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Ch 22: Gauss' Law

Chapter 22, Problem 19

A hollow, conducting sphere with an outer radius of 0.250 m and an inner radius of 0.200 m has a uniform surface charge density of +6.37×10−6 C/m2. A charge of −0.500 μC is now introduced at the center of the cavity inside the sphere. (a) What is the new charge density on the outside of the sphere?

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Welcome back everybody. We have a point charge that is placed inside of the center of a hollow metallic spherical shell. And we're told a couple of things about this situation. We are told that the hollow shell initially carries a charge density of 18 micro columns per meter squared or 18 times 10 to the negative six columns per meter squared. We're told that it has an outer radius Of cm or .35 m. And we are tasked with finding what the new surface charge density will be after this point charge has been introduced. Now, in order to figure this out, we're gonna need to use this equation right here that the charge is equal to the charge density times the surface area. Now, in order to figure out our new charge density, we have to figure out what the new charge is after this little charge is introduced. Well before even tackling that, we have to figure out what the inish in charge of our shell was. Now in order to figure out the initial charge of our cell. We're just going to use this formula right here. So, our initial charge density of our shell is 18 times 10 to the negative six times the surface area of a sphere. This is just going to be four pi times the radius, which is 40.35 squared When you plug all of this into your calculator, you get that the charge of the shell is 2. times 10 to the negative fifth column before the internal charge is introduced in order to calculate our new net charge. Now we're going to have the charge of the shell minus the charge of our point charge which is right here. Fifth, negative 15 micro columns. So this is equal to 2.77 times 10 to the negative fifth minus uh 15 times 10 to the negative six columns which gives us a new charge for the sphere of 1.27 times 10 to the negative fifth columns. Now we are ready to go ahead and calculate our new charge density in order to do that. I'm just going to use this formula and divide both sides by a you'll see that the areas cancel out and we get that our new charge density is equal to our new charge divided by that same surface area. So we have 1.27 times 10 to the negative fifth columns divided by the same area of four pi times 0.35 squared. And when you plug all this in your calculator, you get that. Our new charge density is 8.25 times 10 to the negative six columns per meter squared or 8. micro grams per meter squared, which corresponds to our final answer. Choice of a Thank you all so much for watching. Hope this video helped. We will see you all in the next one
Related Practice
Textbook Question
Some planetary scientists have suggested that the planet Mars has an electric field somewhat similar to that of the earth, producing a net electric flux of −3.63×1016 N·m2/C at the planet's surface. Calculate: (a) the total electric charge on the planet;
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Textbook Question
Some planetary scientists have suggested that the planet Mars has an electric field somewhat similar to that of the earth, producing a net electric flux of −3.63×1016 N·m2/C at the planet's surface. Calculate: (b) the electric field at the planet's surface (refer to the astronomical data inside the back cover);
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Textbook Question
Some planetary scientists have suggested that the planet Mars has an electric field somewhat similar to that of the earth, producing a net electric flux of −3.63×1016 N·m2/C at the planet's surface. Calculate:(c) the charge density on Mars, assuming all the charge is uniformly distributed over the planet's surface.
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Textbook Question
A hollow, conducting sphere with an outer radius of 0.250 m and an inner radius of 0.200 m has a uniform surface charge density of +6.37×10−6 C/m2. A charge of −0.500 μC is now introduced at the center of the cavity inside the sphere. (b) Calculate the strength of the electric field just outside the sphere?
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Textbook Question
A hollow, conducting sphere with an outer radius of 0.250 m and an inner radius of 0.200 m has a uniform surface charge density of +6.37×10−6 C/m2. A charge of −0.500 μC is now introduced at the center of the cavity inside the sphere. (c) What is the electric flux through a spherical surface just inside the inner surface of the sphere?
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Textbook Question
A flat sheet of paper of area 0.250 m2 is oriented so that the normal to the sheet is at an angle of 60° to a uniform electric field of magnitude 14 N/C. (a) Find the magnitude of the electric flux through the sheet. (b) Does the answer to part (a) depend on the shape of the sheet? Why or why not?
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