# Method of images: two conducting planes

Required math: calculus

Required physics: electrostatics

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Sec 3.2, Problem 3.10.

Another example of the method of images is the problem of a point charge ${+q}$ located at point ${\left(x,y\right)=\left(a,b\right)}$ in the first quadrant (at ${z=0}$), between two conducting planes that cover the ${xz}$ and ${yz}$ planes, thus these two conducting planes meet at right angles.

Following the procedure for the simpler problem of an image charge next to a single conducting plane, we can first place images of ${-q}$ at locations ${\left(x,y\right)=\left(-a,b\right)}$ and ${\left(a,-b\right)}$. If we stopped there, the potential would be

$\displaystyle V=\frac{q}{4\pi\epsilon_{0}}\left[\frac{1}{\sqrt{\left(x-a\right)^{2}+\left(y-b\right)^{2}+z^{2}}}-\frac{1}{\sqrt{\left(x+a\right)^{2}+\left(y-b\right)^{2}+z^{2}}}-\frac{1}{\sqrt{\left(x-a\right)^{2}+\left(y+b\right)^{2}+z^{2}}}\right] \ \ \ \ \ (1)$

This clearly isn’t zero on either of the conducting planes, so we need to add another image. If we try an image of ${+q}$ at ${\left(x,y\right)=\left(-a,-b\right)}$ then the potential is

$\displaystyle V=\frac{q}{4\pi\epsilon_{0}}\left[\frac{1}{\sqrt{\left(x-a\right)^{2}+\left(y-b\right)^{2}+z^{2}}}-\frac{1}{\sqrt{\left(x+a\right)^{2}+\left(y-b\right)^{2}+z^{2}}}-\frac{1}{\sqrt{\left(x-a\right)^{2}+\left(y+b\right)^{2}+z^{2}}}+\frac{1}{\sqrt{\left(x+a\right)^{2}+\left(y+b\right)^{2}+z^{2}}}\right] \ \ \ \ \ (2)$

Now ${V=0}$ on both the planes ${x=0}$ and ${y=0}$ as can be seen by direct substitution.

The force on the original charge can be found from the force from the 3 images:

$\displaystyle \mathbf{F}=\frac{q^{2}}{4\pi\epsilon_{0}}\left(-\frac{1}{4a^{2}}\hat{\mathbf{x}}-\frac{1}{4b^{2}}\hat{\mathbf{y}}+\frac{1}{4\left(a^{2}+b^{2}\right)^{3/2}}\left(a\hat{\mathbf{x}}+b\hat{\mathbf{y}}\right)\right) \ \ \ \ \ (3)$

The third term is the force between the actual charge and the image at ${\left(x,y\right)=\left(-a,-b\right)}$. The magnitude of this force is ${q^{2}/\left[4\pi\epsilon_{0}\left(4a^{2}+4b^{2}\right)\right]}$ and we’ve resolved this along the two coordinate axes.

The work required to bring in ${q}$ from infinity can be found from the general formula for work applied to the images:

$\displaystyle W=\frac{1}{2}\sum_{i=1}^{n}q_{i}V(\mathbf{r}_{i}) \ \ \ \ \ (4)$

where ${V\left(\mathbf{r}_{i}\right)}$ is the potential due to all the charges in the collection except ${q_{i}}$. However, this work assumes that we’re looking at all space, whereas the conducting planes cut the space under consideration down to a quarter of all space, so we need to divide the result by 4.

Plugging in the values we get

 $\displaystyle V(a,b)$ $\displaystyle =$ $\displaystyle \frac{q}{4\pi\epsilon_{0}}\left[-\frac{1}{2a}-\frac{1}{2b}+\frac{1}{2\sqrt{a^{2}+b^{2}}}\right]\ \ \ \ \ (5)$ $\displaystyle V\left(-a,b\right)$ $\displaystyle =$ $\displaystyle \frac{q}{4\pi\epsilon_{0}}\left[\frac{1}{2a}+\frac{1}{2b}-\frac{1}{2\sqrt{a^{2}+b^{2}}}\right]\ \ \ \ \ (6)$ $\displaystyle V\left(-a,-b\right)$ $\displaystyle =$ $\displaystyle \frac{q}{4\pi\epsilon_{0}}\left[-\frac{1}{2a}-\frac{1}{2b}+\frac{1}{2\sqrt{a^{2}+b^{2}}}\right]\ \ \ \ \ (7)$ $\displaystyle V\left(a,-b\right)$ $\displaystyle =$ $\displaystyle \frac{q}{4\pi\epsilon_{0}}\left[\frac{1}{2a}+\frac{1}{2b}-\frac{1}{2\sqrt{a^{2}+b^{2}}}\right] \ \ \ \ \ (8)$

Thus

 $\displaystyle W$ $\displaystyle =$ $\displaystyle \frac{1}{4}\frac{1}{2}\frac{q^{2}}{4\pi\epsilon_{0}}4\left[-\frac{1}{2a}-\frac{1}{2b}+\frac{1}{2\sqrt{a^{2}+b^{2}}}\right]\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{q^{2}}{8\pi\epsilon_{0}}\left[-\frac{1}{2a}-\frac{1}{2b}+\frac{1}{2\sqrt{a^{2}+b^{2}}}\right] \ \ \ \ \ (10)$

This technique can actually be applied to a configuration where we have two conducting planes meeting at other angles. Since the image charges have to cancel in pairs, we can derive a formula for the case where we can divide up space into an even number ${n}$ of sectors. We’ve just seen how it works for ${n=4}$, but for general even ${n}$ the argument would go like this.

Suppose the sector bounded by the planes has one plane at ${y=0}$ (that is, the ${xz}$ plane) and the other at an angle of ${2\pi/n}$. Let’s place the test charge at a location ${\left(x,y\right)=d\left(\cos\alpha,\sin\alpha\right)}$ where ${\alpha}$ is the angle the radius vector to the charge makes with the plane ${y=0}$ and ${d}$ is the distance of the charge from the origin. We can then place the first image by drawing a line from the charge perpendicular to the plane at angle ${2\pi/n}$ and extending it an equal distance on the other side. This image will have charge ${-q}$ and be at an angle of ${2\times\frac{2\pi}{n}-\alpha=\frac{4\pi}{n}-\alpha}$. The next image is found by drawing a line from the first image perpendicular to the plane at angle ${4\pi/n}$ and extending it an equal distance on the other side. This charge is ${+q}$ and is at angle ${\frac{4\pi}{n}+\alpha}$. We continue in this fashion until we arrive at a charge of ${-q}$ at angle ${2\pi-\alpha}$, which is the image of the original charge in the ${y=0}$ plane. This will give a total of ${n}$ charges (one is the test charge and the other ${n-1}$ are the images). The potential is then

 $\displaystyle \frac{4\pi\epsilon_{0}}{q}V$ $\displaystyle =$ $\displaystyle \sum_{m=0}^{\frac{n}{2}-1}\frac{1}{\sqrt{\left(x-d\cos\left(\frac{4\pi m}{n}+\alpha\right)\right)^{2}+\left(y-d\sin\left(\frac{4\pi m}{n}+\alpha\right)\right)^{2}+z^{2}}}-\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle$ $\displaystyle \sum_{m=1}^{\frac{n}{2}}\frac{1}{\sqrt{\left(x-d\cos\left(\frac{4\pi m}{n}-\alpha\right)\right)^{2}+\left(y-d\sin\left(\frac{4\pi m}{n}-\alpha\right)\right)^{2}+z^{2}}} \ \ \ \ \ (12)$

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