Question

If the radius of the gaussian surface enclosing a charge is halved, how does the electric flux through the Gaussian surface change?

Answer 1

Gauss law

1. It states that the electric flux within a closed surface is proportional to the electric charge enclosed by the surface.
2. Mathematically:

$\Phi =\frac{Q}{{\epsilon }_0}$

Where,

$Q$ = charge within the given surface,

${\epsilon }_0$ = the electric constant.

$\Phi$= total flux.

Change in electric flux:

1. Given that the radius of the gaussian surface enclosing a charge is reduced to half.
2. As Flux does not depend on the radius of the Gaussian surface. It depends only on the surface.
3. Thus the net charge enclosed by the surface will also be the same.

Hence, flux remains constant irrespective of any change in radius.