question archive A proton with an initial speed of 400000 m/s is brought to rest by an electric field
A proton with an initial speed of 400000 m/s is brought to rest by an electric field
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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
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A)
moves against a electric field
so, moves into a region of higher potential
B)
Work done by electricfield W = k.e.of proton K
q *V = (1/2) * m *v2
1.6 *10-19 * V = 0.5 * 1.67* 10-27 * 4000002
Potentialdifference V = 1.336* 10-16 / 1.6 *10-19 = 835 V
C)
k.e. of proton ineV = V(numerically) = 835 eV
