A spherical conductor of radius 12 cm has a charge of 1.6 × 10^{−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 centre of the sphere?

#### Solution

**(a) **Radius of the spherical conductor, r = 12 cm = 0.12 m

Charge is uniformly distributed over the conductor, q = 1.6 × 10^{−7} C

Electric field inside a spherical conductor is zero. This is because if there is a field inside the conductor, 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` = Permittivity of free space

`1/(4piin_0) = 9 xx 10^9 "N m"^2 "C" ^-2`

∴ `"E" = (1.6 xx 10^-7 xx 9 xx 10^-9)/(0.12)^2`

= `10^5 "N C"^-1`

Therefore, the electric field just outside the sphere is `10^5 "N C"^-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"/(4piin_0"d"^2)`

= `(9 xx 10^9 xx 1.6 xx 10^-7)/(18 xx 10^-2)^2`

= `4.4 xx 10^4 "N"/"C"`

Therefore, the electric field at a point 18 cm from the centre of the sphere is `4.4 xx 10^4 "N"/"C"`.