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Yakvenalex [24]
2 years ago
13

A parallel-plate capacitor with a 4.9 mm plate separation is charged to 57 V . Part A With what kinetic energy, in eV, must a pr

oton be launched from the negative plate if it is just barely able to reach the positive plate?
Physics
1 answer:
Dovator [93]2 years ago
4 0

Answer:

57 eV

Explanation:

d = separation between the plates = 4.9 mm = 0.0049 m

\Delta V = Potential difference between the plates = 57 Volts

q = magnitude of charge on proton = 1 e

K = Kinetic energy of the proton

Using conservation of energy

Kinetic energy lost = Electric potential energy gained

K = q \Delta V\\K = (1 e) (57 )\\K = 57 eV

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