Question

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 proton be launched from the negative plate if it is just barely able to reach the positive plate?

Answer 1

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

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