Please assist me with the attached electric field [map] problem.
8.6) part a is done at the end.
b) EI*4pi*r^2 = Q/eps
=> EI = Q/(4pi*eps*r^2) (radially outward) --Answer
c.) EII*4pi*r^2 = (Q-2Q)/eps
=> EII = Q/(4pi*eps*r^2) (radially inward) --Answer
d.) Because, the charge in the inside region will be = Q - 2Q = -Q
Therefore the charge ...
This solution demonstrates how to solve equations for electric field, as well as includes an attached .jpg file with a diagram of electric field and induced charge.