# Homework Help: Electrostatics problem using Coulomb's law

1. May 16, 2015

### spaghetti3451

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

This problem is taken from 'Introduction to Electrodynamics' by David Griffiths.

(a) Twelve equal charges, $q$, are situated at the corners of a regular 12-sided polygon (for instance, one on each numeral of a clock face). What is the net force on a test charge Q at the center?

(b) Suppose one of the 12 $q$'s is removed (the one on "6 o'clock"). What is the force on Q? Explain your reasoning carefully?

(c) Now 13 equal charges, $q$, are placed at the corners of a regular 13-sided polygon. What is the force on a test charge $Q$ at the center?

(d) If one of the 13 $q$'s is removed, what is the force on $Q$? Explain your reasoning.

2. Relevant equations

3. The attempt at a solution

(a) Zero, because each of pair of opposite charges exert forces on $Q$ of equal magnitude in opposite directions.

(b) The force on $Q$ is given by Coulomb's law, and acts in the direction of "6 o'clock."

(c) This is where I've got stuck. I picture one charge $q$ at the 12 o'clock position, and the others spread symmetrically throughout the rim of the clock face. Then, I have a gut feeling that the net force is due only to the charge at the 12 o'clock position. Using the above picture of the position of the charges, the horizontal components of the forces cancel by symmetry. However, I'm unable to account for the cancellation of the vertical forces due to the other 12 charges.

Thoughts?

Last edited: May 16, 2015
2. May 16, 2015

### Orodruin

Staff Emeritus
Did you try applying symmetry arguments?

3. May 16, 2015

### spaghetti3451

Yes. Using symmetry arguments, I was able to cancel out the horizontal components of the forces.

But, how I can cancel out the vertical components of the forces using symmetry arguments eludes me.

4. May 16, 2015

### Orodruin

Staff Emeritus
What happens if you turn the whole contraption by an angle 2pi/13?

5. May 16, 2015

### spaghetti3451

The system's behaviour is invariant under a rotation of $\frac{2 \pi n}{13}$ - I get that.

But how that relates to the problem at hand is a bit perplexing.

6. May 16, 2015

### Orodruin

Staff Emeritus
So if you assume that the resulting force points in any direction, what would happen to it during the 2pi/13 rotation?

7. May 16, 2015

### spaghetti3451

The configurations of the system under a rotation of $\frac{2 \pi n}{13}$ all have a net force of the same magnitude pointing in the same direction.

How might it be possible to determine the direction of that force and its magnitude?

8. May 16, 2015

### Orodruin

Staff Emeritus
Exactly, but what happens to any force on the charge if you rotate the charge distribution responsible for the force by any angle?

9. May 16, 2015

### spaghetti3451

The direction of the net force will change in general, but its magnitude will remain the same, under an arbitrary rotation.

10. May 16, 2015

### Orodruin

Staff Emeritus
Yes, so now you have two pieces of information. The direction of the force will change by an angle 2pi/13, but the force will remain the same because the charge configuration is the same. What is the only way of accommodating both these criteria?

11. May 16, 2015

### spaghetti3451

I see. It's by having the effects due to 12 charges being equal to zero, and the effect of only one charge contributing to the net force.

12. May 16, 2015

### spaghetti3451

Wait! Actually, the net force is zero. That's because the configuration of charges remains invariant under a rotation of $\frac{2 \pi n}{13}$, yet the direction of the net force ought to change by an angle of $\frac{2 \pi n}{13}$. The only way this can happen is for the net force to be zero.

I think this is the correct argument, isn't it?

13. May 16, 2015

### Orodruin

Staff Emeritus
Correct.

Another way to deduce it is to take your argument that the "horizontal" component is zero. You can make the same argument with any other direction which is rotated by a multiple of 2pi/13. You will then know that the projection onto two linearly independent directions is zero, meaning that the force must be zero.

14. May 16, 2015

### spaghetti3451

Wow! That's also an interesting proof! Thanks!