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PHGN361 Exam 1: NAME 1. (a) Using the integral form of Gauss's Law, find the electric field due to a sphere of charge, with radius R and charge density Ar, where A is a constant. Find the field both

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An insulating sphere of radius 2 meters contains 15 micro Klum's of electric charge uniformly distributed throughout the volume of the sphere what is the electric field 1.5 meters away from the standard of the sphere how can we do this well let×39’s beginning by drawing a picture, so this is going to Bethe sphere, and it's going to have radius which we×39’ll call R and the charges distributed throughout this year if you have a conducting sphere or aspherical conductor the positive charge will spread out on the surface of the spherical conductor which means that there will be no electric field inside the conductor, but we have an insulated din sphere, so the charge is distributed throughout the sphere which means that there exists an electric field inside the sphere so let's draw an imaginary Gaussian surface represented by this yellow sphere and R is going to represent the distance between the center and the Gaussian surface so pecan calculate the electric field at that point now let's use Gauss's law to come up with an equation for the electric field inside the sphere so Glass×39’s law states that the electric flux which is the product of the electric field times the area of the Gaussian surface is equal to the charge enclosed by that surface divided by epsilon sub naught SOI can calculate the charge that is enclosed a simple way to do that is house the volume ratio of the Gaussian surface compared to the actual sphere soothe charge enclosed by the Gaussian surface in yellow is equal to the total charge times the volume of the Gaussian surface which IN×39’ll write as VG divided by the volume of the entire surface which I'll leave as SSO QT I'm just going to write Q for that×39’s going to be the total charge of the insulating sphere the volume of the Gaussian surface is going to be 4/3 PI Cube, so we need to use the radius of the Gaussian surface or the Gaussian sheathe volume of the actual sphere is 4/3pi but using the green R or capital Rs owe can cancel 4/3 pi and so now we have the expression for the charge enclosed by the Gaussian sphere so let's replace to this particular Q with Q times R cube over R cube now the area of the Gaussian surface is simply the surface area chichis 4 pi times R squared, so this is going to be equal to Q times R cube divided by big R 2 so notice that we can cancel woof the blue hours so what we have left at this point is the electric field times 4 pi is equal to Q times R divided by r cube so now let's divide both sides by 4 pi and let×39’s not forget epsilon that should be in the bottom as Wells the electric field is equal to one over 4pi times epsilon sub naught times R divided by R cubed now one of the four pi epsilon sub naught that×39;equal to K, so we can write the equationlike this the electric field is equal TOK times Q that's the total charge in thesphere times the radius of the Gaussian surface inside the sphere divided by the radius of the actual sphere cubed so let's go ahead and use this formulate...

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