Answer
a. $k\frac{Q}{r^2}$
b. 0
c. $k\frac{Q}{r^2}$
d. The presence of the shell affects the field due to Q alone only within the shell material (i.e., in the bulk).
Work Step by Step
Refer to figure 16–34 in the text.
a. The field is that of the point charge, $E=k\frac{Q}{r^2}$.
b. The electric field is zero inside the conducting material.
c. Once again, the field is that of the point charge, $E=k\frac{Q}{r^2}$.
d. The presence of the shell affects the field due to Q alone only within the shell material (i.e., in the bulk).
The presence of the charge Q electrically polarizes the shell, i.e., it induces charge on both surfaces. A charge of -Q is uniformly distributed on the inside surface of the shell, and an equal but opposite induced charge of +Q is uniformly distributed on the outside surface of the shell.
