Inside a uniformly charged spherical shell, electric field is ...... and electrostatic potential is .....

Options

A. equal, zero
B. equal, equal
C. zero, equal
D. zero, zero

Answer

A

Explanation

No charge is enclosed by a Gaussian surface inside the charged spherical shell and so
`phi= (q)/(in_(0))=0`
`:. ointvecE.vecda=0`
`:. E =0 (because phi da ne 0)`
Now, potential difference between a point inside charged shell and a point on its surface is
`V_(i)-V_(s)=-int_(s)^(i)vecE.vecdl`
`:. V_(i)-V_(s)=0 (because E=0)`
`:. V_(i)=V_(s) = (kQ)/(R)`
Everywhere inside the charged spherical shell, electrostatic potential is same, equal to that on the swface.