1. Jun 28, 2004

### Theelectricchild

Would the magnitude of the net electric force on a charge +Q in a case where there is another charge -Q a distance x away from +Q be equal to the force on +Q by -Q in a case where there is a neutral metal rod between them? (and same distance x away from each other)

I would be inclined to say that there would be NO difference, that is, the net force would be equal in both cases, but the question seems deceptively easy--- can i confirm this, and are there special cases when this is not true?

I should note that the problem does not state how the charge is distributed on the neutral rod, that is, if the positive charges are on one side (IE left has the positive and the negative charges are on the right side--- and having an equal amount would make the rod neutral)--- I know metal is a conductor--- would that mean that the charge is distributed uniformly even in a neutral state?

I am quite new to electricity in general so please ask for clarification if something didnt make sense!

Last edited: Jun 29, 2004
2. Jun 28, 2004

### jack_haimer

use coulomb's law

use coulomb's law and just replace the dielectric constant!!

3. Jun 28, 2004

### Theelectricchild

Hi thanks for your response, but could you please explain that a bit? Replace it with what?

Heres my reasoning now that ive thought it over a bit longer:

Wont the negative charge that is in the metal rod wish to "come closer" to the positve Q charge while the positive charge in the metal rod comes closer to the -Q charge? IE the distribution of charge throughout the metal rod is not uniform--- and thus would play a role in showing that the force between Q and -Q would be greater in this case?

4. Jun 29, 2004

### Theelectricchild

I just constantly have a gloomy cloud over my head :(

5. Jun 29, 2004

### Theelectricchild

oh well thanks for your help anyway

6. Jul 4, 2004

### eJavier

As I see it, if you have 1 plus charge and 1 minus charge, you'll have an electric field between them. If you place a metal rod between them, then the plus charges of the rod will be pulled towards the - charge and the negative charges of the rod will go the opposite way. You'll have one end of the rod with + charge and the other with - charges. That will affect the net force on the Q charge

Greetings
Javier

7. Jul 5, 2004

### turin

Does the metal rod touch the charges?

8. Jul 17, 2004

### rayjohn01

I suspect that the rod charges separate as you say ( meaning electrons will accumualte at one side ), if the rod was a plate ( easiesr to think about) then if it was thin there would be no net field disturbance , but if it was thick then it reduces the effective q1/q2 distance -- field lines must be perpendicular to the plate.
You could graph that by splitting the field q1/q2 down the middle and moving thm apart.Ray