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Force induced on a point charge by two conducting planes

  • Thread starter Strawberry
  • Start date
21
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1. Homework Statement

A point charge q is located in the xy plane near two grounded conducting planes intersecting at right angles as shown in the Figure. The z axis lies along the line of intersection of the planes. Find and justify (mathematically show) the force acting on this charge q (both direction and magnitude should be specified).


1. Think about how much charge would be induced on the conducting planes.
2. What would be the potential on the conducting planes (This is the boundary condition) and what can you do to maintain such a condition?
3. If a different method was used for your calculation, please describe the method.

A .doc file containing the question and a diagram is attached.

3. The Attempt at a Solution

I'm assuming a negative charge is induced on the conducting planes, but beyond that I seriously have no clue. Absolutely any help whatsoever would be greatly appreciated.

I found a website with a related problem by google searching "force point charge conducting plane." It's the davidpace one ( I can't link to websites yet). Can I do basically the same thing with each conducting plane individually and use vector addition to find the resultant force?
 

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Answers and Replies

21
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Okay, after looking this over more, I'm thinking that replacing the conducting plates with three point charges would make an equivalent system. Using the plates as axes, there would be mirrored negative charges across the x, y, and y= -x axes. So charges like a square, with one corner being an opposite charge. After that, the force calculation would just be a system of point charges. Is this the right idea?
 
alphysicist
Homework Helper
2,238
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Hi strawberry,

That is the right idea; however I believe you need two corners to be oppositely charged that of the original point charge. Do you see why? Which two corners would you choose?
 
21
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Thank you very much for your response alphysicist. I took your advice and I think I have the right answer.

The equivalent charge system is like a square with opposite corners having the same charge, I think.

(-q) - - - q
|
|
|
q - - - -(-q)


Taking V to be the sum of V for each charge at the origin, (0,a) and (a,0) I found V to be 0 along the original conducting plates. After that I used F = k [(q1*q2)/r^2] r to find the force of each equivalent charge on the initial charge. I ended up with three forces, then added them together.

My final result was F = k* (q^2 / a^2) * [ sqrt(2) - 4 ]/4 * i + [ sqrt(2)-4)/4 ] * j ]
 
alphysicist
Homework Helper
2,238
1
Hi Strawberry,

It looks to me like you are missing a factor of (1/4). Am I just reading your final result wrong?
 
21
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Well I have everything divided by 4, should it be divided by 16?
Is this more clear?
F = k* (q^2 / a^2)/4 * [ sqrt(2) - 4 ] * i + [ sqrt(2)-4) ] * j ]
 
alphysicist
Homework Helper
2,238
1
I'm getting something different from that. For example, for the x direction, I am getting

[tex]
F = \frac{k q^2}{(2a)^2}
[/tex]

for the particle at the upper left corner, and

[tex]
F = \frac{kq^2}{(\sqrt{2}\cdot 2a)^2} \cos(45^{\circ})
[/tex]

for the x component from the particle at the lower left corner. I then subtract those. Is that what you got? I don't think that gives the answer for the x component that you stated. Or am I thinking of it wrongly?
 

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