A spherical charged conductor has surface charge density `sigma`.The intensity of electric field and potential on its surface are `E` and `V` .Now radius of sphere is halved keeping the charge density as constant .The new electric field on the surface and potential at the centre of the sphere are

Options

A. 4E, V
B. E, V/2
C. E, V
D. 2E, 4V

Answer

B

Explanation

`E = (sigma)/(epsi_(0)) and V = (sigma R)/(epsi_(0))`