Proton kinetic energy and work in electric field

[baller2353]

Physics Grade 12 HELP! Electricity Unit

I have attached the question paper with this post

Homework Statement

v=2 x 10 ^ 5 m/s
Volatage=250 or 500?

Constants Given:
Coulombs constant = 9 x 10^9
Charge of proton/electron = 1.6 x 10^-19/-1.6 x 10^-19
Proton rest mass 1.673 x 10^-27 kg
Electron Rest Mass= 9.11 x 10^-31 kg

Homework Equations

EK=EE?
Or Ek+W=0

The Attempt at a Solution

Ek=0.5mv^2
=0.5*(1.673 x 10^-27) * (2x10^5)^2
=3.346 x 10^-17 J

=3.346 x 10^-17 J + W=0
W=-3.346 x 10^-17 J

STUCKKKK HELP MEEE

 

[cepheid]

Welcome to PF!

You're right that if the electric field is constant (which is true in between two long parallel plates), then the electric potential changes linearly with distance. Therefore, at the halfway point, the voltage will indeed be half of the voltage across the full distance. So, in order to make it across half the gap, the proton has to pass across a potential difference of +250 V. How much potential energy must it gain in order to do so? Hint: what is the definition of electric potential?

Recall that any potential energy gained corresponds to kinetic energy lost. How does the energy needed to get across compare to the kinetic energy of the proton?

 

[baller2353]

thank you i figured it out:D