The placing of charges in electrostatics

In summary: Hello,Your intuition was correct in that the third charge needs to be negative and that it has got to be placed closer to the smaller charge. The third charge can be placed at any distance, d, from the two charges.
  • #1
IndiaNut92
14
0

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!
 
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  • #2
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
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
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
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.

:)
 

1. How do charges affect each other in electrostatics?

In electrostatics, charges interact with each other through the electric force. Like charges repel each other, while opposite charges attract each other. This force is described by Coulomb's Law, which states that the force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

2. Can charges be placed at any distance from each other in electrostatics?

No, charges must be placed at a certain distance from each other in order for electrostatic interactions to occur. If the distance between charges is too large, the force between them becomes negligible. On the other hand, if the distance is too small, the force between them becomes too strong and may result in an electric discharge.

3. What is the effect of adding more charges in electrostatics?

Adding more charges to a system can affect the strength and direction of the electric force between the charges. Like charges will repel each other more strongly, while opposite charges will attract each other more strongly. Additionally, adding more charges can also increase the overall electric field in the system.

4. How does the placement of charges affect the electric potential in electrostatics?

The placement of charges affects the electric potential in electrostatics by creating areas of higher or lower potential energy. Charges placed closer together will have a higher electric potential between them compared to charges placed further apart. Additionally, the overall electric potential in a system can be affected by the placement and distribution of charges.

5. Can charges be placed in a vacuum in electrostatics?

Yes, charges can be placed in a vacuum in electrostatics. In fact, most electrostatic experiments are performed in a vacuum to prevent interference from air molecules. The principles of electrostatics still apply in a vacuum, as charges can still interact with each other through the electric force.

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