# How to determine angle of net force-electric charges of right triangle

1. Dec 8, 2013

### Coco12

1. The problem statement, all variables and given/known data

The .274 and .88 was found using the equation of electrostatic force .

2. Relevant equations

K=q1q2/d^2
3. The attempt at a solution
Would I just tan inverse of .274N and .88N which would be 17 degrees. However my question is: how to determine the reference pt: would it just be 17 degrees or 180-17 degrees?

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2. Dec 8, 2013

### BOYLANATOR

Draw the force vectors tip-to-tail and it should be clearer.

3. Dec 8, 2013

### Coco12

It would be in the first quadrant so would it be 90 minus 17 or??

4. Dec 8, 2013

### BOYLANATOR

It doesn't really matter as long as you're clear. State "x degrees clockwise from BA axis" or "y degrees counter clockwise from BC"

5. Dec 8, 2013

### Coco12

Is this a foolproof way? Look at the signs of the y value and the x value when you and that should tell you in what quadrant?

6. Dec 8, 2013

### BOYLANATOR

I'm not sure what you're asking, if you use the vector form of Coulombs Law and keep the signs in on the charges then you get the magnitude of the resultant force as well as the direction (assuming you have defined an x and y axis). If you haven't been introduced to the vector form of the equation then what you have done is fine.

7. Dec 8, 2013

### Coco12

I mean like since the signs of both the charges are positive and since tan is opposite over adjacent ( .274/ .88) that would mean it is in the first quadrant since both the x and y values are the same

8. Dec 8, 2013

### BOYLANATOR

Yes but remember there are two solutions to tan(positive number) - the first quadrant and the third. The charge of the test particle tells you which one it is.

Last edited: Dec 8, 2013
9. Dec 8, 2013

### Coco12

What do u mean by that?

10. Dec 8, 2013

### BOYLANATOR

I made an edit above. The force on q1 due to q2 is given by Coulombs law and for two like charges the direction of the force on q1 is in the direction of q2 to q1, i.e. repulsive. If you keep everything as vectors the right direction pops out of the equation.

Using the non-directional F = $\frac{q1 q2}{k r^{2}}$ you need to consider the geometry of the set up. No direction will pop-out.