# HW3solutions - Physics 2213 b HW#3 Solutions Spring 2009 Q23.7 E dl Va Vb For a closed path a=b so Va Vb so the line integral is zero The

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Physics 2213 HW #3 – Solutions Spring 2009 c a b Q23.7 b ab a E dl V V  . For a closed path, a=b, so VV , so the line integral is zero. The electrostatic field is conservative : It would take zero net work to move a test charge through an electric field on any path that brings it back to where it started. You may recall that gravity is also a conservative force and so has this property. 22.37. (a) Apply Gauss’s law to a Gaussian cylinder of length l and radius r , where , a r b  and calculate E on the surface of the cylinder.   2 E E rl  encl Ql (the charge on the length l of the inner conductor that is inside the Gaussian surface)   encl 00 gives 2 E E rl    0 . 2 E r  The enclosed charge is positive so the direction of E is radially outward. (b) Apply Gauss’s law to a Gaussian cylinder of length l and radius r , where r > c .   2 E E rl encl (This charge comes from the length l of the inner conductor that is inside the Gaussian surface; the outer conductor carries no net charge).   encl gives 2 E E rl   0 . 2 E r The enclosed charge is positive so the direction of E is radially outward. (c) E = 0 within a conductor. Thus E = 0 for r < a ; 0 for ; 0 for ; 2 E a r b E b r c r     0 for . 2 E r c r  The graph of E versus r is sketched to the right. Inside either conductor E = 0. Between the conductors and outside both conductors the electric field is the same as for a line of charge with linear charge density lying along the axis of the inner conductor. (d) inner surface: Apply Gauss’s law to a Gaussian cylinder with radius r , where . b r c We know E on this surface, so we can calculate encl . Q This surface lies within the conductor of the outer cylinder, where 0, so 0. E E   Thus by Gauss’s law encl 0. Q The surface encloses charge l on the inner conductor, so it must enclose charge l on the inner surface of the outer conductor. The charge per unit length on the inner surface of the outer cylinder is . outer surface: The outer cylinder carries no net charge. So if there is charge per unit length on its inner surface there must be charge per until length on the outer surface. The electric field lines between the conductors originate on the surface charge on the outer surface of the inner conductor and terminate on the surface charges on the inner surface of the outer conductor. These surface charges are equal in magnitude (per unit length) and opposite in sign. The electric field lines outside the outer conductor originate from the surface charge on the outer surface of the outer conductor. 22.47. Apply Gauss’s law Use a Gaussian surface that is a sphere of radius r and that is concentric with the charged shells.

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## This note was uploaded on 04/14/2009 for the course PHYS 2213 taught by Professor Perelstein,m during the Spring '07 term at Cornell University (Engineering School).

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HW3solutions - Physics 2213 b HW#3 Solutions Spring 2009 Q23.7 E dl Va Vb For a closed path a=b so Va Vb so the line integral is zero The

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