ch22-p086 - 86. (a) The electric field is upward in the...

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() 2 2 66 1 3 2 22 00 2 14 9 (4.24 10 m s) 4.24 10 m s 1.40 10 m s sin sin 2 3.51 10 m s 6.43 10 s. vv a d t a θθ ×− × × −− == × The negative root was used because we want the earliest time for which y = d . The x coordinate is ( ) 69 2 0 cos 6.00 10 m s 6.43 10 s cos45 2.72 10 m. xv t × × ° = × θ This is less than L so the electron hits the upper plate at x = 2.72 cm. 86. (a) The electric field is upward in the diagram and the charge is negative, so the force of the field on it is downward. The magnitude of the acceleration is a = eE / m , where E is the magnitude of the field and m is the mass of the electron. Its numerical value is a = ×× × 160 10 911 10 351 10 19 31 14 . . .. C2 . 0 01 0NC kg ms 3 2 ch c h We put the origin of a coordinate system at the initial position of the electron. We take the x axis to be horizontal and positive to the right; take the y axis to be vertical and positive toward the top of the page. The kinematic equations are 2 0 1 cos , sin , and sin . 2 y x vt y vt a t v v a t = First, we find the greatest
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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