In the figure a charged particle (either an electron or a proton) is moving righ

Question

In the figure a charged particle (either an electron or a proton) is moving rightward between two parallel charged plates separated by distance d = 3.20 mm. The plate potentials are V1 = 71.0 V and V2 = 46.0 V. The particle is slowing from an initial speed of 96.0 km/s at the left plate. (a) Is the particle: 1) an electron or 2) a proton? Give the number of the correct answer. (b) What is its speed just as it reaches plate 2?

(a) In the figure a charged particle (either an electron or a proton) is moving rightward between two parallel charged plates separated by distance d = 3.20 mm. The plate potentials are V1 = 71.0 V and V2 = 46.0 V. The particle is slowing from an initial speed of 96.0 km/s at the left plate. (a) Is the particle: 1) an electron or 2) a proton? Give the number of the correct answer. (b) What is its speed just as it reaches plate 2?

Explanation / Answer

Consider the work-energy theorem.

the particle is a proton, because it is slowing down, meaning that it is losing kinetic energy as it moves to greater potential.

The amount of work done on the charge from -71V to -46V is:
W = (-71V+46V)*q
W = (-25V)*q
q = 1.6*10^-19C
W = (-25V)*(1.6*10^-19C)
W = -4*10^-18J

This is the amount of kinetic energy that the proton will have lost when it gets to the other plate.
initial K = (1/2)mv^2
= (1/2)*(1.673*10^-27kg)*(96*10^6m/s)^2
= 7.709*10^-18J

final K = 7.709*10^-18J - 4*10^-18J
= 3.709*10^-18J

final velocity:
v = sqrt( 2*K/m ) = [2*3.709*10^-18/(1.673*10^-27)]^0.5
= 66.587 km/s

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