Question

A solid conducting sphere with radius R carries a positive total charge Q. The sphere is surrounded by an insulating shell with inner radius R and outer radius 2R. The insulating shell has a uniform charge density rho.
Part A

Find the value of rho so that the net charge of the entire system is zero.

Express your answer in terms of the variables Q, R, and appropriate constants.Part B

If rho has the value found in part A, find the magnitude of the electric field in the region 0
Express your answer in terms of the variables Q, R, r, and appropriate constants.Part C

If rho has the value found in part A, find the magnitude of the electric field in the region R
Express your answer in terms of the variables Q, R, r, and appropriate constants.Part D

If rho has the value found in part A, find the direction of the electric field in the region R
If rho has the value found in part A, find the magnitude of the electric field in the region r>2R.

Express your answer in terms of the variables Q, R, r, and appropriate constants.Part F

As a general rule, the electric field is discontinuous only at locations where there is a thin sheet of charge. Explain how your results in previous parts agree with this rule.

Answer 1

Answer:

Explanation:

Volume of the insulating shell is,

$V_{shell}=\frac{4}{3}\pi(R^3_2-R^3_1)$

Charge density of the shell is,

$\rho=\frac{Q_{shell}}{\frac{4}{3}\pi(R^3_2-R^3_1)}$

Here, $R_2 =2R, R_1 =R \,and\, Q_{shell} =-Q$

$\rho=\frac{Q_{shell}}{\frac{4}{3}\pi((2R)^3-R^3)}=\frac{-3Q}{28\piR^3}$

B)

The electric field is $E=\frac{1}{4\pi\epsilon_0}\frac{Qr}{R^3}$

For 0 <r<R the electric field is zero, because the electric field inside the conductor is zero.

C)

For R <r <2R According to gauss law

$E(4\pi r^2)=\frac{Q}{\epsilon_0}+\frac{4\pi\rho}{3\epsilon_0}(r^3-R^3)$

substitute $\rho=\frac{-3Q}{28\piR^3}$

$E=\frac{2}{7\pi\epsilon_0}\frac{Q}{r^2}-\frac{Qr}{28\piR^3}$

D)

The net charge enclosed for each r in this range is positive and the electric field is outward

E)

For r>2R

Charge enclosed is zero, so electric field is zero