Homework Help: Point charge equilibrium w/ line charge problem

1. Oct 7, 2011

oddlogic

1. The problem statement, all variables and given/known data
There is a charge located at the origin of magnitude 5nC. There exists a vertical line of charge, 2 meters in length, that runs from (2m,-1m) to (2m,1m).
a) What uniform linear charge density must this line of charge have in order that the net force on a charge placed at (5.0m, 0m) has no net force acting on it?
b) What would be the voltage at this point?
c) aside from x = +/- (infty), where (if at all) would the voltage be zero?

Q1= 5nC. I called the line charge Q2. This could be very wrong. Q3 is the unknown.

2. Relevant equations
F = (Q1Q2)/r^2
(delta)V = Ke(integral) dq/r
dq = (lambda)(dx)

3. The attempt at a solution
I treated the line charge as a point charge by saying:
F13 - F23 = 0 (the force of 1 on three and the force of the line charge acting on 3 must be zero)

This meant that the total charge for Q2 was -1.8nC. To find rho (charge density) I just took 1.8nC and divided by 2, since that is the length of the line of charge.

part b is a little trickier but I think it must be something akin to integrating from -1 to 1 for the integral of (Ke)(dq)/(y), but that doesn't seem quite right since we have Q3 to think about and also because "r" (which in this equation is "y") changes as we go from -1 to 1, so r must be something besides just "y". Obviously, dq is now (lambda)(dy). Ke and lambda are constants, so it should just be dy/r, whatever r is. could r = [(1-y)^2 + 4]?

I have seen (and know how to solve) point charge problems on an X axis that look a lot like this, but the fact that the Q3 is unknown (and my method for solving for Q2 could be wrong) and the y integration is kind of throwing me. Help? (but not too much)

Thanks,

2. Oct 8, 2011

qtm912

First of all it is important to recognise that you have just one unknown and that is q2. Think of q3 as a unit test charge. You do not need its value in this problem.

Also I think you method of calculating q2 is not quite right -- you need to perform an integration and cannot treat the line s a point charge.

Tell me how you get on.

3. Oct 8, 2011

oddlogic

I saw that yesterdayafternoon. So we use the superposition formula and account for all charges, in this case 2, and integrate q2. The integral should look like this?
(Lambda)*(int) dy/r^2, where r is the square root of ([1-y)^2+2^2], yes?

Thanks,