Please explain how to do this. When released from rest at point A, a proton flie

Question

Please explain how to do this. When released from rest at point A, a proton flies to point B. The potential at point A is 60 V. The potential difference between points A and B is 20 V (plus or minus). What is the potential at point B? Explain. What is the change in potential energy in Joules as the proton in goes from A to B? Is this an increase or decrease in potential energy? What is the change in potential energy in eV? What is the external work done as the proton goes from A to B? What is the change in the proton's kinetic energy in eV in going from A to B?

Explanation / Answer

Given that

potential at point Va=60 v

potential at point Vab=20 v

basing on the concept of electric potential

now we find the potential at point B

potential at point B=>Vb=Vab+Va=20+60=80 v

the proton goes from A to B is decreases

=>the change in potential energy =the external work done=q[Vb-va]=1.6*10^-19*20=32*10^-19 J

=>the change in kinetic energy=32*10^-19/1.6*10^-19=20eV

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