A Charge of `4 xx 10^(-8) C` is distributed uniformaly on the surface of a sphere of radius 1 cm. It is covered by a concentric, hollow conducting sphere of radius 5 cm. (a) Find the electric field at a point 2 cm away from the centre. (b) A charge of `6 xx 10^(-8)C` is placed on the hollow sphere. Find the surface charge density on the outer surface of the hollow sphere.

Explanation

(a) Let us consider figure. Suppose, we have to find the field at the point P. Draw a concentric spherical surface thought P. All the points on this surface are equivalent and by symmetry, the field at all these points will be equal in magnitude and radial in direction.
The flux through this surface `=oint vecE d vecS`
`oint E dS=E vec dS`
` =4pix^2 E`,
where `x=2 cm= 2 xx 10^(-2) m`.
From Gauss's law, this flux is equal to the charge q contained inside the surface divided by `epsilon_0`. Thus,
`4pi x^2 E= q/epilon_0`
or `E =q/4piepilon_0 x^2`
`=(9 xx 10^9 N m^2 C^(-2)) xx 4 xx10^(-8)C/4 xx 10^(-4) m^a`
`= 9 xx 10^5 N C^(-1).
(b) See figure Take a Gaussion surface through the material of the hollow sphere. As electric field in a conducting material ia zero, the flux `oint vecEdvecS` Through this Gaussian surface is zera. Using Gauss's law, the total charge enclsed must be zero. Hence, the charge on the inner surface of the hollow sphere is `-4 xx 10^(-8) C`. But the total charge given to this hollow sphere is `6 xx 10^(-8) C`. Hence the charge on the outer surface will be `10 xx 10^(-8) C`.