Homework Answers
Since the outer shell is conducting, +Q is induced on it.
The total charge on the outer shell is -3Q+Q = -2Q
outer shell radius b = 3 a =3 x 4 = 12 cm = 0.12 m
Area of the outer shell is 4Pi b^2
The surface charge density is total charge divided by area of the shell
s = -2Q/(4 Pi b^2) = -2 x 2/(4 x 3 x 0.12^2) = - 23.15 nC/m^2
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