## Answers

A.) The total charge can be treated as a point charge as if thesphere were compacted to a point (by gauss' law). The magnitude ofthe charge can be found bythe electric field equation: E=kQ/R^2.So,

890=(9x10^9)Q/(.75)^2 => Q=5.56x10^-8. .

This charge is distributed uniformly on the outer surface, as italways is on a conductor.

b.) The charge on the is equal to -Q, since the net charge is zero.It is also distributed uniformly.

c.) The charge on the inside is equal to Q, inducing the charge onthe inside of the shell, which in turn induces the charge on theoutside. We can't sayexactly how it is distributed due to gauss'law, but we at least know it is symmetric since the field isuniform.

Hope this helps!

.

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