# Part A What potential difference is needed to stop an electron that has an initial velocity v=5.6×105m/s? Express your answer using two significant figures.

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The potential difference between two points in an electric field is equal to the work done to move a unit electric charge between these points. When an electric charge is released into a potential difference, the field makes a work and the energy is converted into kinetic energy of the particle. So, we can express the following equation:

ΔV = W/q

W = ΔV*q

Based on this analysis, we can say that; to stop the electron, it must be subjected to an electric field, whose work on the electron charge is, at least, equal to the kinetic energy of motion of the electron. Since the work is equal to the potential difference multiplied by the electric charge, we have:

W = ΔV*qe = (mev^2)/2

Where:

qe = 1.602*10^-19 C, is the electric charge of the electron

me = 9.109*10^-31 kg, is the rest mass of the electron

v, is the velocity of the electron.

ΔV, is the potential difference required to stop the electron.

Then we have:

ΔV(1.602*10^-19) = [9.109*10^-31(5.6*10^5)^2)]/2

ΔV = 89.26*10^-2 = 0.89 V