If the deer herd needs to recover, then how about, I don't know, stop shooting them? I've long ago stopped believing the lies of hunters as conservationists, when you see the stuff that goes on such as the above. Hunters are in the woods to kill, any other reason they come up with is merely an...
So Q = V 4 \pi \epsilon_0 R and so \sigma_{out} = \frac{Q}{A} = \frac{V 4 \pi \epsilon_0 R}{4 \pi R^2} = \frac{V \epsilon_0}{R}.
On the inner surface the charge is simply q, so \sigma_{in} = \frac{q}{4 \pi R^2}.
I really don't know how to proceed. Do I add the potential of the negative charge and V and calculate the electric field from that and then use Gauss' law to find Q = \sigma A?
Homework Statement
Inside a metal conducting shell of radius R there is a negative charge q at a distance a from the center M. The shell is brought up to a potential V. What is the surface charge on the inside and outside of the conductor?
The Attempt at a Solution
I'm not sure if this...