why in the above image flux through the sphere and the cube in which sphere is present is equal i.e. charge/epsilon knot? when i calculated flux through the sphere it came after integration(round...

why in the above image flux through the sphere and the cube in which sphere is present is equal i.e. charge/epsilon knot?

when i calculated flux through the sphere it came after integration(round trip) = Q/4*pie*epsilon-not r^2 multiplied by 4*pie*r^2(ie the surface area of the sphere and after cancellation) = Q/E 

therefore my question is how this same expression came for a cube enclosing the same charge too?

Asked on by ashu1998

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to remember Gauss's law which states that the net flux through any closed surface is proportional to the net charge enclosed.

Hence, the net electric flux through the area element is the following, such that:

`d Phi_E= int int bar E* bar dA = 1/(4pi )*Q/(r^2)(r^2 sin d theta d phi)`

Evaluating the total flux through the entire surface, yields:

`Phi_E = bar E* bar dA = Q/(4pi epsi_0) int_0^pi sin theta int_0^(2pi) d = Q/(epsi_0)`

Expressing the Gauss's law, yields:

`Phi_E = int int bar E* bar dA = (q_enc)/(epsi_o)`

q_enc is the net charge inside the surface

Hence, the charge is independent of the shape of the surface that encloses the charge, in the given case, the cube that encloses the sphere.

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