Two point charges, Q₁ = ― 6.7 μC and Q₂ = 2.6 μC, are loc...

Question

Two point charges, Q₁ = ― 6.7 μC and Q₂ = 2.6 μC, are located between two oppositely charged parallel plates, as shown in Fig. 21–74. The two charges are separated by a distance of 𝓍 = . Assume that the electric field produced by the charged plates is uniform and equal to E = 53,000 N/C . Calculate the net electrostatic force on Q₁ and give its direction.
<IMAGE>

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with electrical charges. So in this problem, we have two point charges. Q one is equal to negative 8.5 micro columns and Q two is equal to 3.4 micro columns and they're positioned between two oppositely charged parallel plates. The separation between the two charges is Y is equal to 0.35 m. And the electric field generated by the uh charge plates is uniform and has a magnitude of E is equal to 60,000 newtons per column. We need to determine the magnitude of the net electrostatic force acting on Q one and specify its direction below the question. We're given a diagram of what was described in the problem. We're also given four possible choices as our answers. Choice A is 1.5 newtons towards the negative plate. Choice B is 1.6 newtons towards the positive plate. Choice C is 1.5 newtons towards the positive plate. And choice D is 1.6 newtons towards the negative plate. And we're gonna have two forces acting on Q one, we're gonna have uh the force due to the charges on the plate and then we'll have the force due to the charge Q two. So we can write that as an equation. So we'll have F, which will be our net force acting on Q one that's going to be equal to the force on the plates which will label it as FP plus F two, which is gonna be the force due to charge Q two. Now, for the force of the plates, the only information we were given here is the electric field. So we need to recall our relationship between the electrical force and the electric field. And so in this case, that means that our FP is going to be equal to our charge Q, which in this case is gonna be Q one multiplied by the electric field E. And so here we need to figure out the direction of the electric field and the electric field in this case is going to be going from the positive charge plate towards the negative charge. And so my electric field is going to be going downwards and at this point, we need to define a coordinate system so we can figure out a positive and negative direction and we're gonna have the positive Y direction going upwards. Now, the other force that we're going to be looking at is the force due to charge Q two. And here I'm just going to be using Coolum salal. And here I'm just gonna look at the magnitude of that force. So we'll have the magnitude of F two is gonna be equal to K multiplied by the absolute value of Q one multiplied by Q two divided by Y squared. So K here is a constant, the Q one and Q two are the two charges. And the Y here is the separation distance between the two charges that we are given in the problem. Now, this gives me the magnitude of F two, but I do need to find the direction. And here I noticed that I have a positive and negative charge that means that they're going to attract. That means that Q one is going to be attracted to Q two. So that means that F two is going to be pointing downward. So here I noticed that both my electric field and my force are going to be pointing in the Y direction. So I really only need to find the Y component of my force. So here I'll just write my fy and this is gonna be equal to. And for my FP, I wanna have the Q one multiplied by the electric field, but this is gonna be going in the negative Y direction. So I'm gonna put a negative sign out in front of that term. Then I'm gonna have my F two. And here this is also going in the negative Y direction. So I'll have a minus K multiplied by the absolute value of Q one multiplied by Q two divided by Y squared. So it's at this point that I can plug in all of my values. Do I have FY, which is the total force that I'm looking for that's gonna be equal to minus. And for Q one, I was told that that was a negative 8.5 micro colons. But I need to convert that into columns by multiplying by 10 to the negative six. So that will give me the charging columns. I need to multiply this by the electric field which is the 60,000 newtons per column. Then I'll subtract from that the force F two. So that's gonna be the minus the K which is a constant, which is nine multiplied by 10 to the nine. And that has units of newton meters squared per Q squared. And that gets multiplied by the absolute value of Q one which in this case is gonna be the 8.5 multiplied by tint at the negative six columns that gets multiplied by Q two, which is the 3.4 micro coons. So I need to convert that into coons by multiplying by 10 to the negative six that's cool them. And then this gets divided by Y squared and Y was a 0.35 m. So that quantity is squared. So now I just plug in everything on the right hand side into my calculator and I get FY is equal to and negative 1.6 newtons. So we actually have to interpret our numerical answer here. So the magnitude here in this case is going to be the 1.6 newtons. The negative sign indicates that I'm going in the negative Y direction. So that means that the net force here is going to be pointing downwards towards the negative plate. And so when I take both of those into consideration, this corresponds to answer D. So just a quick little recap of what we've done here, we needed to find the net force acting on charge Q one. And to do that, we needed to find the force due to the charges on the plate, which we had to use our formula for the force in terms of the electric field. And we need to also include the force due to Q two. And there we just had to use Coolum Law. And so we had to just add these two forces together making sure that we get the correct directions for those various force vectors. And that allowed us to calculate our total net force on Q one. So I hope that this has been useful and I'll see you in the next video.

Key Concepts

Coulomb's Law

Electric Field

Net Force

Watch next