Question: 2 Charges Dropped in an Electric Field (Chapter 17, #68)

I am confused on what the equation for conservation of energy should be for this problem....Should we consider potential energy due to the height?

For example I assumed the equation to be: mgh(before)+PE(before)=KE(after)

The question states: "Near the surface of the Earth there is an electric field of about 150V/m which points downward. Two identical balls with mass m=0.540kg are dropped from at height of 2.00m, but one of the balls is positively charged with q₁=650μC, and the second is negatively charged with q₂=-650μC. Use conservation of energy to determine the difference in the speed of the two balls when they hit the ground."

Whoo-ee! That's a nice problem, it incorporates several different ideas, just the kind I hated as a student, and see the value of now when I teach. :)

Your energy equation is mostly correct, but we need to figure out the signs of the energies.

Let's first draw a diagram, because I find that problem a mouth full. (I will put in the electrostatic forces, but not gravitational... we all know that points downward anyway).

So we have the positive charge (drawn on the left) experiencing a downward force due to the electric field, thus as it falls (down) it's electric potential energy will be decreasing, just as it's gravitational potential energy will be decreasing. However, for the negative charge (drawn on the right), the electrostatic force is pointing upward, opposing the falling motion, so it's electric potential will increase as it falls.

We will have to look at the conservation of energy each charge separately. Let's start with the positive charge:

PE(electric)+PE(gravitational)=KE
qEh+mgh=(1/2)mv₁²

(note that I used h for the distance the charge travels through the electric field).

Now let's look at the negative charge:

PE(gravitational)=KE+PE(electric)
mgh=(1/2)mv₂²+qEh

When plugging in the numbers, disregard the sign of the charge. I have included it already by determing whether the electric potential energy increases or decreases. Alternatively you could have used the energy equation for the first charge and plugged in the sign of the charge to get the same thing. Personally, to avoid confusion with signs, I think it's better to work through the logic of what's going on than to plug in signs, but that doesn't make the other way wrong.

So, using the energy equations for each ball, you can solve for the final velocities. I have confidence that you can all plug in numbers, so I will leave it at that. Hope this helps, and if you have any further questions feel free to post a comment.