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Physics 2200 Lab Problems Set 1the energy density of any electric field is (a) is proportional to the magnitude of the electric field (b) is proportional to the squ...

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Physics 2200 Lab Problems Set 1the energy density of any electric field is (a) is proportional to the magnitude of the electric field (b) is proportional to the square of the magnitude of the electric field (c)is inversly proportional to the magnitude of the electric field (d) is inversly proportional to the square of the magnitude of the electric field

Physics 2200 Lab Problems Set 1the energy density of any electric field is (a) is proportional to the magnitude of the electric field (b) is proportional to the square of the magnitude of the electric field (c)is inversly proportional to the magnitude of the electric field (d) is inversly proportional to the square of the magnitude of the electric field



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(a) For the preceding problem, find the energy per unit volume between the plates; this is the electric energy density. (b) Show that the energy density is equal to $\epsilon_{0} E^{2} / 2,$ where $E$ is the uniform electric field between the plates. (The expression $\epsilon_{0} E^{2} / 2$ is actually the energy density in any electric field, not just this configuration.)

Here in this problem we have given to charge, This is positive charge Cuban. And this is negative. Charge and distance them is de And we have a point. A which is distance d away from Charge one and be point which is between them. Any point c, it is again distance d away from negative charge. So this is the point. This is see point. So we will have to find electric field a point a b m c. Now if you see at point a electric field read point a due to positive charge will be in this direction and due to negative charge it will be in this direction So we will write. This is supposed even and this is easy to So the direction off electric field met will be equal toe even minus it'll on the direction is minus. I kept so we know that the value off even is K Cuban. Why are we square here? We have taken more. That means we will take We will talk about magnitude. So this is magnitude. Sign here. K Q two right are two square. This is negative. I can Yeah. Now we will put the value off K. We know that value off kids 19 to 10 to the power nine and Cuban is given one pick. Ochola and Alvin is five centimeters, so I cannot into meter minus again. 911. 9 value off Q two. So magnitude is two into 10 to the power minus 12, divided by two in tow, five into 10 to the power minus two because it is distance truly away. By calculating this, we will get the value off electric field. It is around 1.80 Newton for Coolum and the direction is minus. I kept. So this is electric field at point A. Now at point, we, if we see this, is going to be electric field due to positive charge will be in this direction and you too negative charge will be also in this direction. So we will and even plus, he took now collective field at point will be equal toe que Cuban my Armanis square plus que que toe by our two square and the direction is on The positive X axis or I kept here are one will be equal to D Y two meter and our two is again. Five by two into 10 to the government is 2 m. So we will put the value off one or two and Cuban, Kyoto and the value of K. So we get Yes, the value off electric field at point B will be equal toe around 43. Want to Newton Markkula, and direction is positive I kept. So this is the value of electric field. After putting the value off Evan R. Two here in this case, Alban is 2.5 into 10 to the power minus two and are to also is 2.5 and two tender. The power minus two meters No at point C If we see the point C, this is why and see electric field do toe positive charge will be in this direction. An electric field due to negative charge will be in this direction. So again the net direction will be in negative. XX is so we will This is e to and this is even so you think find not electric field so well right in it. At one c will be called toe que Cuban by item in the square. This is even and minus K Kyoto by our two square again, we will talk about the sign and this is the direction I kept. Now here are one will be equal toe two into five into 10 to the power minus 2 m and are you will be equal to five and 2 10 to the power minus 2 m. Again, we will put the values and we get This is equal to 6.29 Newton Markkula and the direction is minus. I get so we can write e at once. See, six point 29 I kept Nathan for cola. Now in part B, we have to find we have to sketch the electric field lines. So this is rescue and this is negative Tokyo Electric field and will be like this.

Welcome to caution number twenty three point three six. Chapter twenty three Electric Potential. So we have given that there are two. Pause it Lee charge Parallel plates to these are supposedly charge para bread. So one has Q and this has minus Q charge. We're saying that this has Sigma charge, and this has minus sigma charge because they're very long plaits. So surface densities, eh? Better quantity to discuss rather than cute self. So this is surface identity and were given that the distance between those two plates is the Is it? So we need to find out the electric field in between them and the potential difference between those two blades. And there is one more question. Party. We can discuss it later, so we have this formula. So this is the electric field due to one plate, two to one click. Having seen much I density and and cap is the direction towards the Norman off the plate. So let's say we have seen much identity off any bread, so electric field direction his outward If. Q. If sigma is Percy or curious posted, if save my is negative, that means that's why this issue my negative. So they have to create is into the wait. So let me intro one more electric three lines so that every point elective feel is coming to the and for splicing my charge collective three lines out to the normal. So these are normal. Tow this plane off the panel plate. So we have electric field due to stick this in my charge. It works this way. And same similarly we have electric field. You two minus sigma Charge again. This way. So we have e plus Sigma Better plus even minus Sigma Tau which is equals to erect a total. This is due to the less charge plate and this is due to the minus charge. So this becomes Sigma over to Absalon. Direction is let's see, we have high camp again, using for minus sigma We have plastic model or to absent, Not I can't, which becomes attractive children we can forget about. I can't because they say the direction to find we can take jet cap cake or anything I actually want to take. So we have a total equals two e. T. We have defined here. Which is a question, Seema, what to signal with absolute matures This result. So we will use this formula to find an electric field. So Sigma is given here, which is forty seven Nano column per meter square. Oh, it close to forty seven to ten days. We're minus nine column because, nah, noise tradition when it's not so we have but a which is equal to its, you know, absolutely so. Forty seven to ten days, so minus nine column. Well, we're absolutely, you know the way. So one by four by Absalon is it calls to nine into ten, twelve, nine. So this gives us absolute value because too one over four by into nine into tennis supporting this value here and using the calculator so elected field comes out to be five three, one Jill Newton were cooler. So this is the electric field which is in between them. Remember that this is independent of the distant, so whatever they have to feel, as at this point is also had very far away from the plate also. So this is independent the distance now coming back to the B. But so potential defence is it calls to the They could feel times the distance between them So this is the formula for potential potential difference between those two plates due to paella plates. So we have be part because two really close to E. T. D. Is your fight too. You won Joe times you d the zero distance which is zero point zero two two meter. We have to use the science here. So putting here zero point zero two using your complete that you can find out. So this is a closed too One one seven words That should be easy. Now going to the seat back Secret is based on the argument. So see, buddies what happens to the manager, the field and to the potential difference if surface star densities the same but they is double. So if these double is independent of distance so it remains the same so for sea. But seeing my eyes same d becomes Jody now electric field. I do too. Electric feen due to battler plates is independent off distance so hee the demands sing. So is because to get the same value five t one Julian five three one zero. But now, looking at the attention we have e t d. So rico's to e t D. So your distance has doubled now. So the potential difference we will go from three. Two Trevino Rays goes to e to to Dean. So which means new prime, the new or the new is itcause two to eighty deep physicals to to be, which is a constant, too into one, one, seven more. Two one one seven. So this is a cause to do three four. So the credential attention has doubled. So that's your answer for this question. I hope you understand. So thank you.

Fall. So in this problem, we're looking at two opposite Lee equal and opposite. We charge plates that are parallel to each other. So we're interested in finding the magnitude between these plates the electric field as well as the potential difference between the two plates. So let's quickly review which formulas that we're going to need in order to solve this problem. So first, in order to solve for the magnitude of the electric field, we're gonna have this formula just pretty simple. It's just the charge density denoted by Sigma. Then we have absolute not which is a constant. So what we're going to dio is now review the formula for potential difference between two plates and that's simply given by the magnitude of the electric field times the distance between the two plates that we're looking at. So now, for part A, which asks about the magnitude of the electric field, we're just gonna apply this formula. So we have 47.0 nano Kellems, remember when we're actually doing the calculations? Um, what we're gonna use is not nano columns, but we're gonna use cool. Um, so this will go to 47 times 10 to the negative nights columns. And then we're gonna have our constant epsilon not and go ahead and plug that into calculator. We're going to find that the magnitude of the electric field is 5310 Newtons per cool. Um, so we found the magnitude of the electric field And now let's look at what the potential differences. So we'll just apply this formula. And we've already found eat from part A. So it's gonna be just simply multiplying magnitude of the electric field by the distance, which is 0.2 m. Since we were given excuse me, 0.0 to 2 m, We were given that the distance 2.2 centimeters, and that will give us a potential difference of 117 votes. Yes. So we have one last part of this problem to look at and were asked to consider if the distance doubles between the plates. Uh, what effect does that have on the electric field as well as the voltage? So for the electric field, we're gonna find that there's actually no change because, as you can see from this equation here, there's no dependency on D. So he is going to stay exactly the same. And next we're asked Thio the answer If the potential difference is going to change and here we can see relationship between D and D there, directly proportional. So as D increases, so will we. So if the distance doubles Thebes, potential difference will also double hope. Right? And so that's how the electric field on voltage won't and will be affected by the distance doubling.

So in this problem, we're looking at two equal and opposite. We charged plates on. We have some distance deep between them, which we know is 2.2 centimeters. So we want to find out. Is the magnitude of the field between these two plates and the potential difference between them as well? So what we need to know is, First of all, how are we gonna find the magnitude of the electric field? So we have this equation that we can use. It relates the charge density to the electric field and by way of the vacuum primitive ity. So this constant epsilon not here we take sigma, which is 47 nano columns per meter square. That's our charge density. We're dividing that by Epsilon A and that will give the magnitude of the electric field for the potential difference between two opposite. We charged plates, uh, to find the potential between them. We simply take magnitude of the electric field, and we multiply it by the distance between them. So now that we've established these two formulas that we need, let's go ahead and apply them. So to find the electric fields, the magnitude of the electric field between these two plates. Gonna take sigma that were given right here. So we have 47 times 10 to the negative ninth. Cool lumps. Remember, whenever we're doing a math like this, we wanna make sure that our units are consistent. So, uh, that we're not getting a value that is off by some order of magnitude. So we're gonna want to use Kulov's instead of NATO columns. And we divide that by our constant salon. Not on that will give us a magnitude of 5310 Newtons per Coolum. So we found now the magnitude of the electric fields and we can move on to the potential. So to find the potential as we established earlier, all we have to do is take this magnitude from part day on, multiply it by d the distance which once we converted 2 m eyes 0.0 to to meters and want to go ahead on calculate that we'll find that the potential difference between the two plates is 117 volts. So now, in the third part of this problem, we're asked to solve where were asked to answer rather what would happen if we were to double the distance. So in the in the case that we double the distance between the two plates, let's first take a look and see what happens to, uh, the electric fields. So as we can see here, the magnitude of the electric field actually has no dependency on the distance. So we're actually not going to see any change. And he will remain the same in the case that we separate these plates and for V, we actually do see a dependency on D. So we see they are directly proportional. So if we were to multiply D by two on this side, the same would happen to potential difference. So the potential difference would actually also double if we double the distance in this case, yeah.


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