# Electric force problem GRE#10

1. Sep 14, 2004

### quantumworld

here is a question that I stumbled upon, and I would like some insight on it

two positive charges q and 2q coulombs are located on the x-axis
at x = .5a and 1.5a, respectively. There is an infinite, grounded conduction plane at x =0. What is the magnitude of the net force on the charge q?

my problem is that I am unable to account for that infinite grounded conductor, I am unable to imagine which kind of force is it going to
exert on q.

Many thanks :shy:

2. Sep 14, 2004

### ehild

There are some negative charges on the grounded conductor plane accumulated by induction. The effect of this charge distribution is the same as if we had the mirror images of the charges in front of the plane behind the plane, and with opposite sign.

ehild

Last edited: Jun 29, 2010
3. Sep 14, 2004

### K.J.Healey

So is it (1/2)*k*q^2 ? I have to take the GRE's in a month, and the Physics Subject one in November. Im dreading the physics one...

I got this answer by mirroring both across the plane with opposite charges, then added the 3 forces together.

4. Sep 14, 2004

### quantumworld

THank you Healey, and ehild !

I do remember something about this mirror image, and now I got it right, but it is kindof a magical solution, instead of dealing with the mess of charges on the conducting plate, we could treated as a mirror image, I wonder how did they come up with that theory, but I am glad they did

Last edited: Sep 14, 2004
5. Sep 14, 2004

### ehild

Draw arrows pointing into the direction of the forces the individual charges exert on the charge "+q". The "+2q" charge repels it, the negative charges on the opposite side attract it. And do not forget to include the distance between the charges in the formula for the force.

ehild

6. Sep 14, 2004

### K.J.Healey

Ahh, forgot about which way theyre pushing. Gotta get more complicated and use vectors.

k*([(-2q)(q)/2^2](-i) + [(-q)(q)/1^2](-i) + [(2q)(q)/1^2](i)) = (7/2)kq^2

7. Sep 14, 2004

### Tide

It's not magic. A (perfectly) conducting surface must be an equipotential. You probably know from your introduction to fields that if you have two charges of opposite sign that there is a plane between them that is at a single potential.

The problem really becomes interesting when you try the image charge approach to a nonplanar conducting surface such as a sphere. It turns out that it requires an infinite series of image charges to get the equipotential!

8. Sep 15, 2004

### ehild

This is almost all right but a^2 is missing from the denominator.

ehild

9. Sep 15, 2004

### K.J.Healey

I dont see it, please show me.

10. Sep 15, 2004

### ehild

There were "a*s in your first post.

ehild

11. Sep 15, 2004

### K.J.Healey

uhhh, i didnt post this problem. And i assumed those were angstroms. I like how you quoted someone elses post then put my name on it :tongue2:

12. Sep 15, 2004

### ehild

Sorry.... I just did not look at your name, I did not recognize that it was not quamtumworld any more but you.... Or it was not you who originally posted the problem. Can you forgive a scatterbrained old lady?
By the way I do not think they are angstroms. Charge "q" and 2q, position "0.5a" and 1.5a. "q" and "a" can be anything.

ehild