A source consists of a point charge 30 nC at the center of a

A source consists of a point charge 30 nC at the center of a spherical CONDUCTING shell of inner radius a = 0.02 m and outer radius b = 0.03 m which has a charge -10 nC, as shown in the figure. Assume that if the shell is a conducting medium in electrostatic equilibrium, then E_r = 0 for a

Solution

2.(a)

From Gauss Law,

E.A = Qinside / e0

and electric field inside conductor is always zero.

and charge resides on outer surfaces.

Suppose charge on inner surface is Qinner.

for a < r < b

E = 0 so Qinside = 0

Qinside = Qcentre + Qinner = 0

30nC + Qinner = 0

Qinner = - 30 nC

and charge on conductor resides on surface that why

Qinner + Qouter = -10nC

-30nC + Qouter = -10nC

Qouter = 20 nC

b) for r < a

Qinside = 30 nC

E.A = Qin / e0

E ( 4 pi r^2) = (30 x 10^-9) / (8.854 x 10^-12)

E = 269.6/r^2

for r > b

Qin = 30nC - 10nC = 20 nC

E ( 4 pi r^2) = (20x 10^-9) / (8.854 x 10^-12)

E = 179.75/r^2

c) V = integral of E.dr

for infinity to b \"

V = -179.75/r and r is from infinity to 0.03m

V = 179.75/0.03 =5991.83 Volt

and V(b) = V(a) = 5991.83 Volt
now from a to a/2.

V(a/2) - V(b) = - 269.6/r

V(a/2) - 5991.83 = 269.6 [ 2/0.02 - 1/0.02]

V(a/2) = 5991.83 + 13480

Va/2 = 19471.8 Volt Or 19.5 kV