- #1

Jelsborg

- 9

- 1

- Homework Statement
- Calculate the E field outside concentric spherical shell with point dipole in center

- Relevant Equations
- Gauss' law/Field of a dipole

In a recent test we were asked to calculate the electric field outside a concentric spherical metal shell, in which a point dipole of magnitude p was placed in the center.

Given values are the outer radius of the shell, R, The thickness of the shell, ##\Delta R## and the magnitude of the dipole moment p

I am very tempted to say that the electric field outside the conductor is 0 due to the plus and minus side of the inner dipole inducing equal amounts of positive and negative charge on the inside of the shell.

This would leave no residual charge to reside on the outer surface of the shell, which by invoking Gauss' law means that $$Q_{enc}=0 \rightarrow E=0$$

I am not entirely sure of this argumentation and would like some input as I am aware that you would normally compute the electrical field generated far from a dipole by the formula $$E_{dip}=\frac{p}{4 \pi \epsilon_{0}r^{3}} \cdot (2 cos{\theta}\hat{r}+sin{\theta}{\hat{\theta}})$$

So my 2nd thought would be that somehow the field from the dipole moment inside the empty space in the sphere would make the sphere reallocate plus and minus charges in a way that effectively makes the sphere itself seem like a dipole when viewed from far away.

However, my guess would be that this is flawed since this would violate the condition that the conductors boundary be an equipotential, and also violating the condition that ##E=0## inside the conductor (right?)

Can someone provide feedback on my line of thinking and correct me where/if I'm wrong?

Thanks

Given values are the outer radius of the shell, R, The thickness of the shell, ##\Delta R## and the magnitude of the dipole moment p

I am very tempted to say that the electric field outside the conductor is 0 due to the plus and minus side of the inner dipole inducing equal amounts of positive and negative charge on the inside of the shell.

This would leave no residual charge to reside on the outer surface of the shell, which by invoking Gauss' law means that $$Q_{enc}=0 \rightarrow E=0$$

I am not entirely sure of this argumentation and would like some input as I am aware that you would normally compute the electrical field generated far from a dipole by the formula $$E_{dip}=\frac{p}{4 \pi \epsilon_{0}r^{3}} \cdot (2 cos{\theta}\hat{r}+sin{\theta}{\hat{\theta}})$$

So my 2nd thought would be that somehow the field from the dipole moment inside the empty space in the sphere would make the sphere reallocate plus and minus charges in a way that effectively makes the sphere itself seem like a dipole when viewed from far away.

However, my guess would be that this is flawed since this would violate the condition that the conductors boundary be an equipotential, and also violating the condition that ##E=0## inside the conductor (right?)

Can someone provide feedback on my line of thinking and correct me where/if I'm wrong?

Thanks