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A hollow; conducting sphere with an outer radius of 0.240 m and an inner radius of 0.200 m has & uniform surface density of +6.37 10 echage A charge of -0.900 p...

Question

A hollow; conducting sphere with an outer radius of 0.240 m and an inner radius of 0.200 m has & uniform surface density of +6.37 10 echage A charge of -0.900 pC is now introduced into the cavity inside the sphere

A hollow; conducting sphere with an outer radius of 0.240 m and an inner radius of 0.200 m has & uniform surface density of +6.37 10 echage A charge of -0.900 pC is now introduced into the cavity inside the sphere



Answers

A hollow, conducting sphere with an outer radius of 0.250 $\mathrm{m}$ and an inner radius of 0.200 $\mathrm{m}$ has a uniform surface charge density of $+6.37 \times 10^{-6} \mathrm{C} / \mathrm{m}^{2} .$ A charge of $-0.500 \mu \mathrm{C}$ is now introduced into the cavity inside the sphere. (a) What is the new charge density on the outside of the sphere? (b) Calculate the strength of the electric field just outside the sphere. (c) What is the electric flux through a spherical surface just inside the inner surface of the sphere?

04 per day. We know that the initial charge, essentially the initial positive charge on the outer surface of the conductor would be equal to a uniform's surface Charge density times the surface area. This is equaling Sigma four pi r squared. And so this is gonna be equaling for pie. Multi applied by the universe form surface charge density 6.37 times 10 to the negative six Ah, cool arms per square meter multiplied by the radius squared 0.250 meters Quantity squared and we find that the initial charge is gonna be equaling 5.0 times 10 to the negative six cool apps now and ah, the a negative 0.500 micro Coolum charges introduced into the cavity So now the outer charge is gonna be equaling two. Now it would be 5.0 micro cool arms minus 0.500 Micro cool arms. This is giving us 4.50 micro cool ums. And so now the surface charge density has changed and this would be the charge Q divided by the surface area and so this would be equaling for pie multiplied by here. 0.250 meters quantity squared. And then for the numerator 4.50 times 10 to the negative Sixth Cool arms. This is giving us 5.73 times 10 to the negative Sixth cool arms per square meter. This would be our answer for part a four part B. Ah, we're going to use Gauss is lop. Oh God, kAOS is law. We're here. The electric field is equaling the electric flux divided by the area area A And so this is gonna be equaling the charge Divided by Absalon, not time's A and this is equaling the charge. Divided by Absalon not times four pi r squared. And so we can then find the electric field. By substituting for part B, this would be equaling for the denominator 8.85 times 10 to the negative 12. This would be sent a cool I'm squared per newton per meter squared. This would be multiplied by four pi r squared. So point 250 meters quantity squared and then for the numerator we have 4.50 times 10 to the negative sixth cool arms. This is equaling 6.47 times 10 to the positive Fifth Newtons per Coolum. This would be your answer for part B four parts. See, now we're going to use gas. Is law again to find the electric flux? So the electric flux is going to be equaling the total enclosed charge divided by Absalon? Not and so we can simply substitute. This would be equaling negative points 500 times 10 to the negative. Sixth. Cool arms. This would be divided. Bye, Absalon, Not 8.85 times 10 to the negative. 12th. Cool. I'm squared per Newton per meter squared. And so we find that the electric flux is equaling. Negative 5.65 times 10 to the fourth Newton meters squared per cool up. This would be our final answer for Part C and our final answer for part B. That is the end of the solution thing.

Hi. In the Guinean problem, there is ah hollow Mcd Ali charged sphere having in a and how to radio as given for outer radios. This is Capitola is able to 0.250 m and the inner radius is smaller. Is equal to 0.200 m. The surface charge density off the outer Steris outer surface surface. George density off outer surface is sigma is equal to plus 6.37 into tended bar minus six. Ghulam for meter square No. A point charge having a value off cute is equal to minus 0.500 My curriculum has been kept within this hollow cell. 0.500 micro column. This negative charge will induce ah positive charge at the inner surface off this hollow sphere. And as the process off, charging by induction always creates equal and opposite charges on the opposite faces off the conductor over which we are inducing discharge. So as a charge off plus 0.5 micro Coolum is in used at the inner surface, so the same charge, but with negative sign will be induced at its outer surface also so a negative charge zero minus 0.500 Michael Cullen will be induced at the outer surface off this hollow sphere. No. In the first part of the problem, we have to find the new charge density at the outer surface, for which, first of all, we will find the increase to charge density at the outer surface so increased and it will be negative off course. Increase the negative charge Density at the outer surface is given by Delta Sigma. The change in density is equal to the charge induced over the outer surface. Let it take you dash divided by the rages divided by the surface area off outer surface for five yards square. So putting these values here this is minus 0.500 multiplied by attended for minus six Golan 45 in four into 3.14 were declared by 0.250 meter square. So finally this increased charge density comes out Toby minus 0.637 Ghulam meters squared. Hence the new charge density surface charge density at the outer surfaces, so milk charge density surface charge density off course Sigma Dash is able to Sigma plus Delta Sigma. So the sigma dashes will, too. Initially it waas plus 6.37 multiplied by 10 days. Four minus six Ghulam for meter square. And for this increased. Yes. This is okay. Yeah. On escape. This is 0.67 into 10. Dish bar minus six. Cool. Um, per meter square. So that increased one will be the old one, which is this minus as it is having a negative sign toe minus 0.637 into tended bar minus six school. Um, but meter squared. So the final answer for the new surface or density will be fine. 0.37 5.73 5.73 into 10. Dish for minus six column for me to re square. And it becomes the answer for the first part off the problem. No. In the second part of the problem, we have to find electric clubs at a point just inside this follows here. And that electric flux will be purely because off the charge kept within this follows fear. Using courses law, the flux will begin by one. By absolute, not times the charge enclosed And as you know, the charge. Enclosed is only this minus 0.500 micro column, so the flux will be minus 0.500 in potential minus six. Column divided by permitted bitty 8.854 into 10. Dish par minus 12. Lam Square Burn. You can meet a square. So finally, the answer for this electric flux comes out to be minus five point 65 into 10 days of Parma, plus four Newton meters square part Hola. And this is the answer for the second part off the problem. Thank you.

So to begin this problem, you must take a look at the gravitational force acting on this fear. So we know that gravitational force is equal to f equals M g, which is the mass, the object times, its gravitational acceleration. So if we substitute the key, the mass every given and 9.81 for the gravitational acceleration on Earth, we can compute the force. You can compute the gravitational force on the sphere. So if you do that, we're gonna get five times 10 to the power negative, seventh K, g times 9.1. So if you do that, we're gonna get that The force of gravity gravitational force is equal to 4.9 time sent to the negative six Newton. So now we must calculate the electrical force atmosphere, which is which can be calculated by this equation. So F is equal to the charge times the electric fuel strength. So the electric field off the electric field strength off the sheet. Since we don't know this one that was not given to us, we must calculate it. So to calculate that, let's do that over here. So the electric field is equal to this over to times the primitive ity of free space. So if you plug in the values ever given, you can find that we can find out that this so this so we can find out that this is equal to Since these numbers are all given, we know the surface charge density is given in the question which is equal to eight peka peka Cool Mr Pereira Meter square. So if you do that, that's eight PICO is times 10 to the so it's only let me do that. Here, let me do it over your eight times 10 to the negative 12 since its PICO converted back into columns per meter square over to times a primitive ity of free space, which is 8.85 times tend to the negative 12. So if you compute this, we're getting it that the electric field strength is equal to 0.50 point for by 18 Newtons per cool standard units Always good. So now if you go back into the earth electric electrical force, uh, equation over here on this side, we can now use the electrical fuel strength that we completed over here into this equation. So we know that the charge which has given to us is three micro columns, uh, positive three micro plumes. So it's 10 times negative. 63 times at 10 to the power pack of six times the, uh, the electric few friends that we just calculated, which is using a point for 518 So if you just put that into a calculator, we're gonna get that Hold on. Let me just compute this one real quick. I have not read on this. Let me just make the multiple sign. Very clear. So it's three times sent to the negative. Six times zero point 4518 Excuse us. A value off 1.3 55 or list 15 times 10 to the negative six Newtons. So you know that the net force is in a downward direction. So if you seem that Bicol zero at the sheep so the gravitational potential at that point is equal to mg times y or height, so potential, the potential energy at that point is you equals M g times and I which I don't think we need to write down. So for the electric force work done from moving from moving from point A to B is this or if he work done is equal to work done from the point A to B, he is equal to charge times potential difference a minus bi. So that's the potential difference between the 22 points. So if we isolate for the variable cues to find the potential difference, we get that we can just divide both sides by cute. So if you just do that, you can get rid of this. Came over here. And now we can solve for this potential difference which is equal to sorry. So now we can solve for the potential difference. So here, uh, the K is its initial Kontic energy. Hello himself. So So so at point E to be at the sheet and point B is at the distance. Why from the sheet. So if you take that, this potential difference is equal to electrical few friends at why so then the potential energy is you be sequel to negative e y you like. So now if you apply conservation of energy law, it states that the initial potential energy UK plus the initial kind tick potential energy. The initial kind of energy is equal to the final potential energy plus the final Kontic energy Assuming that mirror, this is taking place in ideal conditions and no energy is lost. So we know that the Cayenne tick energy at the start is equal to zero because it starts at rest. So there is no que in this scenario. So if you saw for Katie by moving the the potential energy at point B to the other side of you can get that you a vine issue be he is equal to the kind tick energy and couldn't be so if he further uh, expand on this expression, we're going to get that potential energy at a which we already discussed was MG times in height. So the mg times Why? Why one minus y two? So this is this is a starting height and this is the final height minus. So the potential energy you be completely electrical so minus each you and it's the same thing. So it's why one minus y two consists the difference. So now if we can just substitute since you already know all the values for these return just substituted straight into the calculator. So if you do that or we can go a step further and simplifying this expression so we can do that by factoring out the common common expressions that you're Why one minus y two since Italian. Both. So if you just factor them out, we're gonna get the M G minus C Cube times. Why one? I asked why to is equal to the kind of dick energy at Point B me to find the Cayenne tick energy to find the velocity. So let's keep going. So we know we're in the right direction. So now if you just substitute the values that we have for MGI, which ISS 4.9 times 10 to the power bi calculated this before. So it's the first one 4.9 times 10 to the power negative six. So this gravitational force that we calculated up here mg negative but isn't negative. Six? Yep, minus e que, which we calculated right here. So it's one point 35 time sent to the negative six times that difference, which is just equal to, uh, 0.4 meters. This is straight from the question. So now if you put this into a calculator. We're gonna get that. Hold on. Let me just do this one real quick since I haven't read on this. So 1.35 times sent to the making of six and all of this is one compliant by 0.4. If you do that, Reagan, get that the kind Take energy at point B is equal to, um one point for two times 10 to the negative six and you guessed it. It's type unit juice. So now that we have the kinds of energy at point B, we can use the other former for kind take energy, which is one half MV squared and solved for the we in this case. So if you isolate all the variables for B, you're gonna get an expression that looks something like this. So two times cayenne tick energy over the mass and just square with that whole thing since area square, Uh, for the key to the velocity in this top expression. So now, since we know all these values are just what they're in so two times one point for two time center negative six over the mass, which is five times into negative. Seven. Double check that. Yep. The message. Five times into the negative. Seven on this whole thing is in a square root to know if you compute this in our calculator, we're gonna get that figure. Get the final answer for the velocity. So it's two times the Kontic energy 1.42 times 10 to the negative six over five times sent to them 10 to the power of negative seven. Hold on. I did not write negative there. So 97. And this is equal to 2.1 2.7 Sorry. 2.7 years per second squared. That's it for this one. Thank you for listening.

For this problem on the topic of electric fields, we have a hollow metal sphere that has in an outer radius of 20 and 30 centimeters respectively. And in the figure we can see that a solid metal sphere which has a radius of 10 centimeters, is located at the center of the hollow sphere at a point P. A distance of 15 centimeters from the sphere or from the center we have an electric field. You want to be one times 10 to the four newton speculum that is directed radial inward at a different point Q. That is 35 centimeters from the center. We have an electric field that is one times 10 to the four newton speculum that actually readily outward. We want to find the total charge on the surface of the atmosphere as the as well as the total charge on the inner surface of the hollow sphere and the outer surface of the hollow sphere. Now we let delta be delta equal to 10 centimeters. Be the radius of the solid sphere, the distance between the solid sphere and the inner part of the hollow sphere, as well as the thickness of the hollow sphere. R. P. Here is 15 centimeters and this is the distance from the center to point P. And R Q. Is 35 centimeters. Which is the distance from the center two point Q. Now the calcium surfaces RP encloses the charge on the inner sphere. So E one times four pi R. P squared is equal to the charge enclosed over absolute or not. And so the charge on the inner sphere Q. He's able to absolutely not times four pi E one R P squared. So if we substitute our values into this, this is eight 0.85 times 10 to the minus 12 in sa units times four pi times the electric field which is minus 10,000 newtons speculum. And the distance R p 0.15 m squared. This gives us the charge on the inner sphere, B minus 25 no no columns for part B. For the electric field inside the shell to be zero. The charge on the inner surface of the shell must be equal to the negative of the charge on the inner sphere. So E. Is equal to the enclosed charge of uh four pi absolutely not. R squared is equal to zero or Q. One is equal to minus Q. And so the charge on the inner surface of the shell, Q one is therefore positive. 25 no no columns for part C. The calcium surfaces are Q. Is equal to 35 centimeters from the centre, encloses the inner charge and the charge in the shell. So E two times four pi R q squared is able to the enclosed charge over absolutely not. And this is Q plus the charge on the shell divided by absolute or not. So this means Q plus the charge on the shell. Q. Shell is equal to absolutely not terms for pie E. Two, our Q squared. And the charge on the shell is the sum of the charge on the inner and outer surfaces of the shell, which is key one plus que not. And so we charge on the outer surface of the shell. You're not is equal to the charge on the shell minus Q. One which is the charge on the shell minus minus Q. Which is the charge in the shell plus Q. Which is absolutely not full. Pie times. E to times are Q squared. And so if we substitute our values into this, we get Q. O. To be 8.85 times 10 to the minus 12. That's cool on squared, But newton meters squared times four pi times one times 10 to the powerful Newton speculum time, zero point 35 m squared, which gives the charge of the outer shell to be zero point 136 micro columns. Okay?


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