contestada

A point charge Q at the center of sphare of radius R produces an electric flux of ΦE coming out of the sphare. If the charge remains the same but the radius of the sphere is doubled, the electric flux coming out of it will be
(A) ΦE/2
(B) ΦE/4
(C) 2ΦE
(D) 4ΦE
(E) ΦE

Respuesta :

Answer:

(E) [tex]\Phi_{E}[/tex]

Explanation:

According to the Gauss law the electric flux which is coming out of the surface is equal to the [tex]\frac{1}{\epsilon_{0} }[/tex] times the total charge enclosed by the surface.

Mathematically we can write it as,

[tex]\Phi_{E} =\frac{Q}{\epsilon_{0} }[/tex]

Now the initial condition of the flux is as,

[tex]\phi_{1}=\frac{Q}{\epsilon_{0} }[/tex]

The point charge is constant. The radius will not effect the electric flux.

Therefore, electric flux will remain same.

So, the electric flux that is coming out from the surface of sphere is [tex]\Phi_{E}[/tex].

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