Using Method of Imaging for two point charges between Parallel Plate Capacitors

Click For Summary
The discussion focuses on using the method of image charges to solve a problem involving two point charges between grounded parallel plates. The user finds the image charge configuration complex and questions whether their approach is correct, particularly regarding the need for an infinite number of image charges to maintain zero potential on the plates. They express concern about how to tackle the five specific problems related to energy, force, and charge distribution. Another participant suggests writing down the potential at the position of the +Q charge, indicating that the resulting infinite series can be summed using known techniques. The conversation emphasizes the challenges of applying the method of image charges in this scenario.
Yosty22
Messages
182
Reaction score
4

Homework Statement



Use either image charge method of separation of variables to solve:

The distance between two large, grounded parallel conducting plates is 4x. Between them, two point charges +Q and -Q are inserted and have a distance x and 3x from one of the plates. (A line connecting the two charges is perpendicular with the plates).

(a): How much energy is required to remove and separate the two charges to infinity.
(b): What is the magnitude and direction of the force experienced by each point charge?
(c): What is the magnitude and direction of the force experienced by each plate?
(d): What is the total charge on each plate?
(e): If the negative charge is removed and the positive charge remains unmoved, what is the total charge on each plate?

Homework Equations

The Attempt at a Solution



I am trying to use method of image charges to get to this problem because it seems most intuitive. However, when I try to set up the image charge configuration, it quickly becomes messy. To step you through my thinking:

To cancel out the potential due to -Q located a distance x away from the left plate, you must add an image charge of +Q located a distance x away from the plate (But on the other side of the plate, i.e. 2x to the left of the -Q charge). Next, you need to cancel the potential due to the +Q charge located 3x from the left plate. To do this, I added a charge -Q located 3x away from the left plate (6x away from +Q). This will clearly mean that the left plate is at a potential of 0, which is must be to satisfy the boundary conditions. However, by doing this, the right plate is no longer at a potential of 0, so you have to add more charges to the right of the right-most plate. Once you get the right plate at 0 potential, the left plate is no longer at 0 potential, and so on. That is, to solve this, you need an infinite number of mirrored charges.

Is this thinking correct? If so, how should I go about trying to tackle the 5 problems listed above if I have an infinite number of charges? Surely having an infinite amount of charges won't simplify this problem will it?

Am I going about this wrong? I figure method of image charges must be the easier of the two methods (image charge or separation of variables) because in the region between the plates, the Laplacian of the potential which gives the charge distribution is dependent on a Dirac Delta function since the charges in question are point charges. Surely that would make separation of variables the more difficult method here.

If this really does require an infinite amount of imaged charges, How should I go about solving this?

Thanks in advance.
 
I can't help you with all the answers, but I do think your approach is correct.
Try to write down the potential at the position of e.g. the +Q charge. The resulting infinite series has a well known sum; see here
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
8K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 31 ·
2
Replies
31
Views
2K
  • · Replies 20 ·
Replies
20
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K