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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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