Help with electric field problem- Yr 12 physics

[JenL]

Homework Statement

Protons are accelerated by an electric field passing through a hole in the anode and accelerated towards the cathode (to the right)
A proton passes through the anode at 6.2 × 105 m s–1, and passes through the cathode at 8.8 × 105 m s–1. Show that the strength of the electric field is 100 000 V m–1.

Homework Equations

Ek= 1/2mv^2 and Ep=Eqd

The Attempt at a Solution

Usually we get problems like this when the charged particle is forced to the plate with the same charge thus it gains electric potential. Then the charged particle is released and accelerates towards the plate with the opposite charge so electric potential lost= kinetic energy gained when particle reaches the other plate.
But for this question Ep(lost)=Ek(gained) doesn't seem to apply.
I thought it might be Ep at anode was Ek but I'm kinda confused.

 

[vela]

Homework Statement

Protons are accelerated by an electric field passing through a hole in the anode and accelerated towards the cathode (to the right)
A proton passes through the anode at 6.2 × 105 m s–1, and passes through the cathode at 8.8 × 105 m s–1. Show that the strength of the electric field is 100 000 V m–1.

Homework Equations

Ek= 1/2mv^2 and Ep=Eqd

The Attempt at a Solution

Usually we get problems like this when the charged particle is forced to the plate with the same charge thus it gains electric potential. Then the charged particle is released and accelerates towards the plate with the opposite charge so electric potential lost= kinetic energy gained when particle reaches the other plate.
But for this question Ep(lost)=Ek(gained) doesn't seem to apply.

Why not?

I thought it might be Ep at anode was Ek but I'm kinda confused.

The difference in this problem seems to be that the proton has an initial kinetic energy. Is that what's confusing you?

 

[JenL]

Yes it is the initial kinetic energy that is confusing me.
I tried to work it out but the Ep at the anode was less than the Ek gained at the cathode. I know I need to take into consider the initial Ek but I don't understand how it would work.

 

[BvU]

Anodes are positive and cathodes are negative. So the particles travel from high V_A to lower V_C, losing potential anergy and gaining (kinetic) energy to the tune of q$\Delta$V = q(V_A - V_B). Both factors are positive.

A simple energy balance suffices to extract V_A - V_B

 

[vela]

Yes it is the initial kinetic energy that is confusing me.
I tried to work it out but the Ep at the anode was less than the Ek gained at the cathode. I know I need to take into consider the initial Ek but I don't understand how it would work.

It's just conservation of energy: $K_i + U_i = K_f + U_f$, where $U$ is the electric potential energy (not electric potential).