Method of images between parallel plate capacitor

• guyvsdcsniper
You could just as easily have put in 3h/4. If you had put in 3h/4, the distance between the charges would be 3h/4. So the distance between the charges would be the same in both cases.The distance between the charges would be 3h/4. So the distance between the charges would be the same in both cases.f

guyvsdcsniper

Homework Statement
Find the force on p.c. q placed inside a parallel plate capacitor in
vacuum, with grounded plates, separated at distance h. The point charges are at
distance h/4 from the closest plate
Relevant Equations
F=qe
Before I can find the force on q I must balance the charges. This problem starts of with -q and q inside the capacitor. I have added image charges on the opposite side of each plate. Would this work?

Homework Statement:: Find the force on p.c. q placed inside a parallel plate capacitor in
vacuum, with grounded plates, separated at distance h. The point charges are at
distance h/4 from the closest plate
Relevant Equations:: F=qe

Before I can find the force on q I must balance the charges. This problem starts of with -q and q inside the capacitor. I have added image charges on the opposite side of each plate. Would this work?
View attachment 298454
Why is each plate only reacting to the nearest charge?

Homework Statement:: Find the force on p.c. q placed inside a parallel plate capacitor in
vacuum, with grounded plates, separated at distance h. The point charges are at
distance h/4 from the closest plate
Relevant Equations:: F=qe

Before I can find the force on q I must balance the charges. This problem starts of with -q and q inside the capacitor. I have added image charges on the opposite side of each plate. Would this work?
View attachment 298454
I just want to give a heads up that this is my first real problem that isn't as trivial as a p.c. and a single grounded plane so this is really new to me.

I guess I didnt think about the image charges also effecting the plates.

When I consider the effect an image charge has on the opposing plate I came up with the new image charges in green. Although this seems like it would go on forever until eventually the pc are too far away to really have a true effect on the positive q in the capacitor.

I just want to give a heads up that this is my first real problem that isn't as trivial as a p.c. and a single grounded plane so this is really new to me.

I guess I didnt think about the image charges also effecting the plates. View attachment 298459
When I consider the effect an image charge has on the opposing plate I came up with the new image charges in green. Although this seems like it would go on forever until eventually the pc are too far away to really have a true effect on the positive q in the capacitor.
No, you misunderstood my comment. Just consider one plate but both charges to start with. What do you get for that?

No, you misunderstood my comment. Just consider one plate but both charges to start with. What do you get for that?

I believe I should get this as a result

I guess I didnt think about the image charges also effecting the plates.
I think your green distance is incorrect..
And that the framers of the problem were kind to you ?

I believe I should get this as a result
View attachment 298468
Right.
Bear in mind what the point of the method is. In the above diagram, you can throw away the plate, replacing it with the image charges. This will not change anything in the region above where the plate was; all the fields, forces and potentials will be the same. Things will change below the plate line, but we don't care about that.
Now introduce the upper plate. What do you get?

Right.
Bear in mind what the point of the method is. In the above diagram, you can throw away the plate, replacing it with the image charges. This will not change anything in the region above where the plate was; all the fields, forces and potentials will be the same. Things will change below the plate line, but we don't care about that.
Now introduce the upper plate. What do you get?
Should it be this? Kind of like what I originally drew?

I think your green distance is incorrect..
And that the framers of the problem were kind to you ?
My professor is pretty nice lol

For the green distance, how would I think about it? I am thinking the green charges negate the red charges, so it would make sense they would be h/4 away

The green distance is 3h/4 unless I misunderstand. Why h/2.?? Put in the distances on the image in #8 so we understand each other. The distances don't depend upon charge (for a plane mirror image) Then take look like @haruspex recommends

The green distance is 3h/4 unless I misunderstand. Why h/2.?? Put in the distances on the image in #8 so we understand each other. The distances don't depend upon charge (for a plane mirror image) Then take look like @haruspex recommends
Actually there was no reason. for me to label the h/2. I am trying to find the force wrt to the +q inside the capacitor. So I just need to know how far is each charge so I can plug in the right r when I take the sum of all forces.

So the bottom green q should be a distance h and the top green q a distance 3h/2? All adjacent charges should be a distance h/2 away from each other right?

Should it be this? Kind of like what I originally drew?

View attachment 298473
No. The lower plate should no longer be there since it is replaced by the two image charges. That leaves you with one plate and four charges. So what do you need to add to replace the upper plate?

No. The lower plate should no longer be there since it is replaced by the two image charges. That leaves you with one plate and four charges. So what do you need to add to replace the upper plate?
Right the plate shouldn't be there anymore I missed that. Its like you said, we are throwing away the plate and replacing it with image charges.

So to replace the upper plate, I should have this?

Essentially the same thing but for it to be correct Ii would need to get rid of the plates?

Right the plate shouldn't be there anymore I missed that. Its like you said, we are throwing away the plate and replacing it with image charges.

So to replace the upper plate, I should have this? View attachment 298491

Essentially the same thing but for it to be correct Ii would need to get rid of the plates?
No, you are still missing it. After replacing the lower plate with two image charges you have four charges in total. So how many images in the upper plate?

No, you are still missing it. After replacing the lower plate with two image charges you have four charges in total. So how many images in the upper plate?
So I would need 4 image charges to get rid of the upper plate?

So I would need 4 image charges to get rid of the upper plate?
Yes, I believe so.. though suddenly I am worried by the resulting asymmetry.

Yes, I believe so.. though suddenly I am worried by the resulting asymmetry.
So like this? I am considering the +q already in the capacitor a charge belong to the upper as well. Or would I need an additional negative charge all the way up top?

Yes, I believe so.. though suddenly I am worried by the resulting asymmetry.
Ah! Jst seen my error. I have not allowed for the redistribution of charges on the first plate when the second is introduced. Let me think some more...

Edit: @quittingthecult, looks like you do get an infinite sequence of charge images. As in your post #17, but keep going symmetrically above and below.
You should find that the repulsion of like charges cancels in pairs, leaving only the infinite sequence of unlike charge attractions to sum.

Last edited:
guyvsdcsniper
Ah! Jst seen my error. I have not allowed for the redistribution of charges on the first plate when the second is introduced. Let me think some more...

Edit: @quittingthecult, looks like you do get an infinite sequence of charge images. As in your post #17, but keep going symmetrically above and below.
You should find that the repulsion of like charges cancels in pairs, leaving only the infinite sequence of unlike charge attractions to sum.
Could you show me how that would look? My professor nor my textbook have gone over infiniite sequences

Could you show me how that would look? My professor nor my textbook have gone over infiniite sequences
Fill in the gap sizes in post #17. you should see the pattern.

For me the easier way to do this is to do one charge at a time. Draw the ladder of images for just the positive charge. Then shift the ladder by d/2 and subtract (for the negative) Of course you can also write this down algebraically