Question

A spherical charged conductor has surface charge density . The electric field on its surface is e and electric potential

Answer 1

note : plz write complete question for getting answer quickly.

your complete question is ----> A spherical charged conductor has surface charge density .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 center of the sphere are :

solution : Let surface charge density of spherical shell is $\sigma$ and radius of shell is R.

then, electric field is found by using Gauss -theorem,

electric flux = E.A = $\frac{\sigma A}{\epsilon_0}$

or, $E=\frac{\sigma}{\epsilon_0}$

here you should noticed that electric field due to spherical shell is independent on radius of it. so, electric field remains constant if we change its radius.

e.g., final electric field of spherical shell=E

electrical potential is given by, $V=E.dr$

so, electric potential in initial time, V= $\frac{\sigma R}{\epsilon_0}$

so, potential of spherical shell is directly proportional to its radius.

so, if the radius of spherical shell is halved then, potential of its also will be half. hence, final potential of spherical shell is V/2