Potential in a sphere in a dielectric

[nagyn]

Homework Statement

A solid conducting sphere of radius R and carrying charge +q is embedded in an electrically neutral nonconducting spherical shell of inner radius R and outer radius 2 R . The material of which the shell is made has a dielectric constant of 3.0.

Relative to a potential of zero at infinity, what is the potential at the center of the conducting sphere?

Homework Equations

V = ∫Edr = kq/r + C

Gauss's law in dielectrics: ∫KEdA = q/ε₀

The Attempt at a Solution

I've made a few assumptions.

The field in a conducting sphere is zero, which means the potential in a conducting sphere is constant. Looking at just the sphere alone, potential at the center relative to infinity should be kq/R (positive because I'm going against electric field?).

With the dielectric in place to still take zero at infinity I need to take into account the potential difference across the dielectric (I think?)

Edielectric = q/KAε₀

V(R->2R) = -∫Edielectric*dr (negative because I'm going in direction of electric field?) = -kq/Kr [2R - R] = -kq/2KR + kq/KR = kq/2KR

I know that a dielectric will decrease the electric potential, so I thought V = Vsphere - Vdielectric = kq*((1/R) - (1/6R)).

This is not correct and I'm not sure where I messed up.

 

[haruspex]

I agree with your work up to this point:

so I thought V = Vsphere - Vdielectric = kq*((1/R) - (1/6R)).

Think through that again. I believe you have made quite a simple mistake in what to subtract from what.

 

[rude man]

Maybe more straightforward if you just integrated the E field from infinity to R.

 

[nagyn]

Maybe more straightforward if you just integrated the E field from infinity to R.

So integral of E field from infinity to R = (integral of E field from infinity to 2R) + (integral of E field through dielectric).

From from infinity to 2R, can I still consider the system as a point particle and say the electric field is kQ/R^2? Or does that change with the dielectric in place?

 

[TSny]

So integral of E field from infinity to R = (integral of E field from infinity to 2R) + (integral of E field through dielectric).

From from infinity to 2R, can I still consider the system as a point particle and say the electric field is kQ/R^2? Or does that change with the dielectric in place?

Answer this by applying Gauss' law to a Gaussian surface of radius r>2R.

 

[nagyn]

EA = Qenclosed/E₀

If I'm understanding correctly, a dielectric doesn't change the total charge enclosed, so Qenclosed should still be Q.

E = Q/AE₀ = kQ/R^2

 

[haruspex]

EA = Qenclosed/E₀

If I'm understanding correctly, a dielectric doesn't change the total charge enclosed, so Qenclosed should still be Q.

E = Q/AE₀ = kQ/R^2

Yes, but I'm not sure why this thread has become so complicated.
A dielectric diminishes the potential difference across it. It exerts no long range field, so cannot affect potential gradients outside itself, as you confirmed by following TSny's hint.
In post #1 you found the dielectric layer reduced the potential difference across it from kq/2R to kq/2KR. What reduction in potential in the sphere is needed to keep the potential at infinity zero?

 

[rude man]

EA = Qenclosed/E₀
If I'm understanding correctly, a dielectric doesn't change the total charge enclosed, so Qenclosed should still be Q.
E = Q/AE₀ = kQ/R^2

good thought!
but don't say "E₀" when you mean ε₀.