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Old Apr 6th 2009, 09:15 PM   #1
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Need Pointers for Basic EM Theory (questions)

Hey all,

Just started taking EM last week and our first homework assignment gave me a bit of trouble...

1.) The first problem I had issues with utilized a set-up of three spheres suspended from a common point by threads of length 1.17m. The mass of each ball is 13.3 grams and the spheres hang in the shape of an equilateral triangle of side length 15.3 cm. (Find charge of each ball if they are identical.)

I tried using Coulomb's law with two of the spheres where the force is equal to 0 as the balls aren't moving, but I'm not sure that this is the correct way to go about solving this problem (seems too easy...) I got -6.88 x 10^8 C for the charge on each ball.

Anyone know if this is the correct way to solve this problem?

2.) The second problem was even a bit harder fro me to reason out. It involves a set-up of 4 rods of length 25.0 cm with a uniform (1-d) charge distribution. They form the base of a pyramid, the top of which is a small sphere of mass 3.46 x 10^-4 grams and has a charge of +2.45 x 10^-12 C. If it is in equilibrium 21.4 cm above the center of the square formed by the rods, what is the charge on each rod?

I wasn't sure how to do this one at all. The professor showed us an example of the (1-d) uniform charge distribution in class as an upright rod affecting a test charge x units to the right of it. I thought maybe this would apply to this problem four times over as each of the rods would exert a force perpendicular to it's direction towards the test charge...is this correct, and would someone mind illustrating how I might compute this?

Thanks in advance,

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Old Apr 6th 2009, 10:28 PM   #2
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Consider one ball and find the force exerted on it by the two others. Suppose you are facing this ball at eye level and you see the other two behind at equal distances on each side of the ball you are looking at. The repulsive forces F1 and F2 acting on it by each of the others have an angle of 60 deg between them because the balls form the base of an equilateral triangleThe angle between F1 and F2 is 60 deg. F1 = F2 in magnitude.The sum of the components of these forces which is pointing directly at you is given by 2F1 cos 30 since F1 = F2. THis force is balanced by the restoring force on the pendulum which is mgsin@ where @ is the angle made by the string with the vertical line which passes thru the centre of the triangle. Since you have the length of the side of the eq triangle, you can find the distance to its centre from the ball using geometry. Let this be d.
2F1 cos 30 = mg(d/L) where L = 1.17 m . F1 = force of coulomb repulsion between each ball with distance between them = 15.3 cm. Convert everything to proper units like metres etc. This should give you your answer. Note that the ball does not move horizontally as you face it as the sin components F1 sin 30 and F2 sin 30 cancel each other.
I think this problem is from resnick halliday. is it not?
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Old Apr 7th 2009, 03:28 PM   #3
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Alright, thanks for the procedure, I was able to puzzle it out from your directions...and yes, it is from Resnick...


Any thoughts on the second one? I wasn't able to get to my prof's office hours today to ask him about it...
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