calc speed of proton that accelerated from rest to potential diff. of 60 V. Calc speed of electron that acc. to same pot. diff answer in m/s?

I think I know what kind of problem you are trying to solve.

One way would be to think of this in terms of energy.

Work done by the electric field in moving the proton by distance x = Kinetic energy of charged particle at distance x

(conservation of energy)

------------------------------------------

Work done = charge of proton x change in potential difference

= 1.602 x 10^-19 C x 60 V

= 9.612 x 10-18 J

now as we see from above, this work done is equal do the gain in kinetic energy of the proton

Work done = Kinetic energy gained

9.612 x 10-18 J = 1/2 x mass of proton x velocity^2

so we can represent in terms of velocity

velocity^2 = (2 x 9.612 x 10-18) / mass of proton (mass of proton is approx 1.67 x 10^-27 kg)

= 1.15 x 10^10

velocity = sqrt(1.15 x 10^10) = 107291.08 m/s

= 110000m/s (2 significant figures)

= 1.1 x 10^5 m/s (in the direction of the electric field)

Now the second part of your question... about the electron...

Lets model it the same way, what is the difference between a proton and an electon?

mass of the electron is alot less than the proton, it is around 9.109 x 10^-31kg)

it has opposite charge, that is the same magnitude as the proton, but it is negative.

This has two consequences on our above solution.

The potential difference is due to an electric field. This field has a direction that is always representative of how a positive charge would "react" when subjected to the field.

This means that if the electron is subjected to the same potential difference, it will travel in the opposite direction.

Now the mass, if we get the last part of our last solution for the proton,

velocity^2 = (2 x 9.612 x 10-18) / mass of proton (mass of proton is approx 1.67 x 10^-27 kg)

lets change this to the mass of the electron

velocity^2 = (2 x 9.612 x 10-18) / mass of electron (mass of electron is 9.109 x 10^-31kg)

velocity^2 = 1.055 x 10^13

velocity = 4.6 x 10^6 m/s (in the opposite direction of the electric field.

this is what we would expect, as the electron is alot smaller than the proton, but the same magnitude of work is done on it, it gains the same potential energy, but with a much smaller mass, it gains a greater velocity.

Does this all make sense?

leave a comment if it doesnt and I will fill in any gaps I missed...

Some things to remember are, you must include a direction when you are describing velocity, as it is a vector quantity. So you need to consider the direction of the electric field, that is why I mentioned it.

Also one of the other answers used the same symbol for potential difference (V) and velocity (V) so dont worry if you are confused...

I double checked my answer so it should be good, ask me if you have any questions about it.

Enjoy your physics!