from Gauss law,
E (4 pi r^2) = Qin/e0
(A) r1 = 1.5 m
Qin = q1
E ( 4 x pi x 1.5^2) = (9 x 10^-9) / (8.854 x 10^-12)
E = 35.95 N/C radially outward
(B) r2 is inside the metal so E = 0
{ field inside conductor is always zero.}
(C) Qinner + Q1 = 0
Qinner = - q1 = - 9 nC
Qinner + Qouter = Q2 = - 6 nC
- 9nC + Qouter = - 6
Qouter = 3 nC
Problem 4-A point charge q 9.0 nC is placed inside a spherical shell made of metal,...
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