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A proton with an initial speed of 400000 m/s is brought to rest by an electric field.
Part A- Did the proton move into a region of higher potential or lower potential?
Part B - What was the potential difference that stopped the proton?
?U = ________V
Part C - What was the initial kinetic energy of the proton, in electron volts?
Ki =_________eV

Respuesta :

Answer:

moves into a region of higher potential

Potential difference = 835   V

Ki = 835 eV

Explanation:

given data

initial speed = 400000 m/s

solution

when proton moves against a electric field  so that it will move into higher potential  region

and

we know Work done by electricfield  W is express as

W = KE of proton   K

so

q × V   =  0.5 × m × v²     ......................1

put here va lue

1.6 × [tex]10^{-19}[/tex] × V   =   0.5 × 1.67 × [tex]10^{-27}[/tex] × 400000²

Potential difference V = 1.336 × 10-16 / 1.6  × 10-19      

Potential difference = 835   V

and

KE of proton in eV is express as

Ki  =   V numerical

Ki = 835 eV

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