Question

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

Answer 1

moves into a region of higher potential

Potential difference = 835 V

Ki = 835 eV

Step-by-step 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 ×
`10^(-19)` × V = 0.5 × 1.67 ×
`10^(-27)` × 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