Twa cppasitely charged, parallel plates are separated by 5.13 mm. A potential di
ID: 2035889 • Letter: T
Question
Twa cppasitely charged, parallel plates are separated by 5.13 mm. A potential ditterence ot Go0. V exists hetween the plates. NOTE: You must enter units for each of the answers below. (a) Calculate the magnitude of the clectric field strength bctwccn thc plates. 1125TO NC (b) Calculate the magnitude of the force on an electron between the plates. 1.8e-14 N An electron is moved from the negative plate to the positive plate, What is the change in electric potential energy of the electron? (Think about whether the clectron's potential energy will incrcase or decrcasc in this process., d) Calculate change in electric potential energy of the electron if it is initially positicned 2.91 mm trom the positive plate and moved to the negative plate. (Again, think about whether the electron's potential energy will increase or decrease in decrease in this process.)Explanation / Answer
(a)
The magnitude of the electric field between the plates is,
E = V / d
E = 600 V / 5.33 x 10-3 m
E = 112570 N/C
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(b)
Force on an electron between the plates is,
F = q E
F = (1.6 x 10-19 C) (112570 N/C)
F = 1.8 x 10-14 N
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(c) Change in potential energy is equal to work done.
Work done on the electron to move it to the positive plate is,
W =- F. s
W = -(1.8x 10-14 N) (5.33x 10-3 m)
W =- 9.6 x 10-17 J
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(d)
Work done on the electron to move it to the negative plate is,
W = F. s
W = (1.8x 10-14 N) (5.33x 10-3 m – 2.91 x10-3 m)
W = 4.358x 10-17 J
In two significant digits,
W = 4.4 x 10-17 J
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