9- Electric potential 2-solutions

# 9- Electric potential 2-solutions - gilvin(jg47854 – 9...

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Unformatted text preview: gilvin (jg47854) – 9: Electric potential 2 – meyers – (21235) 1 This print-out should have 22 questions. 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 and the following graphs: S . r R ∝ 1 r 2 M . r R ∝ 1 r G . r R ∝ 1 r 2 Z . r R ∝ 1 r L . r R ∝ 1 r 2 ∝ 1 r Y . r R ∝ 1 r P . r R ∝ 1 r 2 Q . r R ∝ 1 r 2 X . r R ∝ 1 r Which diagram describes the electric field vs radial distance [ E ( r ) function] for a con- ducting sphere? 1. M 2. L 3. Q gilvin (jg47854) – 9: Electric potential 2 – meyers – (21235) 2 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, vector E is parallel to d vector A at each point. There- fore vector E · d vector A = E dA and Gauss’s law, where E is constant everywhere on the surface, gives Φ E = contintegraldisplay vector E · d vector A = contintegraldisplay E dA = E contintegraldisplay dA = E ( 4 π r 2 ) = q in ǫ , 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 ǫ r 2 = Q 4 π ǫ r 2 , where R < r . (1) This is the familiar electric field due to a point charge that was used to develop Coulomb’s law. The electric field for r < R with the sphere conducting: In the region inside the conducting sphere, we select a spherical gaussian surface r < R , concentric with the 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 = , where r < R . (2) G . r R E ∝ 1 r 2 E 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 8. Y 9. M Explanation: The electric field for R < r with the sphere conducting and/or uniformly non-conducting: In the region outside the uniformly charged non-conducting sphere, we have the same conditions as for the conduct- ing sphere when applying Gauss’s law, so E = Q 4 π ǫ r 2 , where R < r , (1) as in Part 1. 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- face at a radius r where r < R , concentric with the uniformly charged non-conducting sphere. Let us denote the volume of this sphere by V ′ . To apply Gauss’s law in this....
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9- Electric potential 2-solutions - gilvin(jg47854 – 9...

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