A 20-cm-radius ball is uniformly charged to 80 nC. (b) How much c...

Question

A 20-cm-radius ball is uniformly charged to 80 nC. (b) How much charge is enclosed by spheres of radii 5, 10, and 20 cm?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem, charged organic matter is used to cast a sphere of radius 125 millimeters and has a total charge of 60 nano columns. Assuming that there is even charge distribution, calculate the charge enclosed by a spherical surface with a radius of I 20 millimeters and I I 60 millimeters. Awesome. So that's our end goal is we're trying to calculate the charge enclosed by a spherical surface with a radius I 20 millimeters and I I 60 millimeters. So we're trying to find two different answers. So we're given some multiple choice answers here and parts A through C are all in the same units of nano coulombs and it looks like D is in micro coulombs. So let's read them off to see what our final answer pair might be. A is 0.25 and 6.6 B is 66.6 C is 0.25 and 8.5. And D is 7.0 and 6.4. OK. So first off, let us assume that the sphere is cast with organic matter and is evenly charged. Let us also assume that the distribution of charge is spiritually symmetric. The spherical surface of radii 20 millimeters and 60 millimeters are Gaussian surfaces which match the charge distribution symmetry. Therefore, this means that the electric field E is perpendicular to the Gaussian surfaces. Note that the electric field is also equal for all Gaussian surfaces. The charge is enclosed by the spherical surfaces is found by recalling and using the charge equation which states that Q is equal to row multiplied by V where row is the volume charged density of the sphere cast with organic matter. And V is the volume note that we can find row by using the following equation row is equal to Q divided by V. But we need to note that the volume of a sphere is equal to four thirds multiplied by pi multiplied by radius. Cute. Therefore, we need to write that row is he equal to Q divided by four thirds multiplied by pi multiplied by radius cubed. And note that we are given the value of Q which is equal to 60 nano cools, which is the same when we convert it to cools 60 multiplied by 10 to the power of negative nine cools. And that the radius is 125 millimeters which is 125 multiplied by 10 to the power of negative 3 m. Ok. We need to do a quick conversion because we need all the units to cancel out. So it's better to have them in coom coom and meters instead of nano coulombs and millimeters, which will make more sense in a second. So now we can plug in all of our known variables and sulfur row. So row is equal to 60 multiplied by 10 to the power of negative nine columns, divided by four thirds multiplied by pi multiplied by 125 multiplied by 10 to the power of negative 3 m cubed. Awesome. So when we plug that into a calculator, we should get 7.3338 multiplied by 10 to the power of negative six. But let us round the three decimal points. So when we do that, we will get 7.334 multiplied by 10 to the power of negative six columns per meters. Cute. And note that because the sphere is cast with the charged organic matter and is uniformly charged, we can write that Q is equal to row multiplied by V which is also equal to row multiplied by four thirds or I should say or multiplied by pi divided by three multiplied by R cubed which is radius cube. Now we can solve for part I and plug in R equals 20 millimeters. So let's do that. Q is equal to 7.334 multiplied by 10 to the power of negative six cool thems per meters cubed multiplied by four pi divided by three multiplied by our radius value which we need to make sure we convert millimeters to meters. So 20 multiplied by 10 to the power of negative 3 m. Cute is equal to 2.458 multiplied by 10 to the power of negative 10 cos which when we convert this to NANOS, we should get 0.25 nano coulombs. And let's box this in green because this is our final answer. For part I OK. So the charge enclosed by a spherical surface of radius 20 millimeters is 0.25 nano coons. Awesome. So now to solve for part I I, we need to plug in here, we can quickly write. So this is part I so to solve for I I, we just need to plug in R equals 60 millimeters. So following the same process as above, Q is equal to 7.334 multiplied by 10 to the power of negative six columns per meter cubed multiplied by four pi divided by three, multiplied by 60 multiplied by 10 to the power negative three. Do not forget to convert millimeters to meters cubed. So when we plug that into a calculator, we should get 6.636 multiplied by 10 to the power of negative nine Coombs which when we round that up or when we did the round. So when we convert that to nanometers, we'll get 6.6 nano cos and that is our final answer. For part I I array we did it. So the charge enclosed by a spherical surface of radius 60 millimeters is 6.6 nano columns. So looking at so scrolling back up to look at our multiple choice answers, that means the correct answer has to be the letter A A is I is 0.25 Nano Coulombs and I I is 6.6 nano columns. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

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A 20-cm-radius ball is uniformly charged to 80 nC. (b) How much charge is enclosed by spheres of radii 5, 10, and 20 cm?

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