Conceptual: electric potential

[calgal260]

Homework Statement

A proton is released from rest at point A in a constant electric field and accelerates to point B (see part a of the drawing). An electron is released from rest at point B and accelerates to point A (see part b of the drawing). How does the change in the proton's electric potential energy compare with the change in the electron's electric potential energy?

http://www.webassign.net/cj8/c19_q_4.gif

Homework Equations

V=EPE/q

The Attempt at a Solution

The proton experiences a smaller change in electric potential energy, since it has a smaller speed at B than the electron has at A. This is due to the larger mass of the proton.

The proton experiences a greater change in electric potential energy, since it has a greater charge magnitude.

One cannot compare the change in potential energies because the proton and electron move in opposite directions.

The proton experiences a smaller change in electric potential energy, since it has a smaller charge magnitude.

The change in the proton's electric potential energy is the same as the change in the electron's electric potential energy.

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Well if one is going in the same direction as the electric force, wouldn't the work be positive and EPE (electric potential energy) be higher since kinetic energy is decreasing? So would the last choice be correct, given that the magnitudes of change are the same but their directions are different?

I'm confuzzled.

 

[lightgrav]

if something is moving in the direction of the Force, so that positive Work is done on it, why is its KE decreasing?
Do things accelerate toward high gravity PE, or toward low gravity PE?

 

[calgal260]

if something is moving in the direction of the Force, so that positive Work is done on it, why is its KE decreasing?
Do things accelerate toward high gravity PE, or toward low gravity PE?

Well its KE is decreasing because the force exerted by the field contributes to its forward acceleration...

 

[lightgrav]

if it's accelerating forward, that usually means that its speed is increasing.
("released" usually means starting with speed = 0 ...)