Question

A proton (H⁺) is released at rest at a location where the electric potential is measured to be 3000 V.

The energy gained by the proton = _________ J BLANK-1 = __________ eV BLANK-2

The speed gained by the proton due to this potential is ______ m/s?

Answer 1

The potential energy(PE) of a point charge Q at a potential V is given by the product of charge and potential at that point

.i.e PE = Q*V

Here ,

Q = Charge of proton = 1.6*10^-19C

V = 3000 V (as given in the question)

PE = 1.6*10^-19 * 3000 = 4.8*10^-16 J

To convert joule into ev , divide by 1.6*10^-19

PE = 4.8*10^-16/1.6*10^-19 = 3000 ev

Here the coulombic force is acting on the charge particle which is conservative in nature , therefore the mechanical energy(the sum of potential energy and the kinetic energy will be constant) will be conserved i.e initial mechanical energy will be equal to final kinetic energy .

Initial potential energy = 4.8*10^-16J Initial kinetic energy = 0

Final potential energy = 0 Final kinetic energy = k(say)

Initial mechanical energy = final mechanical energy

4.8*10^-16 + 0 = k + 0

k = 4.8*10^-16

kinetic energy = (m = mass , v = velocity)

4.8*10^-16 = (here m will be mass of proton=1.67*10^-27kg and v be the velocity)

On solving ,

v = 758188 m/s