You’ve hung two very large sheets of plastic facing each other with distance d between them, as shown in FIGURE EX23.19. By rubbing them with wool and silk, you’ve managed to give one sheet a uniform surface charge density η1=−η0 and the other a uniform surface charge density η2=+3η0 . What are the electric field vectors at points 1, 2, and 3?

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Hi everyone. Let's take a look at this practice problem dealing with electric fields. So in this problem, we're given a figure that shows two horizontal parallel metal plates with the separation distance D between them. We need to determine the electric field vectors at points A B and C. If connecting these plates to a battery results in one plate acquiring a uniform charge density of sigma, one is equal to a positive two sigma knot and the other plate obtaining a uniform charge density of sigma two is equal to minus sigma knot. Below the problem, we're given a diagram that shows our two plates with plate one positioned above plate two and point A is above plate one point B is between the two plates and point C is uh below plate two. We're also given more possible choices as our answers. For choice A we have E A is equal to sigma knot divided by the quantity of two epsilon knot. That fraction is multiplied by J hat, we have EB is equal to minus three sigma knot divided by the quantity of two epsilon knott. And that fraction is multiplied by J hat. And then we have EC is equal to minus sigma knot divided by the quantity of two epsilon knot. And the fraction is multiplied by J hat. For choice B we have E A is equal to Sigma knot divided by the quantity of two epsilon knot multiplied by J hat. We have EB is equal to three Sigma knott divided by two epsilon knot. And that fraction is multiplied by J hat. And then we have EC is equal to Sigma knot divided by the quantity of two epsilon knot multiplied by J hat. For choice C we have E A is equal to minus three Sigma knot divided by two absolute knot. And that fraction is multiplied by J hat. We have EB is equal to minus Sigma knot divided by the quantity of two epsilon knott. And that fraction is multiplied by J hat. And then we have EC is equal to minus Sigma knot divided by two epsilon knot. That fraction is multiplied by J hat. And for choice D, we have E A is equal to minus Sigma knot divided by the quantity of two epsilon knot. And that fraction is multiplied by J hat. We have EB is equal to minus three Sigma knot divided by the quantity of two epsilon knot. And that fraction is multiplied by J hat. And then we have EC is equal to minus Sigma knot divided by the quantity of two epsilon. And that fractions multiplied by J hat. Now to begin with, we need to find the electric field due to both of these plates. And we need to recall our formula for the electric field due to a uniformly charged plate. And so the magnitude of that electric field it's going to be given by E is equal to the absolute value of sigma divided by two epsilon. Not. So now all we need to do is use this magnitude and figure out the directions of the electric field based on the type of charges that we have on each plate. And then combine these two electric fields to find the total electric field at each point. So we're gonna start off with point A, it will have E A and that's gonna be equal to. And here we're gonna have two electric fields, we're gonna have the electric field from plate one. And since plate one is positive, its electric field is going to be pointing away from the plate one, we'll have it pointing upwards. And for plate two, it's a negatively charged plate. So the electric field lines going point towards that plate. And so at point A, the electric field for plate two is gonna point downwards. So that'll be E two. So we'll have E A is equal to E one plus E two. So now we'll plug in for each of those electric fields. So we'll have E A is equal to. Now, the magnitude of E one is going to be for sigma, that's gonna be our two Sigma night and that gets divided by two epsilon knot. And that is pointing upwards. And so we'll call that our positive Y direction. And so that'll be going in the positive J hat direction. Then we'll have plus four E two. The absolute value of our surface charge density sigma is just gonna be Sigma knot. Since the absolute value gets rid of the negative sign, we'll have sigma knot divided by two epsilon knot and that's going downward. So that's going in the minus J hat direction. So we can just add those two together. And so we'll get E A is going to be equal to, I'll have two sigma knot minus Sigma knot. So it just gives me Sigma knot divided by two epsilon knot in DJ hat direction. So that's our answer for point A for point B. We'll do pretty much the exact same thing. But here, the um electric field due to plate one is actually gonna be pointing downwards since it's pointing away from plate one E two will still point downwards is still pointing towards plate two. And so that means that reb it's just gonna be equal to E one plus E two. And so we'll plug in our formulas for E one and E two. And so we'll have EB is equal to, for E one, we'll have again, the two stigma knot divided by two absolute knot. This time it's going in the negative J hat direction. Since it pointing downwards, then we'll have plus again, we'll have the same magnitude for E two. It'll be sigma knot divided by two. Absolutely not. And this is still pointing in the negative um Y direction. So this would be going in the negative J hat direction. So now I need to just add those two quantities together and I'll get EB is equal to three stigma knot divided by two absolute not. And this is going in the negative J hat direction. And so just gonna simplify that a little bit rewrite it. And so we have EB it's equal to minus three sigma knot divided by two epsilon knott J hat. And so that's gonna be our answer for point B. And finally, we'll look at point C. And so here my E one electric field is still pointing downwards is pointing away from one. But now, since I'm below plate two, my E two is going to be pointing upwards because it's gonna be pointing towards plate two. So again, have to just add these two electric filth together. So I'll have EC is equal to E one plus E two. Again, plug in our formulas for the electric field we will have is equal to, we'll have the two Sigma knot divided by two epsilon knot. And this is going in the negative J hat direction or multiplied by negative J hat. Then we'll have plus we have the stigma knot divided by two epsilon knot for the magnitude of E two. This time it's pointing upwards. So we're gonna multiply this by J hat. And so when we add those two together, we get EC is equal to, you will have minus sigma knott divided by two absolute knot. This is multiplied by J hat. And so that's gonna be our answer for point C. So now that I have all three of my electric fields calculated, I can look at my answer choices and this actually corresponds to answer choice. A while this problem was a direct application of our formula for the electric field. Due to a uniform surface, uh uniform charge density on a plate, we did have to be careful, careful about the direction of the electric fields um for the two different plates. So once we took the direction of the electric fields into account, we could just add them together and just do your normal vector addition. So I hope that this has been useful and I'll see you in the next video.

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Key Concepts

Electric Field

Surface Charge Density

Superposition Principle