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Welcome to digital tea with mr.
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E, where we will see what's brewing in the world of physics education.
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In this particular problem, we have a charge of 15 times 10 to minus 9th kulams.
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That's in the center of a hollow shell.
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And then we have a charge of a negative 22 times 10 to the negative 9th kulams that is evenly distributed on the exterior side of the shell.
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We know that our hollow shell has a radius of 0 .1 meters, and we are tasked with finding the value of the electric field at three locations, which i have noted as r1, r2, and r3.
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So, in each particular case, you would be considering a gaussian sphere at the location of each of the r's.
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So a gaussian sphere would make like a circle around there.
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So if i'm writing the equation for the electric field at the location of r1, i would say that the flux through our gaussian sphere should be e times a, assuming their perpendicular, of course, which is also equal to the charge enclosed over epsilon sub -knot.
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I'll add epsilon sub -not over to her.
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Things that we know over here, 8 .85 times 10 to the minus 12.
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So now, in solving for the e1, we'll say that e1 is equal to the charge enclosed, divided by epsilon sub not times the area.
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And the area for that gaussian surface can be given as 4 pi r1 squared.
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It's important to realize what charge enclosed means.
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Charge and close would be look around that blue gaussian surface and what charge is in the middle.
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It's simply put as 15 times 10 to minus 9th coulams.
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So if we take our 15 times 10 to minus 9th, and we divide it by the quantity of 8 .85 times 10 .m.
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Minus 12 times 4 times pi times the 0 .022 squared.
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That will solve for e1...