# Finding capacitence, used C = (EoA)/d, but didn't work!

1. Oct 9, 2005

### mr_coffee

This seems like a plug n' chug problem but i missed it somehow..
A parallel-plate air-filled capacitor having area 38 cm2 and plate spacing 2.0 mm is charged to a potential difference of 450 V. Find the following values.

(a) the capacitance
wrong check mark pF
(b) the magnitude of the charge on each plate
nC
(c) the stored energy
µJ
(d) the electric field between the plates
V/m
(e) the energy density between the plates
J/m3

So i used C = (EoA)/d
A = .38
V = 450
d = .002m
C = [(8.85e-12)(.38)]/.002
C = 1.68e-9, but the want it in pF, p is e-12, so i moved the decimal over to amek it 1680pF, but it was still wrong, did i mess up converting it from e-9 to e-12?

2. Oct 9, 2005

### Physics Monkey

38 cm^2 = .0038 m^2

3. Oct 9, 2005

### mr_coffee

hm...1 meter = 100 centimeters
38cm/100cm = .38m

4. Oct 9, 2005

### lightgrav

Area is NOT measured in "cm", but in cm x cm .
as in 10cm x 3.8 cm ... Convert both of them!

5. Oct 9, 2005

### mr_coffee

I know area is m^2, not m, or cm^2, not cm...
38 cm^2, so i want to conver that into meters, so .38 m^2. 2.0mm is the distance, i don't square that also.

6. Oct 9, 2005

### Physics Monkey

$$38 \, cm^2 = 38 \, cm^2 * \frac{1 \, m}{100 \, cm}* \frac{1 \, m}{100 \, cm} = 38*10^{-4} \, m^2 = .0038 \, m^2$$

7. Oct 9, 2005

### mr_coffee

ohhhh my bad, sorry i'm slow hah, thanks guys!