Hwk1Solns - Assignment 1 Solutions Chpt 21 8. For ease of...

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Assignment 1 Solutions Chpt 21 8. For ease of presentation (of the computations below) we assume Q > 0 and q < 0 (although the final result does not depend on this particular choice). (a) The x -component of the force experienced by q 1 = Q is () ( ) 1 2 22 00 || 1| cos 45 1 44 2 x QQ q Q Qq Q q F aa a πε ⎛⎞ ⎜⎟ =− ° + = + ⎝⎠ | / | | which (upon requiring F 1 x = 0) leads to /| | 2 2 = , or / 2 2 2.83. (b) The y -component of the net force on q 2 = q is ( ) 2 2 | | | 1 sin 45 2 y qQ qq F a = | | Q q which (if we demand F 2 y = 0) leads to /1 / 2 2 C . The result is inconsistent with that obtained in part (a). Thus, we are unable to construct an equilibrium configuration with this geometry, where the only forces present are given by Eq. 21-1. 18. (a) For the net force to be in the + x direction, the y components of the individual forces must cancel. The angle of the force exerted by the q 1 = 40 μ C charge on 3 20 q = is 45°, and the angle of force exerted on q 3 by Q is at – θ where 1 2.0 cm tan 33.7 . 3.0 cm = Therefore, cancellation of y components requires ( ) 13 3 sin 45 sin 0.02 2 m (0.030 m) (0.020 m) kk °= + from which we obtain | Q | = 83 C. Charge Q is “pulling” on q 3 , so (since q 3 > 0) we conclude Q = –83 C. 1
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(b) Now, we require that the x components cancel, and we note that in this case, the angle of force on q 3 exerted by Q is + θ (it is repulsive, and Q is positive-valued). Therefore, () ( ) 13 3 22 cos 45 cos 0.02 2 m (0.030 m) (0.020 m) qq Qq kk °= + from which we obtain Q = 55.2 μ C 55 C . 28. (a) Eq. 21-1 gives . 10 99 . 8 ) 10 00 . 1 ( ) 10 00 . 1 )( / 10 99 . 8 ( 19 2 2 2 16 2 2 9 N m C C m N F × = × × × = (b) If n is the number of excess electrons (of charge – e each) on each drop then n q e =− =− −× × = 100 10 160 10 625 16 19 .
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Hwk1Solns - Assignment 1 Solutions Chpt 21 8. For ease of...

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