Please help! A solid conducting sphere of radius R 1 has a charge Q on it. This

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Please help!

A solid conducting sphere of radius R1 has a charge Q on it. This sphere is enclosed in a thick insulating shell of charge 4Q uniformly distributed between its inner radius R2(R2>R1) and outer radius R3. What is the charge on the outer surface of the conducting sphere?

Find the magnitude of the electric field everywhere in space. (Calculus is not needed!)

- E (r<R1):

- E (R1<r<R2):

- E (R2<r<R3):

A solid conducting sphere of radius R1 has a charge Q on it. This sphere is enclosed in a thick insulating shell of charge 4Q uniformly distributed between its inner radius R2 (R2 R1) and outer radius R3 What is the charge on the outer surface of the conducting sphere? Q Find the magnitude of the electric field everywhere in space. (Calculus is not needed!) E (r

Explanation / Answer

part 1:

charge on the outer surface of the conducting sphere=charge on the sphere=Q

part 2:

for r<R1,as the sphere is conducting, electric field inside a conductor is 0.

so E(r<R1) =0

part 3:

for R1<r<R2 :

consider a gaussian sphere of radius r.

then charge enclosed=Q

if electric field is E,

using gauss law:

epsilon*E*4*pi*r^2=Q

==>E=Q/(4*pi*epsilon*r^2)

part 4:

for R2<r<R3:
charge enclosed=Q+(4*Q*(r^2-R2^2)/(R3^2-R2^2))

if electric field is E,

then epsilon*E*4*pi*r^2=Q+(4*Q*(r^2-R2^2)/(R3^2-R2^2))

==>E=(Q+(4*Q*(r^2-R2^2)/(R3^2-R2^2)))/(4*pi*epsilon*r^2)

part 5:

for r>R3, total charge enclosed=Q+4*Q=5*Q

then electric field=5*Q/(4*pi*epsilon*r^2)

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