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SAME SURNANE STUDEAT NO: when wc want t0 SIGNATURE: differential charges; poiut-like divide ring of charge into spuce. (3 pts) 1-4) Exphin why we in its surroundin...

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SAME SURNANE STUDEAT NO: when wc want t0 SIGNATURE: differential charges; poiut-like divide ring of charge into spuce. (3 pts) 1-4) Exphin why we in its surrounding ckrlno- Licld al 0 Point / force on the Figure calculaic Its = clectnc E ficld and net electrie b) Dal the nci 'point P Figure Ib, (4 pts) clecininshown in Figure IeIt has unifor c)A straighi wire 0f 2L Iength Iop hulf and Q along boutom half: charge ol +Q nong Datermine thc pcLclecuric licly € at point P: (18 pts)Fogure Ie24

SAME SURNANE STUDEAT NO: when wc want t0 SIGNATURE: differential charges; poiut-like divide ring of charge into spuce. (3 pts) 1-4) Exphin why we in its surrounding ckrlno- Licld al 0 Point / force on the Figure calculaic Its = clectnc E ficld and net electrie b) Dal the nci 'point P Figure Ib, (4 pts) clecinin shown in Figure IeIt has unifor c)A straighi wire 0f 2L Iength Iop hulf and Q along boutom half: charge ol +Q nong Datermine thc pcLclecuric licly € at point P: (18 pts) Fogure Ie 24) Expliin why we don" usc cubic Gaussian Surface



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co Figure 23-35 shows a closed Gaussian surface in the shape of a cube of edge length $2.00 \mathrm{~m}$, with one corner at $x_{1}=5.00 \mathrm{~m}, y_{1}=4.00$ $\mathrm{m}$. The cube lies in a region where the electric field vector is given by $\vec{E}=-3.00 \hat{\mathrm{i}}-4.00 \mathrm{y}^{2} \mathrm{j}+3.00 \hat{\mathrm{k}} \mathrm{N} / \mathrm{C}$, with $y$ in meters. What is the net charge contained by the cube?

Let's say we have a red conducting sphere as you can see in the diagram there in the middle as a radius of A. In a charge of plus Q. And that red conducting sphere is surrounded by empty space, but it's also inside of a larger green conducting sphere. Right? And this fear is hollow. So it has an inner radius of B. And an outer radius of C. Um and this outer sphere, like the Green one, there has no charge on it, no net charge. So let's find the electric field at a few different points. And and to do this, we're gonna be using what's called God's Law, right? And God's Law for electric fields basically says that the integral of the electric field in terms of the area D. A. Is equal to the charge enclosed. So Q. E. N. C. I'm trouble getting to see there right over the constant E. Not right. Which is the primitive itty of free space. So the first thing to know is that the electric field inside of any conductor is always zero. So that means we don't even need God's law for inside the red conductor. This is just going to be zero. And also inside the green conductor. So between the radi I. B and C, those are also going to be zero. We don't even need God's Law for for those because anytime you have any conductor, the electric field inside of is zero. Now between the conductors, this is where we do have to use Galaxies long for Galaxies law. We need a calcium surface. So if we're using between the conductors, let's say we have our read conductor here and we have our inner radius here. So this will be radius B. Right? And then of course we have radius A. Inside. So actually we're going to choose a galaxy in surface in between those. It's just gonna be a circle here and that's going to have a radius R. Right? So what that radius are? We know that. Um because we're dealing with spheres, we we we like know that that D. A. And God's Law. It's just gonna end up as the surface area. Because if we integrate every little segment of like area on the surface of a sphere, that's going to be equal to the surface area of that sphere. So we know that the surface area is four pi R squared. So we can just replace that with four pi R squared Q. And close over any. Not. And then we just divide by four pi R squared, Right? Because we're solving for easy and that's basically to give us an electric field of E equals Q. Enclosed over for pie, E. Not R squared. I just switched the variables in the denominator round because um for pie and like not are constants are obviously going to vary depending on how far your calcium surfaces uh from the from the red line. So this is the electric field between the conductors A. And B. Um Remember it doesn't matter how far you are um from the red sphere it's always just gonna be the charge inside of there because there's no charge in the empty space, all the charges in in that red sphere. Great. So now we can go to outside of the green conductor. So now if we look at the outside of our green conductor we should have a radius of C. We're going to pick a galaxy of surface that's way outside of their. So like this is going to be bigger then the entire diagram. And now this is gonna be our like our now remember that the green conducting sphere has no charge. So even outside of this whole thing, the total charges still enclosed in the red sphere so are enclosed charge is still just cute right? And like actually I'm gonna update this here instead of queuing close, I'm gonna just make that Q. Because that is the same queue that we see inside of there. So what that means is we have the exact same equation as before, right? And even though we are around a bigger sphere are are is also going to be bigger but we still have our surface area for bi r squared over, queuing close over any nut. And then just like we did before we saw for E by dividing by the surface area and then this is going to give us you know the exact same equation where we get Q over for pie, E not r squared right now, if the green one had some charge in it, then the queuing clothes would be the some of those charges, but because zero is the same Q. So let's say that we wanted to graph what this actually actually look like over time. And let's use, I mean it really doesn't matter what um what we use here in terms of the X and y axes. But let's say I wanted to go from, I got a B see and then let's say a value of to see. So to see will be out here somewhere. So first we already know that the field from zero to a Sorry, from yeah, from like 0-80 because Ry is gonna be our electric field in terms of our and this is gonna be our and we know that in between B and C, it's also zero, right? So we know that at like at like a r r electric field is going to be at this max value, right? Because all the charges at a surface, the further than that we get from there is going to decrease. And because r squared is in our denominator is going to a decrease in a curve, not not in a linear way. So we know that it's going to decrease in some manner and then go back down to zero, right? And then it's going to be zero from B to C inside of the uh of the green sphere. And then at like a C. We're going to go up again to that value that we were at like a early thing, kind of draw a dash line until I kind of see where we're like, it would end up and then you just smoothly continue to like decrease, right? It's never quite going to be zero, but it's going to keep my following that trend. And that's how you find the electric field inside to conducting spheres.

For this problem. On the topic of castles law, we have shown a close Gaussian surface that is in the shape of a cube with vigilant, two m One corner of the Cube is at X one equals 5 m and why one equal to four m. And it lies in a region where the electric field vector is given by -3 I -4. Y squared J. Blustery K. Newton's Bakula. We want to know the charge that is contained by this cube. Now none of the constant terms will result in a non zero contribution to the flux. So we'll focus on the wide dependent term only and the non constant term we'll call it E N C is equal to minus for y squared times jihad in si units. Now the face of the cube and y is able to fall has area A. Is equal to four square meters and it faces the positive direction and has a contribution to the flux that is equal to e. Non constant times the area. A Use equal to -4 times for squared times four, Which is equal to -256 Newton meters. The cool um squared the face of the cube located at y is equal to two has the same area and however this one faces the minus J. Had direction and the contribution to the flux minus E N C times A. He is able to minus minus four times two squared times four. And so this contribution to the flux is positive 64. Newton meters. Cool um squared and so therefore the net flux hi Is equal to -256 plus 64. Newton meter. Pretty cool, um squared, which is -100 and 92 Newton meters are cool um squared. And so according to God's law, we have the charge Q. That is enclosed to be absolutely not times five. And so this is if we put in our values 8.85 Times 10 to the -12 in Si units times the flux minus 192 newton meter squared, McCullum squared. And we get the enclosed charge By the Cube to B -1.7 Times 10 to the -9 columns.

Higher driven first part. The positive extra nist church. Uh It's distributed uniformly over the interior surface. Sorry, internal surface let sigma not is surface density. He will be zero before orange, less than fuck he is resident. Are. And a quick feel you can calculate by goes through up and it will be absolutely not. E four pi R square to be four by a square sigma not Eugene in God's sarah. So electric field you will get sigma not upon. Absolutely not absolutely A. of one Our Holy Square. If R. Is greater than A. But less than B. For our movie greater than we we will get sigma not upon absolutely not eight Upon our hearts square foot. Or to be greater than we potential at infinity to zero. And potential at our baby. Sigma not a square upon. Absolutely not art. For art to be greater than we. Okay, and for R. Is greater than a. But less than B potential to be sick or not A square upon. Absolutely not. Absolutely are less B. Here V. Is integration constant. Why continuity? You will get okay potential to be sigma not a square upon absolute, absolute not upon upon ar minus one upon me. Plus sigma not a square upon Absolutely not be for are to be greater than a but less than we four Are each less than a five school to a. That is to be constant. So why continue. You will get potential to be sigma not a square divided by absolutely not into absolute Born upon a -1 upon me. Plastic Menard is square upon. Absolutely not be no second part. Okay positive extraordinary church. It's distributed uniformly over. Internal volume of dialectics. Mhm. A rule note is volume charged and stick And a quick feel is 04. Our age less than eight. And fought R. Is greater than A. But less than we you can write. It is to be absolute absolute absolutely not into accident fought by artist square E. Yeah escort 45 by three R cubed minus a cube are not so electric field you will get Not upon three. Absolute not absolute ar minus. He came upon our script. Okay and fought How is greater than A but less than B. For R. Is greater than B. And a trick fate will be four by trip. Meet you minus a cube divided by it will not upon. Yeah absolutely not fought by our square on simplify he will get a doctorate feel to be VQ minus A Q. A. Rule not upon She absolutely not. Our Square four are greater than B. By integration potential you will get with q minus Q. Apart. She absolutely not our for our to be greater than we. This you can often by integration of a tr. Mhm. Fuck. And why is culture B minus parole not upon three. Absoluteal not absolutely our square fight too plus a cuba are four R. Is less greater than A. But less than we. And why continue to you can major B -8. You Upon three. Absoluteal not. We into a role not it's called the U minus But we will not upon three. Absoluteal not absent The square by two plus cuba be so be. You will get although not upon three. Absolutely not. Absolutely Absolutely. And two B 3 -3 upon B plus. We square way to plus 80 wavy so finally potential will be yeah the U minus. Well not upon three absolute into absolute He square by two plus is where that is the minus rule not a square white to absolute absolute to. Absolutely not. Absolutely For our to be less than eight. So E. R. And fire grab can be plotted. Has given in the problem. That's all. Thanks for watching it.

In this problem I'm drawing the diagram first so just look at it carefully. After that. I will simplify the bedroom but the diagram looks something like this which are yet have gone here. This is a smaller, this is capital are and the charted Q. Inside. And this value is the. Now I can write the value of if or by the square is equal to Q enclosed by in order here which is equal to You. Buy the note plus one by He not integration of our two. It's smaller. I didn't fall by uh report by artist where beard now on integrating and solving I can write that they've had to add. He poured pile Rd squared is equal to two -2 by Alpha. R. D squared by a note. Plus sport by alpa. are described by two. In on the intensity. He does not depend on our when the expression in the parentheses equal to zero. So I can write develop U. Is equal to buy al party square and the value of E. Is equal to well provide to you know, tell the answer.


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