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# Physics/Parallel Circuit

Question
Part 1: Two square, parallel conducting plates of side lengths 50 cm are separated by 0.9mm. What potential difference is casued by charges of +/- 80 C on the plates?

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

I started out by solving for the electric field. I did Electric Field = (9 x 10^9)(80)(-80) / (0.009m^2). I got -7.11 x 10^17. Then I did Voltage difference = (-7.11 x 10^17) (0.009m), but my answer doesn't match the ones in the option.

Part 2: In the above question what charge is required to cause a potential difference of 1500 V between the plates?

a. 3.7 microC  b. 15  c. 7.3  d. 23  e.45

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 plates,
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 permitivity is  8.85x10^-12 farads/meter.

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

Plug in the data and solve for V.

2. Now q is unknown and V = 1500 V. Solve for q.

I hope this helps,
Steve

Physics

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#### Steve Johnson

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