A spherical conductor of radius 12 cm has a charge of `1.6xx10^(-7)C` distributed uniformly on its surface. What is the electric field (a) inside the sphere (b) just outside the sphere (c ) at a point 18 cm from the center of the sphere ?

Explanation

(a) radius of the of the spherical conductor, r=12 cm =0.12 m
charge is uniformly distributed over the conductor, `q=1.6xx10^(-7) C`
electric field inside a spherical conductor is zero. This is because if there is field inside the conductors , then charges will move to neutralize it.
(b) electric field E just outside the conductor is given by the relation,
`E=q/(4piin_(0)r^(2))`
where, `in_(0)` =permitivity of free space
`1/(4piin_(0))=9xx10^(9) Nm^(2)C^(-2)`
`:. E=(1.6xx10^(-7)xx9xx10^(-9))/((0.12)^(2))`
`=10^(5)NC^(-1)`
Therefore, the electric field just outside the sphere is `10^(5) NC^(-1)`
(C) electric field at a point 18 m from the centre of the sphere `=E_(1)`
distance of the point from the centre, d=18 cm =0.18 m
`E_(1)=q/(4pi in_(0)d^(2))`
`=(9xx10^(9)xx1.6xx10^(-7))/((18xx10^(-2))^(2))`
`=4.4 xx10^(4) N//C`
Therefore, the electric field at a point 18 cm from the centre of the sphere is
`4.4xx10^(4) N//C`