680_Physics ProblemsTechnical Physics

680_Physics ProblemsTechnical Physics - 20 Electric Fields...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
20 Electric Fields *P23.53 (a) Each ion moves in a quarter circle. The electric force causes the centripetal acceleration. Fm a = qE mv R = 2 E mv qR = 2 (b) For the x -motion, vv a x x xf xi x f i 22 2 =+ di 02 2 va R x a v R F m qE m x x x =− = = 2 2 E mv qR x 2 2 . Similarly for the y -motion, R y 2 a v R qE m y y = 2 2 E mv qR y = 2 2 The magnitude of the field is EE mv qR x xy 2 2 += ° at 135 counterclockwise from the -axis . P23.54 From the free-body diagram shown, F y = 0: T cos . . 15 0 1 96 10 2 °= × N . So T 203 10 2 . N . From F x = 0 , we have qE T sin . 15 0 or q T E = ° = ×° × = sin . .s i n . . .. 15 0 150 100 10 525 10 525 2 3 6 N NC C C ej µ . FIG. P23.54 P23.55 (a) Let us sum force components to find
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.
Ask a homework question - tutors are online