A pair of closely spaced parallel conducting plates, charged with equal and oppo

Question

A pair of closely spaced parallel conducting plates, charged with equal and opposite electric charges, produces a uniform electric field in the region between them. In designing a cutting-edge device that will revolutionize the electronics industry, you set up such a pair of plates with a separation of 0.979 mm between them and charge them so that the direction of the electric field in their interior region is from plate A to plate B. Your idea requires that electrons, when released from rest at one of the plates, reach the other plate at the speed of 1.37% of the speed of light. (The speed of light is c = 3.00

Explanation / Answer

First the electrons should be released from the negative plate so they are attracted to the positive plate.

Now we find the value for E...

We know that F = qE

Also F = ma and from acceleration we can apply vf^2 = vo^2 + 2ad

Since the electrons start from rest, a = vf^2/2d

Now combine the equations with substitution

mv^2/2d = qE

v = (.0137)(3 X 10^8) = 4.11 X 10^6 m/s

(9.11 X 10^-31)(4.11 X 10^6)^2/(2)(9.79 X 10^-4) = (1.6 X 10^-19)E

E = 49121 N/C (which is 4.91 X 10^4 N/C)

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