# Method of images; explanation

• Niles

#### Niles

The setup is: Suppose that we have a point charge q held a distance d from an infinite, grounded, conducting plate. Let the plate lie in the xy-plane, and suppose that the point charge is located at coordinates (0, 0, d). What is the scalar potential above the plane?

The solution to this problem is here: http://farside.ph.utexas.edu/teaching/em/lectures/node64.html

In the example they conclude that "the total charge induced on the plate is equal and opposite to the point charge which induces it" - so it is -q.

My question is: The plate is a grounded conductor, so the potential at z=0, and then the electric field, must be zero at z=0. How can the electric field be zero at z=0 when the plate is negatively charged?

I hope you can help me. I've spent most of the time in the bus home worrying about this.

What is the electric field inside a conductor?

It's zero (I also wrote that - z=0 is where the conductor is).

EDIT: In order for the electric field to be zero inside the conductor, a positve charge must reside at the bottom of the plate?

It's zero (I also wrote that - z=0 is where the conductor is).
Correct, there's your answer. Even if a conductor has a net charge, the electric field inside it is always zero.

How can that be? In my book (Griffith's), it says that there is as much plus charge as there is negative charge inside the conductor. But in this example we have a negative. The example doesn't mention positive charges on the plate.

In order for the electric field to be zero inside the conductor, a positve charge must reside at the bottom of the plate?
No. Consider a simplfied analogy: What is the electric field at a point lying halfway along the line joining two electrons?

It's zero.

Sorry, I'll stop editing.

Ok, here's where I stand now. The electric field is zero inside the conductor, because the external electric field induces charges inside the conductor, so they create an electric field in the opposite direction.

The induced charges are what we are calculating. But again, how can only negative charges be induced if the electric field is to be zero inside the conductor?

Ok, here's where I stand now. The electric field is zero inside the conductor, because the external electric field induces charges inside the conductor, so they create an electric field in the opposite direction.
Firstly, there are no free charges inside a conductor, they all lie on the surface.
The induced charges are what we are calculating. But again, how can only negative charges be induced if the electric field is to be zero inside the conductor?
When using the method of images, it is usual to assume that the plate has no thickness, therefore to talk about 'inside' the conductor is non-sensical. However, if you want something a little more of a substantial explanation consider the point I raised in post #6: The electric field and a point between two equal charges is zero. Now extend these negative charges so that you have two infinite lines of negative charges. If you placed a positive or negative charge at any point between these two lines would it move?

No, it would not. This is because the negative charges have an electric field that points into the wires, so the net electric field at the midpoint is zero.

Can we assume the conductor has a thickness? That would help me.
When we calculate the induced charge, is it the charge that is inside the conductor or is it the free charges that reside on the surface of the conductor?

And where do these free charges on the surface come from?

Last edited:
Ok, so our charge q pushes away the protons in the conductor and attracts the electrons. This leaves a net negative charge in the area directly below q, and this is the induced charge we calculate? And this induced charge is also what creates (together with the repelled protons) the electric field that make the total eletric field inside the conductor go to zero?

- assuming that it makes sense to say that our conductor has a thickness in this case.

Last edited:
Okay, I'll try and answer all your questions in one swoop. Since our plane is conducting and grounded, it must be at zero potential. The presence of the point charge above the conducting plane induces charge distribution on the conducting plane, such that the potential of the conducting plane remains zero. This induced charge distribution represents the image charge and is drawn onto the conducting plate from the ground. Note that this induced charge lies on the surface of the conducting plane and there is no net charge inside the conductor, the conduction electrons in the plate itself are not affected. If there was free charge present in the volume of the conductor, then this would mean there is also an electric field inside the conductor and ergo a non-constant potential inside the conductor, thus violating the requirement of an equipotential plane.

Yes, it does.

Thank you very much for solving my mystery!

Yes, it does.

Thank you very much for solving my mystery!
A pleasure Hmm, I actually have some "post-questions".

When the positive charges are removed (because the conductor is grounded), the electric field is still zero inside the conductor? If yes, what is the counterpart of the negative charges then?

I hope it's ok I ask this one more question.

When the positive charges are removed (because the conductor is grounded), the electric field is still zero inside the conductor? If yes, what is the counterpart of the negative charges then?
Erm, what positive charges? The positive charges inside a conductor are the nuclei, which are fixed in a lattice and do not move.

Hmm, in turns out then that I have not understood it completely. Please correct me where I am wrong:

We have the charge q and a conducting plate. The charge q has an electric field E_0 and it induces charges in the conductor - the positve charges in the conductor are the nuclei and the negative are the electrons from the nuclei (these are all inside the conductor). These charges produce another electric field E_1, which makes the total electric field inside the conductor zero.

Since the plate is grounded, negative charges would come from the ground so there will be no more positive charges., so now the only induced charge on the conducting plate is a negative charge.

If the above is correct, then how can the electric field inside the conductor be zero still?

We have the charge q and a conducting plate. The charge q has an electric field E_0 and it induces charges in the conductor - the positive charges in the conductor are the nuclei and the negative are the electrons from the nuclei (these are all inside the conductor). These charges produce another electric field E_1, which makes the total electric field inside the conductor zero.
Not quite, the conductor is initially uncharged: there are no net charges on or in the conductor. The external electric field induces a charge density on the surface of the conductor by drawing additional charges from the ground, it is these additional charges which form the charge distribution producing the image charge on the conducting plate. There is still not net charge inside the conductor.

Last edited:
Ok, so my "story" will go like this (I hope it's OK I write it explicitly, but in this way I am sure that I have understood it):

The charge q has an electric field E_0 and it induces charges in the conductor by drawing negative charges from the ground - this negative charge distribution is on the surface of the conductor.

But also the charge q induces charges inside the conductor (the electrons and nuclei) and these charges produce another electric field E_1, which makes the total electric field inside the conductor zero.

Am I correct now?

Ok, so my "story" will go like this (I hope it's OK I write it explicitly, but in this way I am sure that I have understood it):

The charge q has an electric field E_0 and it induces charges in the conductor by drawing negative charges from the ground - this negative charge distribution is on the surface of the conductor.
Good But also the charge q induces charges inside the conductor (the electrons and nuclei) and these charges produce another electric field E_1, which makes the total electric field inside the conductor zero.
NO! As I've said many times previously, there is no net charge inside the conductor! If there were a net charge inside the conductor, then there would be an electric field, which would require a non-constant potential, which we can't have since the conductor must be equipotential!

So my story just ends after the two first lines? No additional thing happens?

EDIT: Ok, I understand it now. This is also exactly what you wrote in #12. I should have read it better.

Last edited:

It says: "Hence, charge -Q is induced on the inside of the outer sphere and +Q on its outside. This free charge on the outside of the outer sphere is neutralized due to the Earth connection."

This does not add up with what you have told me. According to you, the negative charges are drawn from the ground, and not in the interior of the conductor?

Last edited by a moderator:

It says: "Hence, charge -Q is induced on the inside of the outer sphere and +Q on its outside. This free charge on the outside of the outer sphere is neutralized due to the Earth connection."

This does not add up with what you have told me. According to you, the negative charges are drawn from the ground?
Okay, so how does this initially neutral sphere acquire a net negative charge? Where does this negative charge come from?

Last edited by a moderator:
Free charges inside the conductor, I would say.

And the reason for my answer is: If I have a negative charge and I take it near a spherical conductor, and I touch the opposite side of the conductor (the side farthest away from pointcharge) with a grounded stick, the spherical conductor will be positively charged. Hence, the charges must come from the interier.

Free charges inside the conductor, I would say.
But there are no net free charges inside the conductor! In other words, the sphere is initially neutral, that is there are equal amounts of positive and negative charge. So where did this negative charge come from?

In that case they come from the ground.

But David J. Griffiths writes in his book (Electrodynamics, page 97 third edition): "Initially, this will drive any free positive charges to the right, and negative ones to the left."

In that case they come from the ground.
Correct, as I said earlier.
But David J. Griffiths writes in his book (Electrodynamics, page 97 third edition): "Initially, this will drive any free positive charges to the right, and negative ones to the left."
Let us clear something up once and for all: For metals there are no free positive charges, they are fixed in a lattice and therefore cannot move. The conduction electrons of each atom constitute the free negative charges, which can move. Yes, the application of an external electric field would cause the conduction electrons to move, however since our conductor is grounded additional negative charge (electrons) is drawn onto the conductor to ensure that there is still, no net charge inside the conductor. Any net charge will lie on the surface.

Ok, I get it now.

You've spent a great amount of time helping, and I don't know how to thank you. I really appreciate it.

You are indeed a very patient man.

Ok, I get it now.

You've spent a great amount of time helping, and I don't know how to thank you. I really appreciate it.

You are indeed a very patient man.
No problem, I'm here because I enjoy talking about Mathematics & Physics and it's a pleasure to explain concepts to someone who wants to learn and is always asking more questions.

I would much rather explain something to someone who is going to think about it and then question why, rather than just take my word for it and then remember it for the future. The latter people are never going to understand why something happens, instead they're just going to know that it does happen.

Anyway, it was a pleasure Ok, I have one last question - and this truly is the last question! Let's look at the following setup: A point charge Q is placed at the center of a spherical, hollow conductor with inner radius a and outer radius b. Since the conductor has no net charge, a charge of -Q will spread uniformly over the inner radius and a charge Q on the outer radius.

These are the free charges inside (at least the electrons), and it makes good sense that they divide up like that since there are no free net charges inside the conductor. But my questions is - will these charges lie outside on the surface of the conductor?

------

Now let's take another setup: This is like the former one, but now a charge Q' is placed on the surface of the conductor. Then a charge -Q must still reside at the inner radius and a charge of Q'+Q must reside on the outer radius. Sinse there are no net free charges in the conductor, all these charges lie on the surface of it as well?

Let's look at the following setup: A point charge Q is placed at the center of a spherical, hollow conductor with inner radius a and outer radius b. Since the conductor has no net charge, a charge of -Q will spread uniformly over the inner radius and a charge Q on the outer radius.

These are the free charges inside (at least the electrons), and it makes good sense that they divide up like that since there are no free net charges inside the conductor. But my questions is - will these charges lie outside on the surface of the conductor?
Yes, electrons will move away from the outer surface and towards the inner surface, leaving a deficit of negative charge on the outer surface and an excess of negative charge on the inner surface. As you say, this charge will be uniformally disributed over the inner and outer surfaces of the sphere, but there will still be no net charge insider the sphere.
Now let's take another setup: This is like the former one, but now a charge Q' is placed on the surface of the conductor. Then a charge -Q must still reside at the inner radius and a charge of Q'+Q must reside on the outer radius. Sinse there are no net free charges in the conductor, all these charges lie on the surface of it as well?
Correct again.

Okay, so how does this initially neutral sphere acquire a net negative charge? Where does this negative charge come from?

Hi Hootenanny

I'm still thinking about this. Say we have a conducting sphere in an electric field, and all the electrons pile up on the right side and there's a deficit of negative charges on the left side. We place a grounded metal-stick on the right side.

Now, why is it that the electrons cannot go into the ground? I know there's no net charge inside or outside the conductor, but what is the argument that tells me that the electrons cannot go to the ground even if the electric field is very strong?

Hi Hootenanny

I'm still thinking about this. Say we have a conducting sphere in an electric field, and all the electrons pile up on the right side and there's a deficit of negative charges on the left side. We place a grounded metal-stick on the right side.

Now, why is it that the electrons cannot go into the ground? I know there's no net charge inside or outside the conductor, but what is the argument that tells me that the electrons cannot go to the ground even if the electric field is very strong?
What makes you think that they won't go to ground?

If indeed a point charge causes point charge displacement in a grounded plate, energy must be expended displacing the charges. Voltage's energy is stored in capacitors, therefore the grounded plate and point charge form a capacitor.

The solitary point charge has a field that influences every point charge in the plate. What is weird is that the number of charges on each plate of a capacitor is equal. Therefore a single point charge in the plate is displaced; all other charges are not displaced.

What makes you think that they won't go to ground?
I don't know why, actually.. it struck me after I had posted. It's only if we place the grounded metal-stick on the side with the deficit of negative charges that additional negative charges are drawn from ground, since the atoms are fixed in the lattice.