Phys 181b Problem Set 3 Solution

Phys 181b Problem Set 3 Solution - 20. We imagine a...

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20. We imagine a cylindrical Gaussian surface A of radius r and unit length concentric with the metal tube. Then by symmetry enc 0 2. A q EdA rE ε ⋅= π = ³ G G v (a) For r < R, q enc = 0, so E = 0. (b) For r > R, q enc = λ , so 0 () /2 . Er r πε With 8 2.00 10 C/m λ and r = 2.00 R 0.0600 m, we obtain 8 3 12 2 2 2.0 10 C/m 5.99 10 N/C. 2 0.0600 m 8.85 10 C / N m E × == × π× (c) The plot of E vs. r is shown below. Here, the maximum value is 8 4 max 12 2 2 0 2.0 10 C/m 1.2 10 N/C. 2 2 0.030 m 8.85 10 C / N m E r × λ = × π
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uniform, and we neglect fringing effect. Symmetry can be used to show that the electric field is radial, both between the cylinder and the shell and outside the shell. It is zero, of course, inside the cylinder and inside the shell. (a) We take the Gaussian surface to be a cylinder of length L , coaxial with the given cylinders and of larger radius r than either of them. The flux through this surface is 2, rLE Φ= π where E is the magnitude of the field at the Gaussian surface. We may ignore any flux through the ends. Now, the charge enclosed by the Gaussian surface is q enc = Q 1 + Q 2 = – Q 1 = –3.40 × 10 12 C. Consequently, Gauss’ law yields 0e n c rL E q πε = or 12 enc 12 2 2 3 0 3.40 10 C 0.214 N/C, 2 2 (8.85 10 C / N m )(11.0 m)(20.0 1.30 10 m) q E Lr επ −− −× == = π× × × or | | 0.214 N/C. E = (b) The negative sign in E indicates that the field points inward. (c) Next, for r = 5.00 R 1 , the charge enclosed by the Gaussian surface is q enc Q 1 3.40 × 10 12 C. Consequently, Gauss’ law yields n c = or 12 enc 12 2 2 3 0 3.40 10 C 0.855 N/C. 2 2 (8.85 10 C / N m )(11.0 m)(5.00 1.30 10 m) q E Lr π × = ×⋅ × × (d) The positive sign indicates that the field points outward. (e) we consider a cylindrical Gaussian surface whose radius places it within the shell itself. The electric field is zero at all points on the surface since any field within a conducting material would lead to current flow (and thus to a situation other than the electrostatic ones being considered here), so the total electric flux through the Gaussian surface is zero and the net charge within it is zero (by Gauss’ law). Since the central rod
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This note was uploaded on 07/18/2008 for the course PHYS 181 taught by Professor Stephenirons during the Spring '08 term at Yale.

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Phys 181b Problem Set 3 Solution - 20. We imagine a...

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