The potential of a sphere with opposite hemisphere charge densities

[Tony Hau]

Homework Statement

A sphereical shell of radius $R$ carries a uniform surface charge $\sigma_o$ on the northern hemisphere and a uniform surface charge $-\sigma_o$ on the southern hemisphere. Find the potential inside and outside the sphere.

Relevant Equations

$V(r,\theta)=\sum_{l=0}^{\infty}(A_lr^l+\frac{B_l}{r^{l+1}})P_l(cos\theta)$

Here is what the solution says:

As usual, quote the general potential formula: $$V(r,\theta)=\sum_{l=0}^{\infty}(A_lr^l+\frac{B_l}{r^{l+1}})P_l(cos\theta)$$

The potential outside the sphere is: $$V(r,{\theta})=\sum_{l=0}^{\infty}\frac{B_l}{r^{l+1}}P_l(cos\theta)$$, which makes sense to me.

The potential at z-axis is: $$V(r,{\theta})=\sum_{l=0}^{\infty}\frac{B_l}{r^{l+1}}P_l(cos\theta)$$ because $P_l(cos\theta) = 1$ for all $l$s.

However, here comes a strange equation: $$V(r,\theta)=\frac{\sigma}{2\epsilon_o}(\sqrt{r^2+R^2}-r)$$ along the z-axis.

I think this equation may come from the Gauss's law: $\oint \vec E \cdot d \vec A = \frac {Q_{total}}{\epsilon_o}$. I set the Gaussian's surface to enclose the whole hemisphere. As the charged surface emits a uniform E field outside and inside, the equation becomes $E\int 2d A = \frac {Q_{total}}{\epsilon_o}$, where $A = 2\pi R^2$. Obviously this is not the correct idea. Can anyone help? Thanks!

 

[Abhishek11235]

Use the boundary conditions for continuity of potential across surface and discontinuity of electric fields across surface to solve for A and B!

The reason Gauss theorem won't work here is that problem does not possesses spherical Symmetry but Cylindrical symmetry!

Also $P_l(1)=1$ for all l!

 

[kuruman]

Since the charge distribution changes sign for $z<0$, the potential must also do so, i.e. $V(x,y,-z)=-V(x,y,z).$ This means that you can toss out some Legendre polynomials. Which ones? Hint: $z=r\cos\theta.$

 

[Tony Hau]

Since the charge distribution changes sign for $z<0$, the potential must also do so, i.e. $V(x,y,-z)=-V(x,y,z).$ This means that you can toss out some Legendre polynomials. Which ones? Hint: $z=r\cos\theta.$

This is hard but I will make a guess. I think for even numbers of $l$ the Legendre polynomials are tossed.

The $V_{out}$ is given by: $$\sum_{l=0}^{\infty}\frac{B_l}{r^{l+1}}P_l(cos\theta)$$

For the first few $l$s, $$\frac{B_1}{r^2} + \frac{B_2}{r^3}cos\theta + \frac{B_3}{r^4}\frac{3cos^2\theta-1}{2} + \frac{B_4}{r^5}\frac{5cos^3\theta-3cos\theta}{2} \dots = \frac{B_1}{r^2} - \frac{B_2}{r^3}cos\theta + \frac{B_3}{r^4}\frac{3cos^2\theta-1}{2} -\frac{B_4}{r^5}\frac{5cos^3\theta-3cos\theta}{2} \dots$$ because $-z=-rcos\theta$ and $V(x,y,-z)=-V(x,y,z)$.

For this boundary condition, the sum of all even numbers of $l$ of the Legendre polynomials is 0.

 

[kuruman]

For this boundary condition, all even numbers of the Legendre polynomials are canceled on both sides.

Yes.

 

[TSny]

The potential at z-axis is: $$V(r,{\theta})=\sum_{l=0}^{\infty}\frac{B_l}{r^{l+1}}P_l(cos\theta)$$ because $P_l(cos\theta) = 1$ for all $l$s.

So, on the z-axis you may write $$V(r)=\sum_{l=0}^{\infty}\frac{B_l}{r^{l+1}}$$

However, here comes a strange equation: $$V(r,\theta)=\frac{\sigma}{2\epsilon_o}(\sqrt{r^2+R^2}-r)$$ along the z-axis.

To find V on the z-axis, you can break up the charge distribution into rings with the z-axis as the axis of the rings. The potential on the axis of a uniformly charged ring is easy to find. Integrating over the rings will give the total potential at points on the z-axis. However, I get a somewhat different result. I find $$V(r)=\frac{\sigma}{\epsilon_o}\frac{R}{r}(\sqrt{r^2+R^2}-r)$$ on the z-axis for $r>R$.

The reason for finding $V$ on the z-axis is that it gives you a way to find the coefficients $B_l$. If you expand the expression for $V(r)$ on the z-axis as a power series in $1/r$, you can compare it to $V(r)=\sum_{l=0}^{\infty} \large \frac{B_l}{r^{l+1}}$ and identify the $B_l$.

 

[Tony Hau]

So, on the z-axis you may write $$V(r)=\sum_{l=0}^{\infty}\frac{B_l}{r^{l+1}}$$
To find V on the z-axis, you can break up the charge distribution into rings with the z-axis as the axis of the rings. The potential on the axis of a uniformly charged ring is easy to find. Integrating over the rings will give the total potential at points on the z-axis. However, I get a somewhat different result. I find $$V(r)=\frac{\sigma}{\epsilon_o}\frac{R}{r}(\sqrt{r^2+R^2}-r)$$ on the z-axis for $r>R$.

The reason for finding $V$ on the z-axis is that it gives you a way to find the coefficients $B_l$. If you expand the expression for $V(r)$ on the z-axis as a power series in $1/r$, you can compare it to $V(r)=\sum_{l=0}^{\infty} \large \frac{B_l}{r^{l+1}}$ and identify the $B_l$.

Nice explanation!