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Current and current density-solutions (2)

Current and current density-solutions (2) - gilvin(jg47854...

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gilvin (jg47854) - 9: Electric potential This print-out should have 22 questions. 2 - meyers - (21235) 1 1 Multiple-choice questions may continue on the next column or page - find all choices before answering. This homework is due Friday, March 4, at midnight Tucson time. 001 (part 1 of 4) 10.0 points Consider a sphere with radius R and charge Q Q r 1 r 2 L. 0 R r 1 r Y . 0 R r 1 and the following graphs: 1 r 2 S. 0 R r 1 r M. P . 0 R Q. 0 R r 2 r 1 r 2 r 0 R r 1 r 2 G. 0 R r 1 r Z . 0 R r 1 r X . 0 R r Which diagram describes the electric field vs radial distance [E(r) function] for a con- ducting sphere? 1. M 2. L 3. Q
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gilvin (jg47854) - 9: Electric potential 2 - meyers - (21235) 2 gaussian surface r < R, concentric with the 4. X 5. Y 6. Z 7. G correct 8. S 9. P Explanation: The electric field for R < r with the sphere conducting and/or uniformly non-conducting: Because the charge distri- bution is spherically symmetric, we select a spherical gaussian surface of radius R < r, concentric with the conducting sphere. The electric field due to the conducting sphere is directed radially outward by symmetry and is therefore normal to the surface at every point. Thus, E is parallel to dA at each point. There- fore E · dA = E dA and Gauss ’s law, where E is constant everywhere on the surface, gives Φ E = E · dA = E dA = E dA ( conducting sphere. To apply Gauss’s law in this situation, we realize that there is no charge within the gaussian surface (q in = 0), which implies that E = 0 , where r < R . (2) E 1 r 2 E G. 0 R r 002 (part 2 of 4) 10.0 points Which diagram describes the electric field vs radial distance [E(r) function] for a uniformly charged non-conducting sphere? 1. X 2. P correct 3. L 4. S 5. Z 6. G 7. Q = E 4 π r 2 ) = q in ϵ 0 , 8. Y where we have used the fact that the surface area of a sphere A = 4 π r 2 . Now, we solve for the electric field E = q in 4 π ϵ 0 r 2 = Q , where R < r . (1) 9. M Explanation: The electric field for R < r with the sphere conducting and/or uniformly non-conducting: In the region outside the 4 π ϵ 0 r 2 This is the familiar electric field due to a point charge that was used to develop Coulomb’s law. uniformly charged non-conducting sphere, we have the same conditions as for the conduct- ing sphere when applying Gauss’s law, so E = Q , where R < r , (1) The electric field for r < R with the sphere conducting: In the region inside the conducting sphere, we select a spherical 4 π ϵ 0 r 2 as in Part 1.
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gilvin (jg47854) - 9: Electric potential 2 - meyers - (21235) 3 The electric field for r < R with the sphere uniformly non-conducting: In this case we select a spherical gaussian sur- as r → 0. Therefore, the result eliminates the problem that would exist at r = 0 if E varied 1 face at a radius r where r < R, concentric as r 2 inside the sphere as it does outside the with the uniformly charged non-conducting sphere. That is, if E 1 sphere. Let us denote the volume of this sphere by V . To apply Gauss’s law in this situation, it is important to recognize that the charge q in within the gaussian surface of the volume V is less than Q. Using the volume charge density ρ ≡ Q V
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