# I Electric field created by point charges and conducting plane

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1. Apr 2, 2017

### KV71

I came upon this:
http://physics.stackexchange.com/qu...change-if-we-place-a-metal-plat/323006#323006

question on Physics Stackexchange which I found very interesting.

The configuration is basically two positive point charges q and a conducting plane equidistant from both charges. What I found most fascinating in particular, is one answer that claims

"if the plate is a plane that extends to infinity, there will be two image charges -q at the positions of each of the original positive charges so that the electric field everywhere is zero"

My question is, is this true? If so, could someone explain why or perhaps tell me more about this?

2. Apr 2, 2017

Staff Emeritus
If the plate extends to infinity, it can get as many cancelling charges as it needs, "from infinity".

3. Apr 2, 2017

### KV71

So the field everywhere is zero?

4. Apr 2, 2017

Staff Emeritus
I don't think so, because the charge and its image are not in the same place.

5. Apr 2, 2017

### KV71

I think there are two images and the image for the charge below the conductor is sitting at the position of the charge above the conductor and vice versa.

6. Apr 2, 2017

Staff Emeritus
Fields from the other side of the conductor don't penetrate the conductor.

7. Apr 3, 2017

### vanhees71

It's very easy to see, why the field cannot be 0. Just draw a sphere $V$ around one of the postive charges, and use Gauss's Law,
$$\int_{\partial V} \mathrm{d}^2 \vec{F} \cdot \vec{E}=Q.$$
The key to understand this is that the image charges are not really there, but they are used as a mathematical trick to get the total field consisting of the field of the real charge and the influence charges in the plate, i.e., to fulfill the boundary conditions.

For your example you can easily solve the problem indeed by using image charges. You need to treat only one charge first. Say the infinite plane defines the $xy$ plane of a cartesian coordinate system, and let the charge $Q$ sit on the $z$ axis at $(0,0,a)$ with $a>0$.

You have to solve the Laplace equation for the potential
$$\Delta \phi=-Q \delta(x) \delta(y) \delta(z-a)$$
with the boundary condition $\phi=0$ for $z=0$.

Obviously the solution for $z<0$ is $\phi=0$ and for $z>0$ you write down the solution for the field of the true charge $Q$ at $(0,0,a)$ and account for the boundary conditions by substituting the plate by the image charge $-Q$ at $(0,0,-a)$, because then you solve for sure the Laplace equation and the boundary conditions, i.e., you have
$$\phi(\vec{x})=\begin{cases} 0 & \text{for} \quad z<0, \\ \frac{Q}{4 \pi} \left [\frac{1}{\sqrt{x^2+y^2+(z-a)^2}}-\frac{1}{\sqrt{x^2+y^2+(z+a)^2}} \right] &\text{for} \quad z>0. \end{cases}$$
For the other charge, $Q'$ sitting at $(0,0,-b)$ ($b>0$) you have in an analogous way
$$\tilde{\phi}(\vec{x})=\begin{cases} \frac{Q'}{4 \pi} \left [\frac{1}{\sqrt{x^2+y^2+(z+b)^2}}-\frac{1}{\sqrt{x^2+y^2+(z-b)^2}} \right] &\text{for} \quad z<0,\\ 0 & \text{for} \quad z>0. \end{cases}$$
The total field thus is
$$\phi_{\text{tot}}(\vec{x})= \begin{cases} \frac{Q'}{4 \pi} \left [\frac{1}{\sqrt{x^2+y^2+(z+b)^2}}-\frac{1}{\sqrt{x^2+y^2+(z-b)^2}} \right ] &\text{for} \quad z<0,\\ \frac{Q}{4 \pi} \left [\frac{1}{\sqrt{x^2+y^2+(z-a)^2}}-\frac{1}{\sqrt{x^2+y^2+(z+a)^2}} \right] & \text{for} \quad z>0. \end{cases}$$
Now set $Q'=Q$ and $b=a$, and you see that the field is not 0 also in this symmetric case.

That's a great exercise to understand the role of the "mirror charges"!

8. Apr 4, 2017

### KV71

Thanks for the very informative and well-written post @vanhees71 . Indeed, I liked this problem too because it has a solution that does not strike me immediately but is actually obvious when I gave it some more thought. Also , the owner of the answer that I referenced in my intro post deleted their answer (must have realized it is wrong) and has posted a new answer--you may look at it (it uses essentially the same method as yours but talks in terms of the field and superposition). After reading it, I am amazed that iafter an infinite conducting plate is placed between two charges, "the force on the charges changes sign". Very interesting indeed.

Let me know what you think!

9. Apr 4, 2017

### KV71

@Vanadium 50 I didn't understand what you meant earlier by
but now I understand. The link in my last post #8 helped me. Thanks so much! It turns out you were absolutely right!