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Unformatted text preview: MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Spring 2010 Problem Set 2 Solutions Problem 1 (10 points): Concept Questions. Explain your reasoning. Concept Question 1: A pyramid has a square base of side a, and four faces which are equilateral triangles. A charge Q is placed on the center of the base of the pyramid. What is the net flux of electric field emerging from one of the triangular faces of the pyramid? 1. 0 2. 8 Q ε 3. 2 2 Qa ε 4. 2 Q ε 5. Undetermined: we must know whether Q is infinitesimally above or below the plane? Answer 2: Explain your reasoning: Construct an eight faced closed surface consisting of two pyramids with the charge at the center. The total flux by Gauss’s law is just / Q ε . Since each face is identical, the flux through each face is one eight the total flux or /8 Q ε . Concept Question 2: A charge distribution generates a radial electric field / 2 ˆ r b a e r − = E r r where a and b are constants. The total charge giving rise to this electric field is 1. 4 a πε 2. 0 3. 4 b πε Answer 2: Explain your reasoning: In order to fine the total charge I choose a Gaussian surface that extends over all space. Because the electric field is radially symmetric, I choose my Gaussian surface to be a sphere of radius r and I will take the limit as r → ∞ . The flux is given by / / / 2 / 2 2 2 ˆ ˆ lim lim lim lim 4 4 lim r b r b r b r b r r r r r r r r a a a d e d a e d a e r a e r r r π π − − − − →∞ →∞ →∞ →∞ →∞ ⋅ = ⋅ = = = = ∫∫ ∫∫ ∫∫ E a r r r r ¡ ¡ ¡ When I take the limit as r → ∞ , the exponential term goes to zero, and so the flux goes to zero. Therefore the charge enclosed is zero. Problem 2 (10 points): Nonuniformly charged sphere A sphere of radius R has a charge density ( / ) r R ρ ρ = where ρ is a constant and r is the distance from the center of the sphere. a) What is the total charge inside the sphere? Solution : The total charge inside the sphere is the integral 4 2 2 3 3 4 4 4 ( / ) 4 4 r R r R r R r r r R Q r dr r R r dr r dr R R R ρ π ρ π ρ π ρ π ρ π = = = ′ ′ = = = = = = = = ∫ ∫ ∫ b) Find the electric field everywhere (both inside and outside the sphere). Solution: There are two regions of space: region I: r R < , and region II: r R > so we apply Gauss’ Law to each region to find the electric field. For region I: r R < , we choose a sphere of radius r as our Gaussian surface. Then, the electric flux through this closed surface is 2 4 I d E r π ⋅ = ⋅ ∫∫ I E A r r ¡ . Since the charge distribution is nonuniform, we will need to integrate the charge density to find the charge enclosed in our Gaussian surface. In the integral below we use the integration variable r ′ in order to distinguish it from the radius r of the Gaussian sphere. 4 4 2 2 3 4 4 1 1 4 ( / ) 4 4 r r r r r r enc r r r Q r r r dr r R r dr r dr R R R ρ π ρ π ρ π ρ π ρ π ε ε ε ε ε ε ′ ′ ′ = = = ′ ′ ′ = =...
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This note was uploaded on 04/20/2010 for the course PHYSICS 8.02 taught by Professor Hughes during the Spring '08 term at MIT.
 Spring '08
 Hughes
 Magnetism, Mass

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