
#1
Feb2510, 02:07 PM

P: 61

1. The problem statement, all variables and given/known data
The electric field on the dashed line in the figure vanishes at infinity, but also at two different points a finite distance from the charges. Identify the regions in which you can find E = 0 at a finite distance from the charges. Check all that apply: A)to the right of point C B)between points A and B C)to the left of point A D) between points B and C See Image: http://tinypic.com/r/15g8o6w/6 2. Relevant equations E=kq/r^2 3. The attempt at a solution I'm afraid that all I know is the formula, but I'm not sure how I'm supposed to apply it. Am I supposed to set E=0 or.. what? I feel like all I need is some direction in how to start and then I'll be able to get it. Thanks! 



#2
Feb2510, 02:19 PM

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P: 3,444

You are given four choices. Pick the first one and draw the electric field contributions from each of the three charges at the point as described. Can these arrows conceivably add up to give you zero?




#3
Feb2610, 10:45 PM

P: 61

I guess what I don't understand is how to incorporate multiple charges into the equation. It just has room for one q. The way you're saying, don't I need to try to find the electric field between two of the charges?




#4
Feb2710, 12:11 AM

P: 15

Electric Field Vanishing at Infinity
I would put A and C for that problem since you can interpret an Efield in terms of vectors.
The direction of the Efield would be that in which a Positive Charge would flow towards... Lets look at it now. A) To the right of Point C Lets Analyze it: If you put a positive charge, it will be repelled by the E field from B and C but attracted by that of A *therefore allowing it to eventually reach static equilibrium once its far enough from A in order for b and c to cancel the Efield of A* B) Between points A and B A will attract it but B and C will push it to the left, therefore it would not reach electrostatic equilibrium in the middle of the two charges (since ALL CHARGES ARE OF EQUAL MAGNITUDE) I wasnt that clear... ask me if you dont understand what I said (I tend to ramble a lot :D) I might be wrong with this explanation though, but I would put that on a test 



#5
Feb2710, 08:06 AM

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