# Introduction to Electrostatics -- Positive charges at the corners of a triangle

1. Mar 7, 2015

### molakko

1. The problem statement, all variables and given/known data
Three positive particles of charges 11 μC are located at the corners of an equilateral triangle of side 15.0 cm. Calculate the magnitude and direction of the net force on each particle.

2. Relevant equations
Coulomb's Law

3. The attempt at a solution

2. Mar 7, 2015

### Staff: Mentor

Hi molakko, Welcome to Physics Forums.

You must make some attempt to show what you've tried or at least what you know about the problem before any help can be given. Pick one of the charges and describe the forces acting on it.

3. Mar 8, 2015

### molakko

For every charge 2 forces are acting. They are equal and they are same for every charge, becuase charges are the corners of equilateral triangle and charges have the same value, so we have to calculate magnitude and direction of the net force only for one particle to solve this task.

4. Mar 8, 2015

### BvU

Correct. So what is it that stops you from doing that ? You have found one symmetry that reduces the work by a factor of three.
Pick one of the three particles and calculate the net force on it !

5. Mar 8, 2015

### molakko

I really don't know how to solve that... just help me, please. I need this solution.

6. Mar 8, 2015

### BvU

Coulombs law gives magnitude and direction for the force of particle 1 on 3 and for the force of particle 2 on 3. Add the two vectors and presto !

7. Mar 9, 2015

### molakko

Is that true that magnitude of the force for two particles is equal to 48.4 N, hence magnitude of net force for each particle is equal to 83.831 N? (Assuming that Coulombs const is equal to 9*109)

Last edited: Mar 9, 2015
8. Mar 9, 2015

### Staff: Mentor

No, you need to add the force vectors. Do you know how to do that? Can you post a sketch of the problem showing the force vectors?

9. Mar 9, 2015

### BvU

No -- at least, I get something else. How did you calculate that ?
Small glitch from our good spirit.

molakko: yes, well done. ${q^2\over 4\pi\epsilon_0r^2} =48.4\ N\ \$ and $\ 2\;48.4\;{\tfrac 1 2}\sqrt 3 = 84 \ N$

10. Mar 9, 2015

### Staff: Mentor

Oops, thanks BvU!