Question

(5%) Problem 17: A solid aluminum sphere of radius r1 = 0.105 m is charged with q1 = +4.4 μC of electric charge. It is surrounded concentrically by a spherical copper shell of inner radius r2 = 0.15 m and outer radius r3 = 0.242 m. An electric charge of q2 = -1.1 μC is placed on the copper shell. The situation is static.
(a) Find the magnitude of the electric field, in newtons per coulomb, inside the aluminum ball.
(b) What is the magnitude of the electric field, in newtons per coulomb, inside the copper of the shell, i.e., at a radius r from the center such that r2 < r < r3?

Answer 1

Answer:

Electric field at radius r inside the solid sphere is

$E=\dfrac{q_{1}r}{4\pi \epsilon_{o} r_{1}^3}\ N/C$

Electric field at radius r between inner radius and outer radius of the shell is

E=0 N/C

Explanation:

Given

The radius of the solid sphere is $r_{i}=0.105 \ m$

The charge on the solid sphere is $q_{1}=+4.4\ \mu C$

The inner radius of the shell is $r_{3}=0.15\ m$

The outer radius of the shell is $r_{3}=0.242\ m$

The total charge on the shell is $q_{2}=-1.1\ \mu C$

PART(A)

The magnitude of electric field at radius r where $r$ \\The volumetric charge density of the solid sphere will be

$\rho=\dfrac{q_{1}}{V}\\ \rho=\dfrac{q_{1}}{\dfrac{4}{3}\pi r_{1}^3}$

The charge enclosed by the radius r inside the solid sphere is

rho=$q_{enc}=\rho\times \dfrac{4}{3}\pi r^3\\q_{enc}=\dfrac{q_{1}}{\dfrac{4}{3}\pi r_{1}^3}\times \dfrac{4}{3}\pi r^3\\q_{enc}=\dfrac{q_{1}r^3}{r_{1}^3}$

According to gauss law

$EA=\dfrac{q_{enc}}{\epsilon_{o}}\\E=\dfrac{\dfrac{q_{1}r^3}{r_{1}^3}}{\epsilon_{o}\times 4\pi r^2}\\E=\dfrac{q_{1}r}{4\pi \epsilon_{o} r_{1}^3}\ N/C$

PART(B)

The electric field at radius r where $r_{2}$

The shell is conducting so due to induction of charge

The charge induced on the inner surface of the shell is equal in magnitude of the total charge on the solid sphere but polarity will be changed because a conducting shell has no net electric field inside the shell

So,

The charge on the inner surface of the shell is

$q_{i}=-q_{1}\\q_{i}=-4.4\ \mu C$

Due to conservation of the charge on the shell

The charge accumulated on the outer surface of the shell is

$q_{o}=q_{2}-q_{i}\\q_{o}=-1.1\ \mu C-(-4.4\ \mu C)\\q_{o}=-1.1\ \mu C+4.4\ \mu C\\q_{o}=3.3\ \mu C$

The charge enclosed by the radius r where $r_{1}$

$q_{enc}=q_{1}+q_{i}\\q_{enc}=4.4\times 10^{-6}-4.4\times 10^{-6}\\ q_{enc}=0$

According to gauss law

$EA=\dfrac{q_{enc}}{\epsilon_{o}}\\ E=0\ N/C$