2}, q_{3}$ and $q_{4}$ fixed in space. Then the total flux of the electric field through a closed surface S, due to all the charges, is:

not equal to the total flux through S due to

q_{3}

and

q_{4}

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equal to the total flux through S due to

q_{3}

and

q_{4}

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zero if

q_{1} + q_{2} = q_{3} + q_{4}

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twice the total flux through S due to

q_{3}

and

q_{4}

q_{1} + q_{2} = q_{3} + q_{4}

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Solution

The correct option is B equal to the total flux through S due to $q_{3}$ and $q_{4}$
From Gauss's law, the flux, $ϕ = \frac{Q_{e n}}{ϵ_{0}}$ where $Q_{e n}$ be the charge enclosed by the surface.
Here, $Q_{e n} = q_{3} + q_{4}$
So, net flux through the closed surface , $ϕ = \frac{q_{3} + q_{4}}{ϵ_{0}}$
Thus, it is due to only charges $q_{3}$ and $q_{4}$ which are inside the surface.

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Similar questions

The electric flux through the surface S as due to all the charges as shown in following figure is $\frac{q_{1} + q_{3} - q_{2} - q_{4}}{ϵ_{0}}$

q_{3}

q_{4}

q_{1}, q_{2}, q_{3} a n d q_{4}

q_{1}

q_{2}

q_{3}

q_{4}

q_{1}

q_{2}

q_{1}

q_{2}

S

q_{1} = q_{4} = 3.1 n C, q_{2} = q_{5} - 5.9 n C, q_{3} = - 3.1 n C

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