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ps03sol

# ps03sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department...

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MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Fall 2008 Problem Set 3 Solutions Problem 1: Closed Surfaces Four closed surfaces, S 1 through S 4 , together with the charges –2 Q , Q , and – Q are sketched in the figure at right. The colored lines are the intersections of the surfaces with the page. Find the electric flux through each surface. By Gauss’s Law, the flux through the closed surfaces is equal to the charge enclosed over ε 0 . So, 1 2 3 0 0 2 ; 0; ; S S S Q Q 4 0 S ε ε Φ = − Φ = Φ = − Φ = Problem 2: Gauss’s Law Warm Up – Cylinder Consider a cylinder of radius R and linear charge density λ . (a) What is the electric field outside the cylinder as a function of distance r from its axis? We use a Gaussian cylinder of length L and radius r > R . Then, 0 0 0 0 ˆ 2 2 2 enc d E rL Q L E r r π ε λ ε λ πε λ πε = = = = = ∫∫ E A E r G G G (b) What is the electric potential difference between a point on its surface and a point a distance r > R from its central axis? To get the potential difference we just integrate, from the outer edge of the cylinder to r : ( ) ( ) ( ) ( ) ( ) 0 0 0 ' 2 ' ' 2 ln ' 2 ln r r R r R V V r V R d r dr r R r λ πε λ πε λ πε = = = − = − = = E s G G What is the sign of this? r > R, so the natural log is negative, and thus so is the potential difference. This should make sense, since we are going downhill as we move away from the charged cylinder. PS03 Solution - 1

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Problem 3: A potential for finding charge The electric potential V(r) for a distribution of charge with spherical symmetry is: 2 0 0 2 0 0 Region I: ( )= for 0 2 Region II: ( ) for 2 2 R egion III: ( ) 0 for 2 V r V r V r R R V R V V r R r R r V r r R + < < = < < = > Where V 0 is a constant of appropriate units. This function is plotted above. (a) What is the electric field ( ) r E G for this problem? Give your answer analytically for the three regions above and also plot the magnitude. Be sure to indicate the maximum value on the vertical axis for E. ( ) ˆ r r E = E G r where E r is given below and plotted above at right. Region I: 0 r R < < 0 2 r V V r E r R = − = Region II: 2 R r R < < 0 2 r V V R E r r = − = Region III: 2 r R > 0 r V E r = − = (b) Use Gauss’s Law arguments to determine the distribution of charges that produces the above electric potential and electric field. When making a Gauss’s Law argument, state the Gaussian surface you are using and what you conclude about the charge distribution by applying Gauss’s Law to that surface. First take a Gaussian sphere with radius r > 2R . Since the electric field is zero on the surface of the sphere, the total charge enclosed by this sphere must be zero. Now take a Gaussian sphere with any radius R < r < 2R . We have 2 0 0 2 4 4 inside o closed surface Q V R r V r π π ε = = = ∫∫ E dA R G G w So for any r in the range R < r < 2R 0 4 inside o Q V R
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