Physics, asked by rohithshyamboddikuri, 9 months ago

If the electric flux entering and leaving an enclosed surface of area s respectively are ϕ1 and ϕ2, the magnitude of electric field across the surface will be

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Answered by ItzMADARA

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Answered by ARKRANGER

Answer:

We know that, electric flux ϕ1 (or electric field lines) entering in a closed surface is -ve and electric flux ϕ2(or electric field lines) leaving a closed surface is +ve.

Hence, net electric flux through the closed surface,

ϕ=ϕ2−ϕ1

Now, according to Gauss' theorem, the net electric flux ϕ passing through a closed surface is equal to the 1/ε0 times of the total charge q, enclosed by surface i.e.

ϕ=q/ε0ϕ2−ϕ1

=q/ε0

q=ε0(ϕ 2

−ϕ 1 )

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