hw3 - Physics 21 Fall, 2009 Solution to HW-3 particles 1...

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Physics 21 Fall, 2009 Solution to HW-3 Coulomb’s Law Tutorial This problem asks you to fnd the Couloumb Forces exerted by particles 1, 2, and 3 on par- ticle 0, as shown in the diagram. (a) Consider the two posi- tively charged particles, one oF charge q 0 (particle 0) fxed at the origen, and another oF charge q 1 (particle 1) fxed on the y -axis at (0 ,d 1 , 0). What is the net Force F 1on0 on particle 0 due to particle 1? (b) Now consider the third, negatively charged, particle, whose charge is q 2 (particle 2). Particle 2i sfxedonthe y -axis at position (0 2 , 0). What is the new net Force on particle 0, from particle 1 and particle 2? (c) Particle 0 experiences a repulsion from particle 1 and an attraction toward particle 2. ±or certain values oF d 1 and d 2 , the repulsion and attraction should balance each other, resulting in no net Force. ±or what ratio d 1 /d 2 is there no net Force on particle 0? (d) Now consider the Fourth charged particle, particle 3, with positive charge q 3 ,fxedinthe yz - plane at (0 2 2 ). What is the net Force F 3on0 on particle 0 due solely to this charge? (a) ±rom the expression For F 2on1 on the Equation Sheet, we can write F = 1 4 π± 0 q 0 q 1 ( r 0 r 1 ) | r 0 r 1 | 3 . r 0 r 1 points From the charge point (particle 1 at r 1 = d 1 ˆ j ) to the feld point (particle 0 at r 0 = 0, the origin). Then r 0 r 1 = d 1 ˆ j and | r 0 r 1 | = d 1 , and so the fnal answer is F = 1 4 0 q 0 q 1 d 2 1 ˆ j . (b) The third charged particle on the y axis (particle 2) introduces an additional Force on particle 0 that adds to the Force From particle 1 via superposition. Since the charge is q 2 , the additional Force is F 2on0 = 1 4 0 q 0 ( q 2 )( r 0 r 2 ) | r 0 r 2 | 3 = 1 4 0 q 0 ( q 2 ) ³ d 2 ˆ j ´ d 3 2 = 1 4 0 q 0 q 2 d 2 2 ˆ j .
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hw3 - Physics 21 Fall, 2009 Solution to HW-3 particles 1...

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