You can easily show that Gauss' law holds for the flux of the electric field of...

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You can easily show that Gauss' law holds for the flux of the electric field of a point charge q at the origin if we pick a spherical Gaussian surface centered about the point charge, i.e. = dor = = () d = 9 Ev 2 Here is the total electric flux, do, = E(F) da, is the differential flux of (F) through the differential area da, of a sphere of radius r. Prove that Gauss' law holds for any arbitrary closed surface by showing that the differential flux of the electric field due to a point charge at the origin through a differential area da, = da, of a sphere of radius is the same as the differential flux through the differential area d = ndA and d is a distance R from the origin of an arbitrary closed surface. Do not assume that n is in the same direction as . See Fig. 2 for details. The differential flux of the electric field through the sphere is given by do, = E(7) d, and the differential flux of the electric field through the arbitrary area is doR = () d. If you can show that the differential fluxes are equal, i.e. do, = do then here is how it proves Gauss' law for the arbitrary closed surface: but $ do, = f dor ST dor = SR q Ep SO = doR and dor = thus q = =(R)- d JSR = E(R)-d. Ep JSR ern dA = dA dR = F daR -da =dar Figure 2: Sphere of radius r with point charge q at its center and differential area da, a distance r from the origin. Arbitrary shape with differential area d at a distance R from the origin. Your job is to show that doR dor. Follow the steps to do so. Do not deviate from these steps and produce an alternate proof (such as the one in the video on the proof of Gauss' Law in the Ch.2 Module in Canvas). (a) (2 pts) Find the electric field of the point charge at r and R, so you need to find (F) and (). Recall that the point charge is located at the origin. Next, write the electric field at R in terms of the electric field at r. You want an expression that looks something like E(R) = a (F). You need to determine a. Note that you cannot divide vectors but you can divide their magnitudes. Hint: a depends on r and R. Once you found a, replace the a in the following expression with what you found: E(R) = a (7). (b) (5 pts) You need to figure out how the differential area d = ndA is related to the differential area dar =dar so that you can write dag in terms of da,. Revisit the Ch.1 lecture notes on flux. There we saw how the tilted area (called the new orientation in the notes) was related to the upright area (called the old orientation in the notes). The relationship between the magnitudes of those two areas in the notes is similar to the relationship between dag and dA. We can think of dar as the upright area or "old orientation" and dA as the tilted area or the "new orientation." It might help if you draw dar and dA as viewed from the edge (so they both look like lines) and note that dA is tilted away from dar by an angle Orn. If you know anything about projections, we're trying to find the magnitude of the projection of d in the radial direction, or, if you like, we're trying to find the radial component of dA. (c) (2 pts) Write dar in term of day. To do this you'll need to make use of the fact that they share the same solid angle in common. The solid angle dQ = sin Oded is the angle subtended by the areas da, and dar. Recall that the differential area of a sphere of radius r is da, = r sin 0d0dy = rdQ. We can likewise write the differential area of a sphere of radius R as daR = R sin @dody = RdQ. Notice that you can write da in terms of da, (since they have the solid angle in common) and you'll get an expression like dag = da,. You need to find . Hint: depends on r and R. Once you find an expression for , replace the in the following equation with what you found: dar = dar. (d) (6 pts) Now show do, do where do, = E(F) da, and dR = () - d (e) (10 pts) Now it's time to use the steps you explored in the proof above to show that Gauss' law holds for a point charge that is not at the center of the sphere. Note that since the charge is not at the center, the electric field will not have the same value all over the sphere: in places where the charge is closer to the surface the field will be stronger and in places where is farther away from the surface the electric field will be weaker. Note that there are not any tilted areas to worry about here so this proof should be a bit simpler than the first. You can easily show that Gauss' law holds for the flux of the electric field of a point charge q at the origin if we pick a spherical Gaussian surface centered about the point charge, i.e. = dor = = () d = 9 Ev 2 Here is the total electric flux, do, = E(F) da, is the differential flux of (F) through the differential area da, of a sphere of radius r. Prove that Gauss' law holds for any arbitrary closed surface by showing that the differential flux of the electric field due to a point charge at the origin through a differential area da, = da, of a sphere of radius is the same as the differential flux through the differential area d = ndA and d is a distance R from the origin of an arbitrary closed surface. Do not assume that n is in the same direction as . See Fig. 2 for details. The differential flux of the electric field through the sphere is given by do, = E(7) d, and the differential flux of the electric field through the arbitrary area is doR = () d. If you can show that the differential fluxes are equal, i.e. do, = do then here is how it proves Gauss' law for the arbitrary closed surface: but $ do, = f dor ST dor = SR q Ep SO = doR and dor = thus q = =(R)- d JSR = E(R)-d. Ep JSR ern dA = dA dR = F daR -da =dar Figure 2: Sphere of radius r with point charge q at its center and differential area da, a distance r from the origin. Arbitrary shape with differential area d at a distance R from the origin. Your job is to show that doR dor. Follow the steps to do so. Do not deviate from these steps and produce an alternate proof (such as the one in the video on the proof of Gauss' Law in the Ch.2 Module in Canvas). (a) (2 pts) Find the electric field of the point charge at r and R, so you need to find (F) and (). Recall that the point charge is located at the origin. Next, write the electric field at R in terms of the electric field at r. You want an expression that looks something like E(R) = a (F). You need to determine a. Note that you cannot divide vectors but you can divide their magnitudes. Hint: a depends on r and R. Once you found a, replace the a in the following expression with what you found: E(R) = a (7). (b) (5 pts) You need to figure out how the differential area d = ndA is related to the differential area dar =dar so that you can write dag in terms of da,. Revisit the Ch.1 lecture notes on flux. There we saw how the tilted area (called the new orientation in the notes) was related to the upright area (called the old orientation in the notes). The relationship between the magnitudes of those two areas in the notes is similar to the relationship between dag and dA. We can think of dar as the upright area or "old orientation" and dA as the tilted area or the "new orientation." It might help if you draw dar and dA as viewed from the edge (so they both look like lines) and note that dA is tilted away from dar by an angle Orn. If you know anything about projections, we're trying to find the magnitude of the projection of d in the radial direction, or, if you like, we're trying to find the radial component of dA. (c) (2 pts) Write dar in term of day. To do this you'll need to make use of the fact that they share the same solid angle in common. The solid angle dQ = sin Oded is the angle subtended by the areas da, and dar. Recall that the differential area of a sphere of radius r is da, = r sin 0d0dy = rdQ. We can likewise write the differential area of a sphere of radius R as daR = R sin @dody = RdQ. Notice that you can write da in terms of da, (since they have the solid angle in common) and you'll get an expression like dag = da,. You need to find . Hint: depends on r and R. Once you find an expression for , replace the in the following equation with what you found: dar = dar. (d) (6 pts) Now show do, do where do, = E(F) da, and dR = () - d (e) (10 pts) Now it's time to use the steps you explored in the proof above to show that Gauss' law holds for a point charge that is not at the center of the sphere. Note that since the charge is not at the center, the electric field will not have the same value all over the sphere: in places where the charge is closer to the surface the field will be stronger and in places where is farther away from the surface the electric field will be weaker. Note that there are not any tilted areas to worry about here so this proof should be a bit simpler than the first.