Answer:
Explanation:
Volume of the insulating shell is,
[tex]V_{shell}=\frac{4}{3}\pi(R^3_2-R^3_1)[/tex]
Charge density of the shell is,
[tex]\rho=\frac{Q_{shell}}{\frac{4}{3}\pi(R^3_2-R^3_1)}[/tex]
Here, [tex]R_2 =2R, R_1 =R \,and\, Q_{shell} =-Q[/tex]
[tex]\rho=\frac{Q_{shell}}{\frac{4}{3}\pi((2R)^3-R^3)}=\frac{-3Q}{28\piR^3}[/tex]
B)
The electric field is [tex]E=\frac{1}{4\pi\epsilon_0}\frac{Qr}{R^3}[/tex]
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
[tex]E(4\pi r^2)=\frac{Q}{\epsilon_0}+\frac{4\pi\rho}{3\epsilon_0}(r^3-R^3)[/tex]
substitute [tex]\rho=\frac{-3Q}{28\piR^3}[/tex]
[tex]E=\frac{2}{7\pi\epsilon_0}\frac{Q}{r^2}-\frac{Qr}{28\piR^3}[/tex]
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
