The placing of charges in electrostatics

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Homework Statement


Two point charges, one with charge +q and the other with charge +9q, are placed a distance "d" apart. It is possible to place a third charge so that the net electrostatic force on all three charges is zero. What are the sign, magnitude, and position of this third charge (in terms of q and d)?


Homework Equations


F=(k(q_1)(q_2))/(d^2)


The Attempt at a Solution


Since both of these charges are positive, I believe that the third charge needs to be negative, to "hold" the first two in place. If it were positive than at least one would be pushed away and not held in place.

In regards to magnitude, the Coulomb equation would yield (9kq^2)/(d^2). Since this is obviously depended on distance as an inverse square law, I think that this charge will be placed in the middle of the two positive charges, closer to the smaller one. The problem is I don't know how to determine the exact position of the charge. I'm thinking that if it is placed at say (d/4), then the magnitude would be (9kq^2)/(16d^2).

My problem with this question is twofold. First, I don't know how to find the actual distance, which prevents me from finding the magnitude. Secondly, I have thought about this problem at length and mostly every answer I've gotten so far is from thinking. I don't really have any math to back it up. Do you set the two Coulombs equations equal to each other? How do you go about solving this? Many thanks!
 

Answers and Replies

  • #2
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Hello IndiaNut92,

Your intuition is correct - that the third charge has got to be negative and that it has got to be placed closer to the smaller charge.

Place the third charge at some distance, d, from the one of the charges.
Write down the coulombs equations on any of the three charges, and equate it to zero.
Solve the quadratic equation in terms of d.
From the two values you get, use the second of your intuitions to pinpoint which of two d values would be correct?
You could try solving the coulomb''s equation for any other charge, and this value of d would be common to all of them.
 
  • #3
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Graphene, thanks for the response. Setting the third charge with the larger charge yields (9kqn)/(d^2) where n is the new charge and d is the arbitrary distance. However setting this to zero yields d=0, as you would multiply by zero.

My questions are 1) How do I solve this with two variables (n and d)? And also, 2) What am I doing wrong that is causing me to get d=0? Should I also put the Coulomb law of the pair of original charges equal to zero? This would also yield d=0 though. Thanks for the help!
 
  • #4
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After working on this farther by setting the Coulombs law of the original two charges equal to that of the new charge and the larger one (assuming they both equal 0), I got that r=d*sqrt(n/q), where r is the distance the particle should be placed, d is the original distance between the two particles, n is the charge of the new particle, and q is the charge of the old particle. Does this make sense or did I mess this up? Thanks so much for helping me.
 
  • #5
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pal, what i told you previously wasnt quite the ideal way to solve it. sry.

First, get the force on the third charge to be zero. From this equation, you can find out r. The equation does not involve n.

Next consider one of the outer charges. Put in the value of r, and you will have an equation for n. Solve it.

:)
 

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