Electric potential, spherical conductor

[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!

 

[ehild]

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!

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

 

[ehild]

Hi sparkle123!

I'm afraid he potential at A is not equal to B, nor is C=D.

Ignoring the outer sphere, the potential between R and 2R is kQ/r.
Furthermore, even with the outer sphere, this is still true.
As for the reason, we would have to apply Gauss's law (the first of Maxwell's equations).
Are you familiar with that one?

Outside the outer sphere the electric field of the inner sphere cancels the electric field of the outer sphere...

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

 

[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.

 

[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!

 

[I like Serena]

Yep!

 

[sparkle123]

Thanks! :)