1.

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

Answer»

equal, zero
equal, equal
zero, equal
zero, zero

Solution :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.


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