Point charge equilibrium w/ line charge problem

[oddlogic]

Homework Statement

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.

Homework Equations

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

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,

Brad

 

[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.

 

[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,

Brad

 

[qtm912]

Superposition principle is the correct approach as you point out and q1 is easier. For the integral, the expression in terms of r is correct (ignoring some constants) but not r itself which should be:

y^2 + 3^2

Check that you agree by relooking at the geometry (the coordinates are in terms of (x,y) )

 

[paranormal]

Hello Brad,

Thank you for sharing your attempt at a solution for this problem. Your approach of treating the line charge as a point charge is a good start, but there are a few things that need to be corrected.

First, the total charge for Q2 should actually be +1.8nC, not -1.8nC. This is because the line charge has a positive charge density (since it is a vertical line of positive charges) and the force acting on Q3 should be repulsive, not attractive. So the equation should be F13 + F23 = 0.

Secondly, in order to find the charge density (lambda) of the line charge, you need to take the total charge (+1.8nC) and divide it by the length of the line (2m), not multiply it by the length. So lambda = +0.9nC/m.

For part b, you are on the right track with integrating for the voltage. However, the correct equation is (delta)V = Ke(integral) dq/r, where r is the distance between the point of interest (Q3) and the infinitesimal charge element (dq). In this case, r = sqrt((x-2)^2 + y^2). So the integral should be from -1m to 1m for the y component, and from -1m to 5m for the x component. The final answer should be in units of volts.

For part c, the voltage would be zero at any point along the y-axis (x=0), since there is no x-component in the equation for r. The voltage would also be zero at any point along the line charge (x=2m), since the distance between the point of interest and the line charge would be zero.

I hope this helps clarify the problem and guide you towards the correct solution. Keep up the good work!

Best,