# Coulombic Force on Q from two particles

1. Feb 10, 2012

### nateja

1. The problem statement, all variables and given/known data
Point charge 3.5microC is located at x = 0m, y = 0.30m , point charge -3.5microC is located at x= 0m y= -0.30m . What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q= 4.5microC at x= 0.40m , y= 0m?

2. Relevant equations
These are the equations I've been using:

Coulomb's Law
F = k*(absolute_value(q1*q2)/r^2)

Law of Cosines
c^2 = a^2+b^2-2abcos(∅)

Magnitude of a Force
F = sqrt(Fy^2+Fx^2)

3. The attempt at a solution
1) I drew out the situation on a standard xy-plot. I then made a force diagram with Q as the origin and q2 .5m (used the pythagorean theorem to calculate this) from Q on the x-axis and q1 ∅ degrees from the x-axis and .5m from Q. I used the law of cosines to calculate ∅ (I got 73.7398 degrees)

2) set the sum of forces to:
= (1/4*pi*ε0)*(absolutevalue(q2*Q)/r^2)*cos∅ + (1/4*pi*ε0)*(absolutevalue(q2*Q)/r^2)*sin∅ + (1/4*pi*ε0)*(absolutevalue(q1*Q)/r^2)

3)I then plugged in the values to the equations (I converted microC to C and had ∅)
and got Fx = .724792N and Fy = .543594N

4) I used the magnitude of a force equation and received the value .90599N

Any help would be appreciated!

2. Feb 11, 2012

### BruceW

It should be divided by (4*pi*ε0), not multiplied by (1/4*pi*ε0). Also, the horizontal component from charge 1 is going opposite to the horizontal component of charge 2, so one of the horizontal terms in the equation should have a negative sign in front of it.

EDIT: Even more importantly, you need to remember that you sum the horizontal components, to get the horizontal component of the force, and the vertical component you have already. But when you want to get the magnitude, you don't just add the horizontal and vertical components together. This is a vector we are talking about.

3. Feb 11, 2012

### nateja

Thanks Bruce!

It was a problem with my vectors additions and I drew the force diagram completely wrong. Thank you also for pointing out my error in the notation I put up for the coulomb's law equation. I'll pay more attention to that next time.

4. Feb 11, 2012

### nateja

So one last question about the 2nd part of the question

I calculated the angle between the 2 forces as:

∅= arctan(Fy/Fx) = -53.1301 degrees.

I used the same values for Fy and Fx that I used in part a and I got the answer correct for that. The question asked for the degrees clockwise from the positive x-direction. I figured the Fnet would be in the 4th quadrant (-x direction and +y direction) w/ -53.1301 degrees as the angle from the negative x-axis. So I added 180 degrees to 53 and I got the wrong answer.

the answer is 90 degrees, how??

5. Feb 12, 2012

### BruceW

I'm not sure where you got those angles from. But if you draw a diagram using the axes that were defined in the question, and look at the magnitudes of the distances and charges, then you can see that the answer must be 90 degrees.