# Using the image charges method to find the electric field

The following is a question from a tutorial in my Physics 2 course about conductors and the Method of image charges.

We are given two infinite perpendicular and grounded plains.

The first plain is in the $X-Z$ plain and the second one is on the $Z-Y$ plain.

A point charge $q$ is set at the point $(a,b)$, where $a,b>0$.

1. Find image charges to the problem

2. Find the electric field

This question has a solution in the tutorial, the answer to the first question is given as the following image which also explains the setting of the problem:

The answer to the second question is divided into two parts, for the region $x,y>0$ and elsewhere.

The solution claims that in the region $x,y>0$ the electric field is that of the four charges seen in the image, and I agree (this follows from the uniqueness theorem).

However, I don't understand the last part of the solution, the solution says that the electric field in those regions $x<0$ or $y<0$ is $0$.

Why is the electric field $0$ there ?

I know that since we put charges there we can not use the method of image charges since we actually changed the charge density in the place we are trying to calculate the electric field at, so I don't have a method for tackling this.

Since the answer doesn't justify this claim, and since the answer is $0$, I figured there must be a simple explanation, but I am not able to think of one.

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Let me ask you a question, do you think the negative semi infinite half of these planes matter(i.e. $x<0$ in the $X-Z$ plane)? –  Ali Aug 21 '13 at 3:48

Well, the other two regions are shielded from the given charge by perfect conductors. These shield the electric field and it is zero beyond them.

A bit more rigorously, say you're in the second quadrant ($x<0$, $y>0$). Of course, the image charge at $(-a,b)$ is not present, but it doesn't mean its effect isn't. The image charge is actually a nice representation for the surface charge induced by your original charge at $(a,b)$ on the vertical conducting plane. This surface charge looks like (is indistinguishable from) a point charge a length $a$ behind the conducting plane, so in the first quadrant it looks like a negative charge at $(-a,b)$, but on the second quadrant it sits on top of, and cancels, the original charge.

A similar thing happens (must happen) with the image charges in the third and fourth quadrant. The one at $(a,-b)$ is a representation of a surface charge centered around $(a,0)$, and therefore looks like the same image charge at $(a,-b)$ as seen from the second quadrant. This is cancelled in the second quadrant by a surface charge around the origin, on the positive $x$ and $y$ axes, which is represented (from the first quadrant) by the third-quadrant image charge. Since the total field must cancel, this third induced charge must look like (from the second quadrant) as a positive charge sitting at $(a,-b)$, though I don't know of a simple explanation for this.

Let me expand on this a bit to explain how one finds these surface charges. Consider a positive charge sitting a distance $d>0$ above a single conducting plane:

For convenience set the charge at $x=y=0$. Then you know the electrostatic potential looks like $$\phi(x,y,z)=\left\{ \begin{array} \quad \\ \frac{q}{\sqrt{x^2+y^2+(z-d)^2}}-\frac{q}{\sqrt{x^2+y^2+(z+d)^2}}\quad\text{ if }z>0, \\ 0\quad\text{ if }z<0. \end{array} \right.$$ The potential is zero and continuous at the boundary, but its gradient, the electric field, is not. This discontinuity in the electric field is due to the negative surface charge induced on the conductor, and that surface charge can be found from the discontinuity by using Gauss's Law in its boundary form, $$\hat{\mathbf n}\cdot\left({\mathbf E}_\uparrow-{\mathbf E}_\downarrow\right)= \frac{1}{2\epsilon_0}\sigma$$ where $\hat{\mathbf n}$ is the upward unit normal to the surface, ${\mathbf E}_{\uparrow \left(\downarrow\right)}$ is the electric field just above (below) the surface, and $\sigma$ is the surface charge density.

I'll leave the math to you ;).

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