Recent content by Shukie

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...

Thanks a ton!

Thanks a lot!

Yes, you're right, so on the outer surfaceit's just Q = V 4 \pi \epsilon_0 R and the inner surface q.

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}.

That's \frac{Q}{4 \pi \epsilon_0 r} for r \geq R.

It would be 0 for r < R and \frac{Q}{4 \pi \epsilon_0 r^2} for r \geq R.

Yes, \int \mathbf{E} \cdot d\mathbf{A} = \frac{Q}{\epsilon_0}, but I don't know how to apply it here. The potential V will it affect it right?

Do I need to use image charges to solve this? Because otherwise I have no idea.

So we have: V = - \int_{\infty}^R \mathbf{E} \cdot d\mathbf{r} I don't know how to continue...

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...

Thanks

Are the forces on the mass above the spring then T and -mg and - kx?

Thanks.