A very large, horizontal, nonconducting sheet of charge has uniform charge per unit area σ = 5.00×10−6 C/m2. (a) A small sphere of mass m = 8.00×10−6 kg and charge q is placed 3.00 cm above the sheet of charge and then released from rest. (b) What is q if the sphere is released 1.50 cm above the sheet?

Question

Accepted Answer

Hey everyone. So this problem is dealing with charged particles. Let's see what they are giving us and what they're asking from us. So we know we have a particle with a certain mass and a given charge, it's at rest, this given height five centimeters above a large flat nonconducting plate. That plate has a constant charge density given to us. And they're asking us the charge of another particle where the mass is the same as the first particle that also remains stationary when placed without initial speed. So at rest uh at a distance of two centimeters above the plate. So the first thing that we need to do here is draw our free body diagram. So we have this charge and we have the electric force from the charge density acting in an upward direction. And then we have the weight of the um charge particle acting in a downward direction. So from here, let's recall that the electric force is given by QE where Q is a charge and E is the magnitude of the electric field and weight is given as mass times gravity. So we're going to solve the free body diagram for both of the um particles. So P one and P two and then um see if that tells us what the charge of the second particle is. So for particle one, we have Q one E one equals M one G gravity is a constant. So we can rewrite this as Q one equals M one G over E one. For the second particle, the free body diagram looks exactly the same. The only difference is between these two is the height away from the or above the electric field. And so Q two, it's going to be, or for particle two, it's gonna be Q two B two equals M two G where QE two equals M two G over E two. And so we are already given in the problem that M two equals M one. And we know that the um plate has a un a constant charge density, which means it has a uniform electric field. So E two equals E one. So we can rewrite this as Q one equals M one G over E one and Q two equals M one G over E one. And so from here, we can see that Q one and Q two are the same. And so therefore, Q two must equal Q one which was given to us in the problem as 2.83 times 10, the negative 11 pull ups. And so that is the answer to this problem when we look at our potential solutions. And we see that, that aligns with choice B. So that's all we have for this one, we'll see you in the next video.

Verified Solution

Key Concepts

Electric Field due to a Charged Sheet

Force on a Charged Object in an Electric Field

Newton's Second Law of Motion