# Homework Help: Finding net electric force

1. Jan 29, 2017

### Physicsnoob90

1. The problem statement, all variables and given/known data
Given that q = 20 µC and d = 11 cm, find the direction and magnitude of the net electric force exerted on the point charge q1 below.

2. Relevant equations
F(elect) = k (q1,q2)/d^2

3. The attempt at a solution

1. after converting each unit to its appropriate form,
F(net 1x) = +(k(q(1)[(2.0)q(2)])/d^2) + -(k(q(1)[(3.0)q(3)])/2d^2)

= +[(9.0e^9 N m^2/C^2)(20e^-6 C)(4e^-5 C)/0.0121 m^2] - [(9.0e^9 N m^2/C^2)(20e^-6)(6e^-5)/0.0484 m^2)

2. F(net1x) = 595.041 - 446.281 = 148.8 N towards q2 (it came back wrong though)

2. Jan 29, 2017

### TSny

Check to see if you used the correct charges for your second force.

3. Jan 29, 2017

### TSny

Also, I'm not sure which direction you are taking to be the positive direction.

4. Jan 29, 2017

### Physicsnoob90

i'm using +q(1,2) going right and -q(1,3) going left. I still get -446.281 N in the second force

5. Jan 29, 2017

### TSny

Did you use the correct value of q1 in the second force?

(I see that you are using "toward the right" as the positive direction. Good.)

6. Jan 29, 2017

### TSny

It looks like you are plugging the correct numbers into your second force, but I get half the value that you get for the second force.

7. Jan 29, 2017

### Physicsnoob90

Did you use the same number/equation as well?

F(net(q1,3))= k(q1 *(3.0)(q3))/2(d)^2

= - ( (9.0e^9 N m^2/C^2) * 20e^-6 C * (3.0)(20e^-6 C) )/ 2 (0.11 m)^2 = - 446.3 N

8. Jan 29, 2017

### TSny

9. Jan 29, 2017

### Physicsnoob90

You're right. I got +372 N total as my result (which came back correct!)

Thank You!

10. Jan 29, 2017

### TSny

Great! I now see that when you wrote your equation in the first post, the denominator of the second force should have been written (2d)2 rather than 2d2. But you had the correct numerical value for the denominator.

11. Jan 29, 2017

### Physicsnoob90

i was wondering why my initial numbers were looking correct but still getting the wrong result.