# Help with electric field problem- Yr 12 physics

Tags:
1. Nov 4, 2014

### JenL

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
Ek= 1/2mv^2 and Ep=Eqd

3. 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.

2. Nov 4, 2014

### vela

Staff Emeritus
Why not?

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

3. Nov 6, 2014

### 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.

4. Nov 6, 2014

### BvU

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

A simple energy balance suffices to extract VA - VB

5. Nov 6, 2014

### vela

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