ps02sol

# ps02sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department...

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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): Non-uniformly 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 non-uniform, 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 PHYS 8.02 taught by Professor Dourmashkin during the Spring '10 term at MIT.

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ps02sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department...

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