# Method of Images

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1. May 24, 2016

### VMP

1. The problem statement, all variables and given/known data

We have a coordinate system (x, y, z). Two conducting plates 1 and 2 are parallel and lie in the x-y plane. Plate 1 is at height (x, y, 0) and plate 2 is at height (x, y, 4a), where a is an arbitrary constant.

Between these two plates there are 2 charges +q and -q. Charge -q is at a point (0, 0, a), charge +q is at a point (0, 0, 3a).

What is the induced charge on plate 1 and 2?

2. Relevant equations

$$Q=\int_{0}^{\infty }E\cdot \varepsilon _{0}2\pi rdr$$

3. The attempt at a solution

My first step was to discover how the charges were layed out. I think they are in a row of alternating signs at distance 2a one from another along the z axis.

After that my first method was to say that the induced charge is equal to sum of all image charges of the respective plate:

$$Q_{induced} =\sum_{n=0}^{\infty }(-1)^{n}Q=Q/2$$

My argument is the following:

$$Q_{induced} = +Q-Q+Q-Q+Q-\cdot \cdot \cdot$$(1)
$$Q_{induced} +Q=+Q+Q-Q+Q-Q+Q-\cdot \cdot \cdot$$ (2)

We have two possible outcomes, the sum is either 1 or 0, but if the last term is one,
I can argue that it is infinitely far away, thus the charge it induces is cca 0.

If this is so then we can add equations (1) and (2) which will yield the following result:

$$2Q_{induced}+Q=2Q$$

$$Q_{induced}=Q/2$$

This was the first method, second method was to consider a point at distance r from the
z axis at the height of 4a (This is where the plate is, I'm looking for a functional dependence of E field so that I can find charge per area).

After a bit of algebra I find this sum:

$$E=\sum_{n=0}^{\infty }\frac{(-1)^{n}2kQ(2n+1)a}{((2n+1)^{2}a^{2}+r^{2})^{3/2}}$$

After I plug it in this equation:

$$Q=\int_{0}^{\infty }E\cdot \varepsilon _{0}2\pi rdr$$

I find the following result:

$$Q_{induced}=Qa\sum_{n=0}^{\infty }(-1)^{n}(2n+1)\int_{0}^{\infty }\frac{rdr}{((2n+1)^{2}a^{2}+r^{2})^{3/2}}$$

After integrating I find this result:
$$Q_{induced}=\sum_{n=0}^{\infty }(-1)^{n}Q$$

This is very weird, the problem is that I'm new to physics and I don't have much mathematics under my belt to really confirm that this really works, but I know that induced charge on this plate must be between 0 and 1 and that electrostatics ( the name implies) is static i.e. nature doesn't do it two ways, so I came here for help; Thanks.

Last edited: May 24, 2016
2. May 24, 2016

In your alternating series, I would make it into a geometric series $Q=Q-\alpha Q +\alpha^2 Q-\alpha^3 Q+...$ where $\alpha$ is just slightly less than 1. Then $S=a/(1-r)$ and $r=-1$ (very nearly). The series then converges to $S=Q/2$ rather than giving an indeterminate answer. The $\alpha$ just slightly less than unity should be quite justifiable in any real physical system. (The induced image will never be quite 100% because of finite conductivity, etc., as well as any edge effects from a finite plate size, etc.)

Last edited: May 24, 2016
3. May 24, 2016

### VMP

Thank you!

4. May 25, 2016

Just as a little additional info, I have seen a similar alternating series appear in a couple of other places in physics and mathematics calculations. One case I believe was in optics/diffraction theory where the diffraction from the entire plane is computed using the contributions from adjacent Fresnel zones (that are equal and opposite). If I recall correctly, the introduction of any kind of obliquity factor results in the series summing to half of the first zone result. A second series that is similar is the integral of $\sin{x}$ from 0 to infinity. Introducing a $exp(-\gamma x)$ factor results in a convergence of the integral.

5. May 25, 2016

Just a follow-on comment to the $\sin{x}$ integral mentioned above. (I think you correctly solved the physics in the problem of the images. The evaluation of the alternating series and its convergence is also of much interest here.) I took a closer look at $I=\int \limits_{0}^{\infty}\ e^{-\gamma x} \sin{x} \,dx$. This integral can readily be evaluated using the Euler formula for $\sin{x}$ and the result is $I=1$ in the limit that $\gamma=0$. This is somewhat expected since $\int \limits_{0}^{\pi}\ \sin{x} \,dx= 2$. (It is consistent with our previous discrete case where we used an $\alpha$ factor and found S=Q/2 for the alternating series). Alternatively, in a more algebraic solution, the properties of the exponential along with the periodicity of $\sin{x}$ can be used to rewrite the integral $I$ as the sum of terms $n=1$ to $n=+ \infty$ of $(-1)^{n-1} (e^{-\gamma \pi})^{n-1} \int \limits_{0}^{\pi} e^{-\gamma x} \sin{x} \,dx$. This shows that using the $e^{- \gamma x}$ factor makes the integral of $\sin{x}$ precisely into a geometric series with $r=-e^{-\gamma \pi}$. As $\gamma$ approaches 0, $r=-1$. Once again, as $\gamma$ approaches zero, the result is $I=1$ (consistent with the simpler evaluation of the integral). This solution requires a little more algebra, but it shows that the $e^{-\gamma x}$ factor is equivalent to the discrete case where a factor of $\alpha$ (with $\alpha$ just slightly less than unity) was used to get the alternating series to converge. In both cases, the alternating series is converted into a geometric series that is just inside the circle of convergence. This is a little bit of extra detail, but you might find it of interest.

Last edited: May 25, 2016
6. May 27, 2016

### VMP

Thanks a lot for the feedback. Modeling problems with math can lead to obscure places. Adding the fudge factor was a great idea, it completed the idea (as distance tends to infinity) the induced charge is less and less. Math is made for math and sometimes it seems you've got to adjust it.

7. May 27, 2016