Question
For the configuration from Ch. 24 pr 51, suppose a= 5 cm, b= 20 cm, and c= 25 cm. Suppose the electric field at a point 10 cm from the center is measured to be 3.6 x 10^3 N/C radially inward and the electric field at a point 50cm from the center is of magnitude 200 N/C and points radially outward. From this information, find a) the charge on the insulating sphere, b) the net charge on the hollow conducting sphere, c) the charge on the inner surface of the hollow conducting sphere and d) the charge on the outer surface of the hollow conducting sphere.
Explanation / Answer
(1) electric field at a point point a < r < b Electic field due to insulating sphere E = (1/4_0) Q / r ^2 Given at r = 0.01 m Electric field E = 3.6*10^3 N/C and _0 = 8.85* 10^-12 N.m^2 / C^2 charge on the insulating sphere Q = E * r^2 *4_0 plug the all values Q = (3.6*10^3 ) ( 0.01 )^2 * 4* * (8.85*10^-12) = 4.0*10^-11 C Since electric field direction is radially inwaerd the charge on the insulating sphere is negative charge = - 4.0*10-9 C (b)The net charge on the shell be Q' + Q the electric field at a point r > c E = (1/4_0) *'(Q '+Q ) / r ^2 q = E * r^2 * 4_0 given at r = 0.05 m E = 200 N/C net charge on the shell Q ' + Q = (200 N/C) ( 0.05 m )^2 *4* * (8.85*10^-12) = 5.56*10^-9 C Q ' = 5.56*10^-9 C + 4.0*10-9 C = 9.56*10^-9 C Since field is radially out ward the charge on the shell is positive (c) The insulating sphere induces the charge on the inner surface of the conducting shell hence the charge on the inner surface of the shell is exactly opposite to the charge on the insulating sphere it is Q +4.0*10^-11 C (d) The charge on the outer surface of the conducting shell q = Q' - Q = 9.56*10^-9 C - 4.0*10^-11 C = q = 5.56 nC