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Old Sep 5th 2010, 12:19 AM   #1
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Questions Related to Coulomb's Law and Charges Distributions

I got these 3 questions for my homework. I tried it nearly a whole day but...

**A cube of edge a carries a point charge q at each corner. Show that the resultant electric force on any one of the charges is given by
F = 0.262(q^2)/ε(a^2)
Here is my picture of the problem and my work
http://img4.imageshack.us/img4/441/003bgk.jpg

The only thing I'm in trouble with is how can I find F act on 1 'cause of 6 (F16) ? The book gives me something with radical 3 along with i, j, k; but I just don't get how they are there.

**Where can you place a second point charge q (equal to the charge on the rod) so that q' is in equilibrium (ignore gravity)? Solve this problem assuming that (a) q is positive and (b) q is negative.
I tried to get a picture of the problem and some of the forces drawn
http://img691.imageshack.us/img691/5360/002akh.jpg

Since q' has to be in equilibrium, that means sum of the forces must be 0, so I thought that: dF + dFy + (something) = 0. But the problem is I don't know what is that "something" is. Is it gonna be another dF' of the second charge (q), or what? I'm kind of lost.

** Last question, the book gives me an example of Ring of Charge with q' on the z axis. But my question is what if q' is not on z axis? Like maybe a little bit next to the right of the z axis instead? Is there going to be any change in my calculations?
Here is the picture of the book example and my summarized work
http://img231.imageshack.us/img231/7632/001sqj.jpg

Please help me. I'll really appreciate.
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Old Sep 5th 2010, 10:54 PM   #2
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For the first question, note that it is a cube so that the distance can be calculated by the Pythagoras' theorem.
Distance between 1 and 6 is sqrt(a^2+a^2)

For the second question, you should note that the something is the resultant force of dF + dFy but in OPPOSITE direction so that the forces can cancel out each other. Note that for same kinds of charges, repulsion exists and for like charges, attraction exists. This helps you to determine the direction of the charge to be placed.

For the third question, I guess the calculation would be different if it is not placed at the axis of symmetry as some horizontal forces may not be able cancel each other.So no integration can be used directly.
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