Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I think this is an interesting question, to which I don't really know the answer to. (Also, not a homework question.)

Say you have an uncharged metal sphere constrained to move in the z-axis. There is a charged ring lying in the x-y plane centered at the origin. Two cases: 1) the diameter of the ring is larger than the sphere, so that the sphere can pass through the ring, and 2) the diameter of the ring is smaller than the sphere, so the sphere cannot pass through the ring but can touch it.

What are the equilibrium position(s) of the ball?

For case 1) it is obvious that the center, by symmetry, is an equilibrium point. But are there more? The complication arises because of the finite size of the sphere. As the sphere starts to pass through the ring, the charge of opposite sign to the ring is induced near the ring, but the angle is very shallow, so there is not much attractive force. On the other hand the induced charge of the same sign gets pushed to the far end of the sphere, which causes a strong repulsive force, so the sphere might get repelled as it enters the ring. I'm not sure whether this is correct though. It might very well turn out that there is only 1 equilibrium point...

So aside from solving the system exactly to find the equilibrium points, is there a way to argue how many there are, for the 2 cases, and where they are?

share|improve this question
Might want to check out Earnshaw's theorem. Because the Laplace equation doesn't allow maxima and minima in-between boundaries, there won't be any point of stable equilibrium. –  santa claus Feb 3 '13 at 0:11
true, but the sphere is constrained to move along the z-axis. The unstable directions are the x and y directions but there's presumably some rod that the sphere slides on. –  nervxxx Feb 3 '13 at 0:38
add comment

1 Answer

up vote 5 down vote accepted

Using the method of images, you can calculate the force between the ring of charge and the sphere.

Assume the sphere is on the z axis with it's center on the point $z$, a radius of $R_s$ and the ring's radius is $R_r$ with a charge density $\lambda$. So $z$ denotes the center of the sphere.

To calculate the force, you can replace the sphere with a charged ring (method of images) with a charge density $\lambda'$ placed at a distance $d$ below the sphere's center and a point charge $Q'$ at the center of the sphere to make the sphere electrically neutral:

$$\lambda'=-\lambda\frac{ \sqrt{R_r^2+z^2}}{R_s}$$ $$d=\frac{zR_s^2}{z^2+R_r^2}$$ $$Q'=Q\frac{R_s}{\sqrt{z^2+R_r^2}}$$

Now the problem reduces to finding the force between the main ring and the induced ring and point charge.

The image below shows the choose of parameters:

enter image description here

I assume that $R_{\text{ring}}>R_{\text{sphere}}$. The equations for the case of $R_{\text{ring}}<R_{\text{sphere}}$ are the same, except that the motion is restricted to $z>\sqrt{R_s^2-R_r^2}$.

The field of the main ring differs for points with $r>R_r$ and $r<R_r$:($Q$ is the ring's total charge)

$$\Phi(r,\theta)=\frac{Q}{4\pi\epsilon_0}\sum_{l=0}^\infty \mathrm{P}_{2l}(\cos \theta)\cases{\frac{R_r^{2l}}{r^{2l+1}}\mathrm{P}_{2l}(0)\,\,\,\,\,\,\,\,\,\,\text{$r>R_r$}\\\frac{r^{2l}}{R_r^{2l+1}}\mathrm{P}_{2l}(0)\,\,\,\,\,\,\,\,\,\,\text{$r<R_r$}}$$ and $$\mathbf{E}=-\frac{\partial \Phi}{\partial r}\hat r-\frac{1}{r}\frac{\partial \Phi}{\partial \theta}\hat \theta$$

Because of the rotational symmetry of the induced charges, on which we want to find the force, the force will be in the $z$ direction: (I omitted the lengthy calculations)

$$\mathbf{F}_{\text{total}}=\hat z \frac{Q^2}{4\pi \epsilon_0} \left[\frac{z}{(R_r^2+z^2)^2}-\frac{1}{R_r^2\sqrt{z^2+R_r^2}}\cases{{\times f_{\text{out}}(r,\theta)}\\{\times f_{\text{in}}(r,\theta)}} \right]$$

Where $f_{\text{out}}$ is used when $r>R_r$ and $f_{\text{in}}$ is used when $r<R_r$. These are dimensionless functions that appear when differentiating the above potential:

$$f_{\text{out}}=\sum_{l=0}^\infty\mathrm{P}_{2l}(0)\left(\frac{R_r}{r}\right)^{2l+2}\left( (2l+1)\mathrm{P}_{2l}(\cos \theta)\cos\theta +\mathrm {P}^1_{2l}(\cos \theta)\sin\theta \right)$$

$$f_{\text{in}}=\sum_{l=0}^\infty-\mathrm{P}_{2l}(0)\left(\frac{r}{R_r}\right)^{2l-1}\left(2l\mathrm{P}_{2l}(\cos\theta)\cos \theta-\mathrm{P}^1_{2l}(\cos \theta) \sin \theta \right)$$

and $r$ and $\theta$ are functions of $z$: $$r=\left| z-\frac{R_s^2}{\sqrt{z^2+R_r^2}} \right|$$ $$\cos \theta=\frac{z-\frac{zR_s^2}{z^2+R_r^2}}{r}$$ $$\sin \theta=\frac{\frac{R_rR_s^2}{z^2+R_r^2}}{r}$$

$\mathrm {P}^1_{2l}(\cos \theta)$ s are associated Legendre functions, and appear when differentiating $\mathrm {P}_{2l}(\cos \theta)$ w.r.t. $\theta$.

The reason of summing over even terms (indexes are $2l$ instead of $l$ ) is that $\mathrm {P}_{l}(0)$ is nonzero for even terms, as is equal to: $$\mathrm {P}_{l}(0)= \frac 1 {2^l} \sum_{k=0}^l {l\choose k}^2 (-1)^{l-k}$$

It seems impossible to say anything intuitively, but the above equations can be used to simulate the system (with a rather good accuracy, Since the higher order terms in the above series go as $r^{2l-1}$ instead of $r^l$).

share|improve this answer
Mostafa: I am not sure that the method of images works in this case. For the method of images, you require that the sphere is grounded, i.e. that $V$ at its surface is $0$. Charge is not conserved. For this Laplace problem with pure Dirichlet boundary condition, the solution is the ring as you have described. Unfortunately, in my question, the sphere is not grounded. Thus mathematically, the problem becomes $V = c(r)$ on the surface of the sphere for some function $c$ that depends on the distance $r$, and $Q = 0$ to ensure neutrality of the sphere. This is not a pure Dirichlet condition. –  nervxxx Aug 8 '13 at 4:11
Thus, the solution is not given by what you have described. Physically, what's wrong with your solution is the sphere is always charged. So... that's not right. Good effort though. –  nervxxx Aug 8 '13 at 4:14
No. This method works fine even if the sphere is not grounded, although I forgot to put a point charge at it's place. The sphere is a equipotential surface and it's potential is always a constant (not a function of $r$ and $\theta$). Now, to understand why and how you can use method of images when the sphere has charge $Q$ and is not grounded, you can apply linear superposition: (next comment) –  Mostafa Aug 8 '13 at 5:37
First ground the sphere (in the presence of outer point charge) and you will have the regular image charge, as you said. Now disconnect the ground wire; since the induced surface charges are still in equilibrium, no charge transfers. To come up with the final situation you just have to put a $Q-q'$ point charge at the center of the sphere, and this added charge distributes itself uniformly on the sphere, because the forces by the outer charge $q$ are already balanced by $q'$. –  Mostafa Aug 8 '13 at 5:38
Ok now I agree with your use of the method of images! Putting a charge at the center of the sphere corrects the mistake I was talking about. Update the section which you have labelled incomplete and if I don't see anything I am unhappy with, I will upvote and accept your answer. By the way, the $r$ I was talking about in my first comment is not the radial distance from the center of the sphere, but rather the distance from the sphere to the charged ring. What I meant was, the surface of the sphere is an equipotential but the value of this potential is not a constant. –  nervxxx Aug 8 '13 at 9:02
show 2 more comments

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.