ps05sol

ps05sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department...

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MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Fall 2009 Problem Set 5 Solutions Problem 1: Field on Axis of a Line Charge A wire of length l has a uniform positive linear charge density and a total charge Q . a. Calculate the electric field at a point P located along the axis of the wire and a distance a from the left end (as pictured above) We begin by breaking the line of charge up into small segments of length dx’ a distance x’ from point P. Each of these segments have a charge λ dx’ where Q λ = l . They each create an electric field dE at point P, pointing to the left. We are going to add up all of the contributions from each of these segments (integrate) to find the total field: () 2 11 1 ˆˆ ˆ ˆ ˆ a a ee e e e a a dx Q kk k k k xa a a a a a x λλ + + ⎡⎤ =− = = = ⎢⎥ + ++ ⎣⎦ Ei i i i i l l l ll l b. In the limit that the length of the wire goes to zero, does your answer reduce to the right expression? In the limit that goes to zero, our expression becomes: 2 ˆ e Q k a ⎛⎞ ⎜⎟ ⎝⎠ r l This is what we expect for the electric field a distance a away from a point charge Q. Problem 2: Uniformly Charged Cylinder A solid insulating cylinder of length L , radius R has a uniform volume charge density ρ . (a) Find the electric field inside the cylinder as a function of distance z from its center along its axis. (b) What does this simplify to when both R , z << L ? HINTS: (1) You can easily build a cylinder out of something we calculated the field from in class. (2) Start by finding E at the end of a cylinder then use superposition. PS05 Solution - 1
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Problem 2: Uniformly Charged Cylinder continued From hint (1) we know to build the cylinder out of something, and the easiest thing to build it out of is a stack of disks. We calculated the field of a disk (on axis) to be: () disk 1/2 2 22 ˆˆ 11 oo zQ z R zR σ επ ε ⎡⎤ ⎢⎥ =− = ++ ⎣⎦ Ek r k where the disk has radius R and lies in the xy plane (coaxial with the z-axis). Using hint (2) we will first calculate the electric field at the top of a cylinder of height h . Thus we stack disks of thickness dz and where the charge on each disk is thus : 2 QR ρπ = d z 2 cylinder top 2 00 0 hh h zz z Rdz z z dd R ρ πε == = z ⎤⎡ ⎥⎢ = ⎦⎣ ∫∫ EE k k rr where z is the distance from the top of the cylinder to some disk and ranges from 0 to h . Now just integrate, using for the second integral ;2 uz Rd u z d z =+ = 2 2 1 2 2 2 0 ˆ 2 hR h z uR o du zh u h R h R u ρρ εε + + = = = = + k k r This points in the direction as expected. ˆ + k Now to answer the real question – anywhere along the axis. We can do this by superposition: put one cylinder of length 2 Lz + below the point and another of length 2 above it and that will give us the field at point z (measured from the center of the cylinder) inside a cylinder of length L .
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This note was uploaded on 05/06/2010 for the course 8 8.02 taught by Professor Hudson during the Fall '07 term at MIT.

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ps05sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department...

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