# Homework Help: Point Charges

1. Feb 7, 2010

### NewbAtPhysics

1. The problem statement, all variables and given/known data
q1 = 4 mC at the origin
q2 = 2 mC at x=1.0m y=0.0m
q3 = 1 mC at x=1.0m y= 1.0m
q4 = 3 mC at x=0.0m y= 1.0m

What is the net force on q2? Give magnitude and direction of the force.

2. Relevant equations
E = kq/r^2
E = F/q

3. The attempt at a solution

Ok after plotting the locations of all 4 charges on a graph, you see that it makes up a square. So to find each electric field, i plugged in k, which is 9*10^9, r, which for q1 would be 1 because that's the distance charge q1 is from q2. By doing this for all 4 charges, i get the magnitude of each electric field.

The problem is, I'm not sure what to input for q2 as the distance is 0 (r=0) and you can't divide by 0. I'm also sure unsure whether or not charge q2's electric field has any effect on the other charges.

What I planned to do to solve the problem was to find each electric field and plug it into E = F/q to find the force and add them up to find net force, but I'm stuck on the steps mentioned.

Any help would be appreciated, thanks.

2. Feb 7, 2010

### fluidistic

Welcome to PF!
I think that E is generally defined once F is defined. So instead of calculating E for each charge, simply calculate F for each charge. (ok, that's more or less what you mention in your last sentence).
A big tip: What you want is the net force applied on q2. That is, the sum of the force exerted by q1, q3 and q4 (not q2! Hence you don't have to "divide by zero").

3. Feb 7, 2010

### NewbAtPhysics

That's kinda what I was thinking, but I assumed you needed the force of q2 for some reason, since they specifically tell you what the charge is of q2.

Thanks for help though

4. Feb 7, 2010

### fluidistic

The charge q2 is indeed important.
In Coulomb's force between 2 charges (say q1 and q2), you have $$F=\frac{kq1q2}{r^2}$$.
So q2 will appear in your answer.

5. Feb 7, 2010

### cepheid

Staff Emeritus
As it turns out, a charge is not influenced by its own electric field, only those of others. Thankfully so! Otherwise, electrodynamics would be a mathematically much more complicated theory than it is (non-linear)

Of course it does. Every charge will accelerate. Therefore, if you were being asked to figure out what happens to the system as a function of time, it would be quite complicated. Fortunately, you are only being asked to figure out the force on q2 at this instant, at which the charges have the locations stated.

They give you q2 because once you find the net electric field at q2's position, you need to multiply it by q2 in order to calculate the force!

6. Feb 7, 2010

### NewbAtPhysics

Ohhhh, ok. I stupidly applied the formula for E and instead of multiplying it with charge q2 to find force, I multiplied it to charge q1 once again.

Thanks for the help, appreciate it.