HW3solutions-10 - Physics 2213 HW#3 – Solutions Spring 2010 c a b Q23.5 b a b a V V l d E − = ⋅ so if E is zero everywhere along the path

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Unformatted text preview: Physics 2213 HW #3 – Solutions Spring 2010 c a b Q23.5 b a b a V V l d E − = ⋅ ∫ , so if E is zero everywhere along the path, the integral is zero and the potential difference, b a V V − will be zero. This does not mean that the electric field will be zero everywhere along any path from a to b . It only means that the path integral will be zero along any path, since the electric force is a conservative force and therefore the line integral of the electric force (and therefore the electric field) will be independent of the path. An example is given in the figure. The charged hollow conducting sphere has zero electric field inside, so the path integral along path #1 must be zero. However, because the electric force is a conservative force, the line integral along path #2 must also be zero, even though it goes outside the sphere where E is not zero. Q23.7 b a b a E dl V V = − ∫ . For a closed path, a=b, so a b V V = , 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 π Φ = encl Q l λ = (the charge on the length l of the inner conductor that is inside the Gaussian surface) ( ) encl gives 2 E Q l E rl λ π ε ε Φ = = . 2 E r λ πε = 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 π Φ = encl Q l λ = (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 Q l E rl λ π ε ε Φ = = a b +Q #1 #2 E=0 E ≠ 0- 2 - . 2 E r λ πε = 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 ; for ; 0 for ; 2 E a r b E b r c r λ πε = < < = for . 2 E r c r λ πε = > 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 λ lying along the axis of the inner conductor. (d) inner surface: . b r c < < Apply Gauss’s law to a Gaussian cylinder with radius r , where 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....
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This note was uploaded on 06/08/2010 for the course PHYS 213 taught by Professor Perelstein during the Spring '08 term at Cornell University (Engineering School).

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HW3solutions-10 - Physics 2213 HW#3 – Solutions Spring 2010 c a b Q23.5 b a b a V V l d E − = ⋅ so if E is zero everywhere along the path

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