{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Hwk1Solns

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

This preview shows pages 1–3. Sign up to view the full content.

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 2 2 0 0 | | 1 | cos45 1 4 4 2 2 2 x Q Q q Q Q q Q q F a a a πε πε = °+ = + | /| | which (upon requiring F 1 x = 0) leads to /| | 2 2 Q q = , or / 2 2 2.83. Q q = − = − (b) The y -component of the net force on q 2 = q is ( ) ( ) ( ) 2 2 2 2 2 2 0 0 | | 1 | | | | 1 sin 45 4 4 2 2 2 y q Q q q F a a a πε πε = °− = | | Q q which (if we demand F 2 y = 0) leads to / 1/ 2 Q q = − 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 μ = 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 θ = = ° Therefore, cancellation of y components requires ( ) ( ) 1 3 3 2 2 2 2 | | sin 45 sin 0.02 2 m (0.030 m) (0.020 m) q q Q q k k θ °= + from which we obtain | Q | = 83 μ C. Charge Q is “pulling” on q 3 , so (since q 3 > 0) we conclude Q = –83 μ C. 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document