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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?

Respuesta :

Answer:

57 eV

Explanation:

[tex]d[/tex] = separation between the plates = 4.9 mm = 0.0049 m

[tex]\Delta V[/tex] = Potential difference between the plates = 57 Volts

[tex]q[/tex] = magnitude of charge on proton = 1 e

[tex]K[/tex] = Kinetic energy of the proton

Using conservation of energy

Kinetic energy lost = Electric potential energy gained

[tex]K = q \Delta V\\K = (1 e) (57 )\\K = 57 eV[/tex]

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