A charge `Q` is enclosed by a Gaussian spherical surface of radius `R`. If the radius is doubled, then the outward electric flux will

Options

A. increase four times
B. be reduced to half
C. remain the same
D. be doubled

Answer

C

Explanation

According to Gauss.s law,
`phi_(E) = (Q_("enclosed"))/(epsilon_(0))`
If the radius of the Gaussian surface is doubled, the outward electric flux will remain the same. This is because electric flux depends only on the charge enclosed by the surface.