EM (coulombs law)

1. Sep 3, 2005

matpo39

ok i have been thinking about this problem for a lil bit now and i think i have it correct but i would just like to double check:

(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 at 6 o'clock). what is the force on Q?

for (a) i was thinking that would end up cancelling each other out leaving no net force on Q.

for (b) all the charges would cancel one another except for the charge at 12 o'clock because there would be no charge at 6 o'clock the cancel it out so the force on Q would simply be k*q*Q/r^2 where r is the distance from q at 12 o'clock to Q.

i was just wondering if this was the correct way to think about this problem and if im off the mark a bit with it it would be great if someone could help me out a bit..

thanks

2. Sep 3, 2005

StatusX

That's all correct. In general, the way to think about symmetry in physics problems is that the answer must retain the symmetry of the problem. For example, this problem has symmetry in rotation by 30 degrees, but if the force pointed in any direction in the plane, this symmetry would be gone, and this is no good. The reason is that you might rotate the whole problem by 30 degrees, including the force vector, and now you have the same exact physical setup with a different force. In this case you can also eliminate the possibility that the force points out of the plane of the charges because the problem has mirror symmetry in this plane. The only answer that has all these symmetries is a force of zero. This all seems very intuitive and obvious, of course, but in cases where the symmetry is less clear, understanding exactly how it affects the solution can be very helpful.

Also, another way to do the second part is to use the superposition principle. Just add the setup from the first part to a setup with a negative charge placed at the spot where the charge was removed. Since the first part gives a force of zero, the total force in this new setup is just the force from this negative charge.

Last edited: Sep 3, 2005