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Unformatted text preview: Physics 208, Assignment 2, Solutions, 2/2/09, RT Problem 1. (a) Positive charges fall from high to low potential. But electrons are negative. Hence V 2 > V 1 . _ 1 V 2 V 1 V 2 1 electron charge = -|e| V = 2 V - + (b) The battery holds V constant (equal to its E . M . F . ). Since V =- integraltext 2 1 Edx , if the path length from 1 to 2 increases, the electric field decreases more or less proportionally. (If the electric field were uni- form the proportionality would be exact.) (c) For the electron, the change of potential energy-| e | ( V 2- V 1 ) (which is negative) is compensated by a change of kinetic energy mv 2 / 2 (which is positive); 1 2 mv 2- | e | V = 0 . When the cathode to anode distance is changed, since V is un- changed, v is unchanged. Problem 2. See figure. (a) All of the charge + Q resides on the outer surface. The electric field is radial with value E ( r ) = 0 for r < b , 0 for b < r < c , kQ r 2 for c < r . (b) See figure for location of charge. 1 b c + + + + + + + + + + + + + + + + + + +Q r 2 ~ 1 kQ 2 c kQ c r 1 ~ r 3c 2c c (b) E(r) r 3c 2c c V(r) (d) b b (c) Evaluate V ( r ) by integrating in from (where V = 0 by definition) to r along the horizontal axis; V ( r ) =- integraldisplay r E dr = kQ c for r < b , kQ c for b < r < c , kQ r for c < r . (d) See figure. Note that the vanishing of E in a region implies only that V is constant in the region, and nothing about the value of the constant.-------------------Q 2 r 2 ~ 1 r 2 ~ b 3c 2c c b 1 ~ r-kQ c + + + + + + + + + + + + + + + + + + r E(r) r V(r) (e) +2Q 3c 2c c (f)-2Q (e) With point charge- 2 Q placed at the origin the charges on the con- ducting shell rearrange themselves. The electric field in the metalducting shell rearrange themselves....
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This note was uploaded on 09/01/2009 for the course PHYS 208 taught by Professor Amadeuri during the Spring '08 term at Cornell University (Engineering School).
- Spring '08