A particle with charge +4.20 nC is in a uniform electric field E directed to the left. The charge is released from rest and moves to the left; after it has moved 6.00 cm, its kinetic energy is +2.20x10^-6 J. What is (a) the work done by the electric force?

Video transcript

Hey everyone. This problem is about work and the work energy theorem. So let's see what it's telling us and what it's asking from us. So we know we have an electric field, it is vertically upward. We are given a point charge. We're giving the charge value there and it's initially held at rest, it's freed and travels upward. Uh the kinetic energy of the charge is given to us as positive 3.6 micro jules. And the distance when this energy is, the kinetic energy is calculated is also given to us. And so they're asking us to calculate the work done by the electric force. So this is a little bit of a tricky question, just because of how simple it is and they give you a lot of information that you don't need. So the first thing we need to do is recall the work energy theorem. So that is simply just the work is equal to the change in kinetic energy from the final point, from the original point to the final point. So it's K f minus K nine K not is going to be zero because we know that initially this point charge is held at rest And then they give us the kinetic energy at the final point in the problem. So it is really just that simple work is equal to the final kinetic energy, which is equal to positive 3. micro jewels. And that's it's the rest of the information. And this problem is superfluous. We look at our answers. The answer is d and that's all we have for this video See you in the next one.