¿Cuál es la frecuencia de una nota musical cuyo periodo es 0,005 s?
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Answer:
a. It starts at point B.
vp = 2.53*10⁴ m/s
a. it starts at point A.
ve= 1.08*10⁶ m/s
Explanation:
a) As the proton is a positive charge, when released from rest, it will be accelerated due to the potential difference, from the higher potential to the lower one, so it is at the point B when released.
Once released, as the total energy must be conserved, the increase in kinetic energy must be equal (in magnitude) to the change in the electric potential energy, as follows:
ΔK + ΔUe = 0 ⇒ ΔK = -ΔUe =- (e*ΔV)
⇒
where e= elementary charge= 1.6*10⁻¹⁹ C, VA = 1.83 V, VB= 5.17V, and mp= mass of proton = 1.67*10⁻²⁷ kg.
Replacing by these values, and solving for v, we have:
⇒ vp = 2.53*10⁴ m/s
b) If, instead of a proton, the charge realeased from rest, had been an electron, a few things would change:
First, as the electrons carry negative charges, they move from the lower potentials to the higher ones, which means that it would have started at point A.
Second, as its charge is (-e) the change in electric potential energy had been negative also:
ΔUe = -e*ΔV = -e* (VB-VA)
In order to find the speed of the electron when it is just passing point B, we can apply the conservation of energy principle as for the proton, as follows:
where e= elementary charge= 1.6*10⁻¹⁹ C, VA = 1.83 V, VB= 5.17V, and me= mass of electron = 9.1*10⁻³¹ kg.
Replacing by these values, and solving for v, we have:
⇒ ve = 1.08*10⁶ m/s