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# Physics/Parallel plates

Question
Two square, parallel, conducting plates of side length 50 cm are separated by 0.9mm. What potential difference is caused by charges of +/- 80 C on the plates?

a. 32,500 V  b.17,000 V  c.1100 V  d.105 V  e.550 V

For this I started off by doing V = (80)(.009/1000)  /  (8.85x10^-12)(0.25), but Im not gettiing the right answer.

Hello Sona,

1. Please go to the website
http://en.wikipedia.org/wiki/Capacitive
See the sections: "Theory of operation" and "Parallel-plate model".
Note the 2 formulas for capacitance that apply to a parallel plate capacitor:
C = epsilon*A/d and C = q/V
In those formulas:
epsilon is the applicable permitivity constant,
A is the area in meter^2 of the plate,
d is the distance in meters between them,
the plates have charge q and -q in Coulombs,
and V is the potential difference in Volts between them.
If we assume there is just air between the plates, the permittivity is  8.85x10^-12 farads/meter.

Since C = C, we can write
q/V = epsilon*A/d

Everything above here is from my previous reply. (Except I corrected the spelling of permittivity.)

The units of the permittivity can also be Coulombs/Volt.meter. That equivalent version will work better here.

I don't know why you entered .009/1000, or .000009, for d. You said d is 0.9 mm which equals 0.0009 m. Using that value, I get 32,500,000,000 V. Which is not one of the answers. Could it be that the charge is 80 microC?

I hope this helps,
Steve

Physics

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#### Steve Johnson

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