Problem Set 2 Solutions

Problem Set 2 Solutions - MASSACHUSETTS INSTITUTE OF...

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MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Fall 2007 Problem Set 2 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, 12 3 4 00 0 SS S S 2; 0 ; ; QQ ε Φ= Φ= Φ= Problem 2: Field on Axis of a Line Charge A wire of length has a uniform positive linear charge density and a total charge . Calculate the electric field at a point located along the axis of the wire and a distance from one end: l Q P a a. Give an integral expression for the electric field at point P in terms of the variables used in the above figure () 2 ˆ al e dx k a x λ + =− Ei b. Evaluate this integral. 11 1 ˆˆ ˆ ˆ ee e e lQ kk k k a x al a aal λλ + ⎡⎤ == = = ⎢⎥ ++ i i i + ⎣⎦ c. In the limit that the length of the rod goes to zero, does your answer reduce to the right expression? 2 ˆ In the limit that goes to zero, our expression becomes , as we desire. e Q lk ⎛⎞ ⎜⎟ a ⎝⎠ G PS02 Solution - 1
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Problem 3: Charged Slab & Sheets An infinite slab of charge carrying a charge per unit volume ρ has its boundaries located at x = 2 meters and x = +2 meters. It is infinite in the y direction and in the z direction (out of the page). Two similarly infinite charge sheets (zero thickness) are located at x = 6 meters and x = +6 meters, with area charge densities σ 1 and σ 2 respectively. In the accessible regions you’ve measured the electric field to be: ˆ 06 ˆ 10V/m -6 m < x < -2 m ˆ 10V/m 6 m > x > 2 m x <− = i i E i G m ˆ m x >+ i (a) Use Gauss’s Law to find the charge density of the slab. You must indicate the Gaussian surface you use on a figure. You can use the symbol ε o in your answer. 0 2 2 LL RR S Ad dE EA A ⋅=⋅+⋅= = ⇒= ∫∫ EAEA EA G GG G w A () 23 0 5 V/m C/m ⎡⎤ = ⎣⎦ 00 2 10 V/m 4 m d εε = (b) Use Gauss’s Law to find the two surface charge densities of the left and right charged sheets. You must indicate the Gaussian surface you use on a figure. You can use the symbol o in your answer. Using Gauss’s law with the Gaussian pillboxes indicated in the figure, we have 2 10 (10 10 V/m C/m S qA enc 1 V/m) dA σ σε ⋅= = = = ∫∫ G G w 2 20 10 V/m C/m =− In a similar manner, PS02 Solution - 2
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Problem 4: Charge Slab Consider a slab of insulating material which is infinitely large in two of its three dimensions, and has a thickness d in the third dimension. An edge view of the slab is shown in the figure. The slab has a uniform positive charge density ρ .
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This homework help was uploaded on 04/07/2008 for the course 8 8.02 taught by Professor Hudson during the Fall '07 term at MIT.

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Problem Set 2 Solutions - MASSACHUSETTS INSTITUTE OF...

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