# Electric potential, spherical conductor

1. Aug 28, 2011

### sparkle123

Hello,
in this diagram, the shaded regions are spherical conductors.

What's the potential at A=B?
Ignoring the outer sphere, it should be kQ/R.
When you add the outer sphere, potential at C=D=0 and electric field between B and C is kQ/x^2
so i integrated (kQ/x^2) dx with interval [2R, R] and got -kQ/2r
Is that right? I think the potential at A=B should be positive though, and the answer says so too :(
Thanks!

2. Aug 28, 2011

### ehild

Do not forget that the electric field is a vector, and it points outward in this case. When you calculate the potential, you determine ΔU with the line integral -∫Edr. If you go inward E and dr point to opposite directions, the integrand is negative. You do not get confused when integrating from B to C and get U(B) knowing that U(C)=0.

ehild

3. Aug 28, 2011

### ehild

The potential is the same at every point of a metal. So U(A)=U(B) and U(C) = U(D). I think you meant the electric field, kQ/r2.
The potential is not kQ/r when the outer sphere is present, as this would give non-zero potential outside. You need an additive constant: U=kQ/r + C so as it give zero at r=2R.

ehild

4. Aug 28, 2011

### I like Serena

Sorry, I've deleted my post to avoid confusion.
I was mixing it up with another problem where the spheres were non-conducting.

5. Aug 28, 2011

### sparkle123

Thanks to all of you! :D
just to be sure, so the potential at B is kQ/R without the outer sphere, and with the outer sphere its kQ/2R, and ehild's additive constant C is -kQ/2R to give U(B) = kQ/R - kQ/2R = kQ/2R?
Thanks again!

6. Aug 28, 2011

### I like Serena

Yep!

7. Aug 28, 2011

Thanks! :)