# Method of image charges (semisphere on a metal)

I'm currently trying to study ahead for the upcoming semester since I'm on break and I'm stuck on the method of image charges. I've tried watching some youtube videos on that topic and I thought I understood to some degree and so I tried myself on some exercises in my workbook. Sadly there was only one on that topic since the book only covered briefly on that topic.

There's a metallic semisphere with radius a lying on an infinitely expanded metal plate, which lies in the x-z-plane. At $X= (R \cos\alpha, R\sin\alpha, 0)$ there is a charge q. (1) Determine with the use of the method of image charges the potential $\phi(x)$ outside the sphere and show that it solves the Poisson's equation.

(2) Write down the electrostatic potential $\phi(x)$ in the quadrupole approximation with monopole-, dipole- and quadropole-moment.

(3)In relation to the origin compute the monopole-, dipole- and quadrupole-moment of the system with the charges and image charges.

To be honest I'm totally lost here. I've tried following the video step by step to try and solve this but I can't seem to do it. I don't even know where I should place the image charge. Should it be mirrored across the z-axis? Or mirrored across the semisphere? My guess would be the first one, but I'm not totally sure since I tried it with that and still didn't know how to proceed from that. And concerning (2) and (3): I'm guessing the Quadrupole approximation is concerning the multipole expansion with monopole, dipole and quadrupole? At least that was the first thing I found while trying to search for the Quadrupole approximation.

I know it's rude of me to ask since I can't really provide any work on my part considering worthy but can anyone show me a general approach to these type of problems? I feel really stupid after dealing with this problem. My workbook even says that this is an easy exercise to approach this topic.

Edit:

@nbubis I tried drawing the configuration of the image charges. Is it like this?: About (2) and (3): Should I look at it as two dipoles and then add both potentials of those dipoles to get the potential of the quadrupole? But then again I wouldn't need the quadrupole moment for that, right?

I'm guessing they just want a general formula for quadrupole approximation and the explicit answer of that in (3) then. So if I were to go the multipole expansion route would I just need to compute the monopole-, dipole- and quadrupole moment and add them up to get the potential? And truth be told I'm still kinda lost on the terminology in the multipole expansion provided in Wikipedia. It's hard to understand that from me without seeing an example where the multipole expansion is applied.

## 1 Answer

First understand the method of image charges. The idea behind the method is to bypass actually solving the differential equation with boundary conditions, and instead "cheat" by guessing the correct solution. To this end, we find a configuration of imaginary charges that together with the real ones will make the potential on all surfaces be what is given.

In your case, you have two surface, each with constant potential zero. For a plane, if you have a charge $q$ at $(x,y,z)$ and another one at $(x,-y,z)$, clearly the potential at $y=0$ will be zero. For the sphere, if a charge is at radius $R$ from the sphere with radius $a$, an image charge with charge $-qa/R$ should be placed at radius $a^2/R$, but the same idea remains.

Now, first add the image charge for the sphere. Now the potential on the sphere is zero, but on the plane, we still have a gradient. However, if you mirror both charges off the plane, you should see that the potential on both the sphere and the plane is zero.

Writing this all down, we have: $$\begin{eqnarray} \phi({\bf x}) &=& \frac{q}{\sqrt{(x-R\cos\alpha)^2+(y-R\sin\alpha)^2+z^2}} \\ &-& \frac{q}{\sqrt{(x-R\cos\alpha)^2+(y+R\sin\alpha)^2+z^2}} \\ &-&\frac{qa}{R\sqrt{(x-(a^2/R)\cos\alpha)^2+(y-(a^2/R)\sin\alpha)^2+z^2}} \\ &+&\frac{qa}{R\sqrt{(x-(a^2/R)\cos\alpha)^2+(y+(a^2/R)\sin\alpha)^2+z^2}} \end{eqnarray}$$ Clearly, at $y=0$ we have $\phi=0$. Now, what about on the hemisphere? points on the hemisphere satisfy $z^2 = a^2 - x^2-y^2$, so that substituting leads us to the potential on the hemisphere:

$$\begin{eqnarray} \phi({\bf x}_{sphere}) &=& \frac{q}{\sqrt{-2xR\cos\alpha -2yR\sin\alpha + R^2 +a^2}} \\ &-& \frac{q}{\sqrt{-2xR\cos\alpha +2yR\sin\alpha + R^2 +a^2}} \\ &-&\frac{qa}{R\sqrt{-2x(a^2/R)\cos\alpha -2y(a^2/R)\sin\alpha + (a^2/R)^2 + a^2}} \\ &+&\frac{qa}{R\sqrt{-2x(a^2/R)\cos\alpha +2y(a^2/R)\sin\alpha + (a^2/R)^2 + a^2}} \\ &=&0 \end{eqnarray}$$ By putting the $a/R$ into the square root.

To write down the multipole expansion, you just need to write down the Taylor expansion of the potential around $1/r$, with $r = \sqrt{x^2+y^2+z^2}$. This gives that any expression of the form: $$\lim_{r\to\infty}\frac{1}{\sqrt{r^2+b}}= \frac{1}{r} - \frac{b}{2r^3} + \frac{3b^2}{8r^5} + \cdots$$ You can then use this expansion on the components of the potential to get your result.

• About the image charge for the sphere? Where would I put it? At the origin of the sphere or a distance R-a from the surface of the sphere? R being the distance to the charge from the point of origin. – Rafa Fafa Feb 23 '16 at 22:10
• @RafaFafa As stated in the link in my answer, the image charge for a sphere lies on the line connecting the charge and the center of the sphere, on the other side of the sphere. – nbubis Feb 23 '16 at 22:49
• @RafaFafa - see edit. – nbubis Feb 24 '16 at 8:48
• Thanks for your answer. I think I can imagine the charge configuration. Could you look at my edit and tell me if that's right? And I'm having troule with the multipole expansion on (2) and (3). From a subsection on Quadrupoles in Wikipedia I can see that it's somehow the sume of all the multipoles to get the potential of a particular configuration. Meaning I should just compute the monopole-, dipole- and quadrupole-moment? – Rafa Fafa Feb 24 '16 at 9:59
• @RafaFafa - See edit – nbubis Feb 24 '16 at 13:02