# Homework Help: Force exerted by one disc on the other in a capacitor?

1. Jan 28, 2014

### mot

1. The problem statement, all variables and given/known data
Two circular disks, each of area 4.10×10-4 m2, are situated parallel to one another. The distance between them is small compared with their radii. Both disks are uniformly charged; their charges per unit area are σ = 5.00×10-5 C/m2 for one and -σ = -5.00×10-5 C/m2 for the other. Compute the force exerted by one on the other.

2. Relevant equations
E=σ/ε
F=qE

3. The attempt at a solution
E=σ/ε=5.00×10-5 C/m2/8.85x10-12
=5.649717514x106N/C
Charge on one plate = σ*Area=5.00×10-5 C/m2*4.10×10-4 m2=2.05x108C
F=qE=2.05x108C*5.649717514x106N/C
=0.1158N

I have tried both the positive and negative version of this answer, both are wrong. I have no idea what I'm doing wrong here....the distance between the plates isn't given, so I can't use coulomb's law (but they're not point charges anyways)
Help?

2. Jan 28, 2014

### Staff: Mentor

While the net field between the plates is the sum of the fields created by the charges on both of the plates, the force acting on one plate is due only to the field created by the other plate. That is, the field created by a given plate does not create a force on itself.

So. The field strength to use is that due to a single plate. That field will act on the charge on the other plate.

3. Jan 28, 2014

### haruspex

Sign error in exponent, but looks like this was just an error in copying out.
That's what I get.
When two charged plates are very close, the field is essentially perpendicular to the plates. This means there is no attenuation with distance: the field is the same everywhere between the plates. The exact distance ceases to matter
Right, but the question asks for the force, not the field. The product of the charges will enter into it.

4. Jan 28, 2014

### Staff: Mentor

Yes, but the field due to one plate is due to the charge on that plate. So that charge is accounted for. Then F = Eq is the force on the other plate. The OP used the intermediate step of calculating E, so I did likewise.

Of course one can replace E by the expression involving the charge on a plate. Then product of the two charges emerges.

5. Jan 30, 2014

### mot

Mulled this over for a little while...got it! The field strength due to one plate is half as strong as the one calculated, and then I just used F=qE as normal to get my answer.
Thanks!