# Electric potential energy problem

A charged particle (either proton or electron) is moving rightward between 2 parallel charged plates separated by distance d = 2mm. The plate potentials are V1 = -70.0 V and V2 = -50.0 V. The particle is slowing from an initial speed of 90.0 km/s at the left plate.
a) Is the particle an electron or a proton?
b) What is its speed just as it reaches plate 2?

i used the equation V2 - V1 = -W_E / q (W_E is the negative of the work done by electric field). so i have 20 = -E(.002) and i got E = -10000 N/C. i don't know what to do next. how do i tell if its a proton or electron?

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## Answers and Replies

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phyzguy
Note that the particle is slowing. What does this tell you about the direction of the force vector?

oh so that means the direction of the Electric field vector is opposite of the displacement vector so since the particle moves to the right the electric field points to the left. and since the particle is traveling the opposite direction the electric field is pointing at, does that mean it's an electron? i checked the answer in the book but it said it was proton though.

thanks for not helping me...

The electric field always points in the direction of the force on a positive particle. So if E is negative the force will reduce a positive velocity only if the particle is a positively charged proton. The problem with your reasoning is that the particle is not traveling opposite the electric field BECAUSE of the electric field. It was given an initial velocity by some divine methods ; ) Ever since then the Electric field has been trying very hard to convince the particle to travel in the same direction as itself. It's a subtle difference but it will really help you understand this stuff if you think the difference through.

Now, you've got everything you need for b):

$$W_{E}$$ = $$\Delta$$U = - $$\Delta$$K

$$W_{E}$$ = Fd = $$q_{e}$$Ed = $$q_{e}$$$$\Delta$$V

$$K_{f}$$ = $$K_{i}$$ - $$q_{e}$$$$\Delta$$V

Get the final KE and you're on your way to the finish...

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