The placing of charges in electrostatics

AI Thread Summary
To achieve a net electrostatic force of zero among three charges, a third negative charge must be placed between two positive charges, one with charge +q and the other with +9q. The third charge should be closer to the smaller charge to balance the forces. The position of this charge can be determined by applying Coulomb's law and setting the forces acting on it to zero. The challenge lies in solving for both the magnitude of the third charge and its exact position, which requires equating the forces and possibly solving a quadratic equation. Ultimately, the correct approach involves calculating the forces on the third charge and using those to derive its position and magnitude.
IndiaNut92
Messages
14
Reaction score
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!
 
Physics news on Phys.org
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.
 
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!
 
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.
 
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.

:)
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top