Assignment 1 Solutions Chpt 218. 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 q1 = Qis ()()()() ()122200||1|cos45144222xQQqQQ qQqFaaaπεπε⎛⎞⎛⎞⎜⎟=−°+=−+⎜⎟⎜⎟⎝⎠⎜⎟⎝⎠|/||which (upon requiring F1x= 0) leads to /||22Qq=, or /222.83.Q q= −= −(b) The y-component of the net force on q2= qis ()() ()22222200||1||||1sin 4544222yqQqqFaaaπεπε⎛⎞⎛⎞⎜⎟=°−=⎜⎟⎜⎟⎝⎠⎜⎟⎝⎠||Qq−which (if we demand F2y= 0) leads to /1/ 2Q q= −2C. 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 +xdirection, the ycomponents of the individual forces must cancel. The angle of the force exerted by the q1= 40 μC charge on 320qμ=is 45°, and the angle of force exerted on q3by Qis at –θwhere 12.0 cmtan33.7 .3.0 cmθ−⎛⎞==°⎜⎟⎝⎠Therefore, cancellation of ycomponents requires ()()1332222||sin 45sin0.022 m(0.030 m)(0.020 m)q qQ qkkθ°=+from which we obtain |Q| = 83 μC. Charge Qis “pulling” on q3, so (since q3> 0) we conclude Q= –83 μC. 1
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