a-charge-of-3-q-is-transferred-to-a-solid-metal-sphere-of-radius-r-where-will-this-excess-charge-res

Question

A charge of $-3 Q$ is transferred to a solid metal sphere of radius $r .$ Where will this excess charge reside?
(A) $-Q$ at the center, and $-2 Q$ on the outer surface
(B) $-3 Q$ at the center
(C) $-3 Q$ on the outer surface
(D) $-Q$ at the center, $-Q$ in a ring of radius $\frac{1}{2} r,$ and $-Q$ on the outer surface

Caitlin Mccabe · Accepted Answer

So for the problem we&#x27;re working with today we have been given a charge of negative three que and we&#x27;re told that has been placed onto a metal sphere solid and has a radius of our. So it looks somewhat like this. And the important thing to take note of with this is that all the charges we have been given are negative, and we know that when we have, like, charges that they will repel against each other. So that means that each of the negative charges were working with are going to want to get as far away from each other as possible. The Olympic force between them will propel the two. And so as such, they will want to move away. However, they still must be contained within the surface of this solid metal sphere. They can&#x27;t simply fly off into the sky or leave from it. And so such in order to maximize their distance, will still staying on this fear. They&#x27;re going to be going towards the surface of the sphere in a formacion that is somewhat like this, where they will be able to maximize the distance between each of these charges while still remaining contained within the body. This fear and so as such, the answer to our question will be see, but the negative three Q will be on the outer surface of the sphere as this is a point at which they will maximize their distance between one another while still being contained.

Ze-Han Lee · Answer

So in this problem, we have a solid metal sphere, and it would put charge negative. Three Q onto this fear. Uh, then we can see that basically all the charges will be distributed on the surface off this sphere. That&#x27;s because a ll the charges they have, ah, repulsive and directions on each other. So they will choose. Ah, food. This is for the furthest distance from each other. So that results this situation. So all the charges, they reside on the surface of this fear. All right. There is no charging side.

Mayukh Banik · Answer

here. We have negative charge at the center, and we have God was discharge on the spherical shell. So the direction of the electric field is like this and electric fields. He&#x27;s statically symmetric and Boyens ready of the out Lord outside the system.

Kayla Gephart · Answer

So here we have spheres A, B and C A has a charge of minus two. Q. B has a charge of minus. Q and see has a charge of plus four. Q. So first Spheres A and C are touch together, and so the charges on them are going to spread out as much as they can. And so the positives and the negatives that are equal are going to cancel out. And then sphere is going to get Ah, plus one charge, and severe C is going to get plus one charge. That&#x27;s because the total charge between the two spheres before they were touched is to Q positive. Two Q. So when they&#x27;re touched, they&#x27;re going to gain equal amount of charge, since charges can flow through freely through these metal spheres, then chart spheres C and B are touched, and they are going to both become neutral, since, together they have a net charge of zero. Q. They don&#x27;t have any net charge, and so if they&#x27;re touched, then the positive charge is going to cancel with the negative charge, and they will both be electrically neutral. And so at the end, Sphere B is going to be neutral

Vishal Gupta · Answer

Hi. In the given problem, there is a conducting spherical shell having in the radius as a and outer radios as B. This is the outer radios and this is the inner radius A No. It carries a net charge Cube capital que So all of the charge will spread over the outer surface only as the shell is conducting in nature. So all the charge will be accumulated on Lee at the outer surface. Now a point charge small que has been kept at its center. This small charge Q kept at the center will induce a negative charge here and the inner surface oficial. So balance this negative induce charge a charge plus you will be shifted toe the opposite surface off the shell means outer surface off the shelf. So overall the inner surface will be having a charge minus you at inner surface charge piece minus you. Hence surface charge density will be sigma at inner surface physical toe net charge on the surface, which is minus you. Divided by the surface area, which is full by a square is the radius. And this is the answer for the first part of the question. Now, in the second part of the question, we have to find the surface charge density at the outer surface. So at the outer surface, the net charge is kill les que So the surface charge density sigma oh, at the outer surface becomes the total charge, which is Q plus Q divided by the surface area, which is four by be sweat. And it becomes the answer for the second part off the question. Thank you.

Shareef Jackson · Answer

let&#x27;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&#x27;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&#x27;s find the electric field at a few different points. And and to do this, we&#x27;re gonna be using what&#x27;s called God&#x27;s Law, right? And God&#x27;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&#x27;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&#x27;t even need God&#x27;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&#x27;t even need God&#x27;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&#x27;re using between the conductors, let&#x27;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&#x27;re going to choose a galaxy in surface in between those. It&#x27;s just gonna be a circle here and that&#x27;s going to have a radius R. Right? So what that radius are? We know that. Um because we&#x27;re dealing with spheres, we we we like know that that D. A. And God&#x27;s Law. It&#x27;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&#x27;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&#x27;re solving for easy and that&#x27;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&#x27;t matter how far you are um from the red sphere it&#x27;s always just gonna be the charge inside of there because there&#x27;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&#x27;re going to pick a galaxy of surface that&#x27;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&#x27;m gonna update this here instead of queuing close, I&#x27;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&#x27;s say that we wanted to graph what this actually actually look like over time. And let&#x27;s use, I mean it really doesn&#x27;t matter what um what we use here in terms of the X and y axes. But let&#x27;s say I wanted to go from, I got a B see and then let&#x27;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&#x27;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&#x27;s going to decrease in some manner and then go back down to zero, right? And then it&#x27;s going to be zero from B to C inside of the uh of the green sphere. And then at like a C. We&#x27;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&#x27;re like, it would end up and then you just smoothly continue to like decrease, right? It&#x27;s never quite going to be zero, but it&#x27;s going to keep my following that trend. And that&#x27;s how you find the electric field inside to conducting spheres.

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How far apart are two charges $\left(q_{1}=8 \tim…

Problem 7 Hard Difficulty

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