Physics question

Question

1. Starting from rest, a proton falls through a potential difference of 1200 V. What speed does it acquire?

 

2. Calculate the net work done in moving an electron from point A, where the potential is -50 V, to point B, where the potential is +150 V. (Answer directly in eV).

Answer 1

 Equate Charge Times Voltage To Kinetic Energy.

Then charge_proton(potential difference) = (0.5)(mass_proton)(velocity_proton)².

Place values to obtain (1.602177335E-19 Coulomb)(1200 Joules Per Coulomb)

Equals (0.5)(1.6726231E-27 Kilogram)(Velocity_proton)².

Rewrite Last Equation As 1.922612802E-16 Joule = (8.3631155E-28 Kilogram)(Velocity_proton)².

Simplify To 479470.4732 Meters Per Second (Equivalent To 479.47 Kilometers Per Second) As The Velocity Of The Proton.

_________________________________________________________________________________

Write |-1.602177335E-19 Coulomb| Times [150 Minus -50] Joules Per Coulomb To Obtain 

3.20435467E-17 Joule As The Net Work Done In Moving The Electron.