# Surface charge density and more, for a parallel plate cap.

1. Feb 3, 2010

### free-node-5

1. The problem statement, all variables and given/known data
An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.60 mm.
1. If a 25.0 V potential difference is applied to these plates, calculate the electric field between the plates.
2. What is the surface charge density?
3. What is the capacitance?

2. Relevant equations
field=kq/(r^2)
V=kq/r
so q = Vr/k

3. The attempt at a solution
1. field=kVr/kr^2=V/r=25/.0016=15625N/C (correct)
2. q/A=Vr/kA=25(.0016)/(k*.00076)=5.84795nC/m^2 (wrong by orders of magnitude)
3. C=epsilon*A/d=1(.00076)/.0016=47500000pF (orders of magnitude wrong)

(we seem to have been given a large amount of questions based on material that we didn't really cover very much if at all)

I feel like I should have the right answer based on the results of using these techniques in some of my other problems but seem to be wandering in the dark
Any help would be greatly appreciated

Last edited: Feb 3, 2010
2. Feb 3, 2010

### rl.bhat

εo is not equal to 1.
So check the C value.
Surface charge density = Q/A = C*V/A =(εo*A/d)*V/A