What Is the Capacitance Between Two Plates?

[fenixbtc]

Homework Statement

An air filled capacitor consists of two parallel plates, each with an area of 8.39cm^2, separated by a distance of .81mm. A 12.3V potential difference is applied to the plates. Find the capacitance. Answer in units of pF (pico Farad)

Homework Equations

1. E = kV/d
2. C=q/v
3. q=E0EA E0 being permittivity constant, 8.85e-12

The Attempt at a Solution

using 1. E = 12.3e-3 / 8.1e-4 = 15.185
using 3. q = 8.85e-12 * 15.185 * 8.39e-2 = 1.12807e-11
using 2. C=1.12807e-11 / 12.3 = 9.17136e-13
so wouldn't it be 91.7136 pF?
not sure where i went wrong?

thanks
david

 

[willem2]

A 12.3V potential difference is applied to the plates.

1. E = kV/d


using 1. E = 12.3e-3 / 8.1e-4 = 15.185

why do you have 12.3e-3 for V here instead of just 12.3

 

[rl.bhat]

Capacitance of a capacitor does not depends on the voltage across the parallel plates.

 

[fenixbtc]

Willem: i had used 12.3e-3 because part A of the question asked for the electric field between the plates in kV/m, which 15.185 was the answer. i tried doing this part with 12.3v for the whole thing and it still came out wrong.

rl.bhat: i also tried C=E0A/d E0 being permittivity constant, 8.85e-12 and it still came out with the same values which is being said is wrong. with the work for that being...
(8.85e-12 * 8.39e-2) / (.00081) = 9.16685e-10 to pF 916.685

i have 3 more chances to get it right, with each chance decreasing the points i can get... :/

 

[rl.bhat]

Area should be 8.39*10^-4 m^2. Try this.
For electric field
E = V/d = 12.3/0.81*10^-3 =...?

 

[fenixbtc]

well that is just aggravating. i had the decimal in the wrong place on 3 of those 4 times i got it wrong. it is correct now. thank you!

where i went wrong was with the area. it is 8.39e-4 instead of 8.39e-2 (going from cm to m) because it's two dimensions, right?

once again, thank you.