# Calculating the net force exerted on an electron

1. Apr 4, 2014

### need_aca_help

1. The problem statement, all variables and given/known data
Given two particles with Q = 2.30 µC charges as shown in the figure below and a particle with charge q = 1.38X10-18 C at the origin, answer the following. (Note: Assume a reference level of potential V = 0 at r = ∞.)

(a) What is the net force exerted by the two 2.30 µC charges on the test charge q?
(b) What is the electric field at the origin due to the two 2.30 µC particles?
(c) What is the electrical potential at the origin due to the two 2.30 µC particles?

2. Relevant equations

F = (keQQ) / r2
E = kQ / r2

3. The attempt at a solution

F = (9X109)(2.3X10-6)(1.38X10-18) / 0.82
F = 4.4634X10-14

4.46X10-14 N? No way... it can't be that small...

What am I doing wrong...?

Last edited: Apr 4, 2014
2. Apr 4, 2014

### BOYLANATOR

We can't see the figure. You seem to have only found the force on q from one of the Q particles rather than the two of them.

Try using F = ma on with your answer to see what kind of acceleration the electron would have.

3. Apr 4, 2014

### need_aca_help

The mass of the electron hasn't been provided...

Also I've added the image from the question

4. Apr 4, 2014

### BOYLANATOR

It is freely available on the internet and has a value of around 9.11 x 10-31 kg.

I should say that the test charge q is not an electron in this case so my last comment probably wasn't the clearest.

Nonetheless you can see that a 10-14N force on a 10-31kg particle produces a huge acceleration. So the electric force is actually very strong.

5. Apr 4, 2014

### need_aca_help

Unfortunately 4.46X10^-14 is not the right answer...

6. Apr 4, 2014

### BOYLANATOR

OK now I can see the picture. As I said before you have only worked out the force on q from one of the charges Q. What about the other?

7. Apr 4, 2014

### need_aca_help

I'm not sure how to do it...

Just based on the diagram I'm assuming the charge on the center will not move since the forces are balanced.

8. Apr 4, 2014

### BOYLANATOR

Correct. You could say:

Fq = FQ1 + FQ2

So you could calculate each of the forces separately and then add then together. Hopefully the result should be 0.

9. Apr 5, 2014

### need_aca_help

Is the electric field strength and the electric potential also going to be zero?

10. Apr 5, 2014

### BOYLANATOR

Well, is it?

11. Apr 5, 2014

### need_aca_help

Using E = F / r2 I get zero when I add them up since for the distance I have to use negative for one of the charge.

For electrical potential V = kq / r, I also get zero if I add them of since one of the distance is negative.

12. Apr 5, 2014

### BOYLANATOR

One of those equations should have a direction and one shouldn't.

13. Apr 5, 2014

### need_aca_help

My guess is that electric field strength has a direction?

14. Apr 5, 2014

### BOYLANATOR

Yes. It is really described by a vector equation and the potential is a scalar.

15. Apr 5, 2014

### need_aca_help

Yes electric field has a direction and a magnitude... But its not helping me understand what is going on...

Could you please explain it little further?

16. Apr 5, 2014

### BOYLANATOR

Sure. The electric field caused by Q1 is going in one direction and the electric field from Q2 is going in the other. At the mid-point they cancel and the electric field at q is 0. The potential at point q caused by Q1 is just a number. The potential from Q2 will be the same number (note the r on the bottom of the equation is a magnitude, it doesn't change sign). So the potentials add rather than cancel.

17. Apr 5, 2014

### need_aca_help

VT = 2(kQ / r) = 2(20700 / 0.8) = 2(25875) = 51750

51750 V

Is it right?

Edit: Nope its wrong...

Last edited: Apr 5, 2014