Oppositely charged parallel plates are separated by 4.31 mm. A potential differe

Question

Oppositely charged parallel plates are separated by 4.31 mm. A potential difference of 600 V exists between the plates. (a) What is the magnitude of the electric field between the plates? Incorrect: Your answer is incorrect. You appear to have made an error in converting units or in keeping track of exponents. N/C (b) What is the magnitude of the force on an electron between the plates? N (c) How much work must be done on the electron to move it to the negative plate if it is initially positioned 3.02 mm from the positive plate?

Explanation / Answer

PLEASE RATE:)

SOLUTION:-
E=V/d
= 600/4.3e-3
= 1e5N/C
= 1x10^5 N/C a)
F = E*q
= 1e5 x 1.6e-19
= 1.6e-14 N b)
4.3 - 2.91 = 2.47mm Work
= F*x
= 1.78e-14 x 2.47e-3
= 4.41e-17 J c)

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