A small metal sphere, carrying a net charge of q 1 = -2.90 C , is held in a stat

Question

A small metal sphere, carrying a net charge of q1 = -2.90 C , is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2 = -8.00 C and mass 1.80 g , is projected toward q1. When the two spheres are 0.800 m apart, q2 is moving toward q1 with speed 22.0 m/s (Figure 1) . Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.

Part A

What is the speed of q2 when the spheres are 0.410 m apart?

v= m/s?

Part B

How close does q2 get to q1?

r = m?

Explanation / Answer

q1 = -2.90 uC

q2 = -8 uC

m2 = 1.80 g

ri= initial distance between them = 0.800 m

Vi = initial speed = 22 m/s

A) Vf= fnal speed of q2 = ?

rf = final distance = 0.410 m

initial total energy,Ei = Ki + Ui

Ei = ( 1/2)*m2*Vi2 + ( k*q1*q2 / ri)

Ei = (1/2)*1.80*10-3*222 + ( 9*109 *-2.90*10-6*-8*10-6 / 0.800)

Ei = 0.6966 J

final energy,Ef =  ( 1/2)*m2*Vf2 + ( k*q1*q2 / rf)

Ef = (1/2)*1.80*10-3*Vf2 + ( 9*109 *-2.90*10-6*-8*10-6 / 0.410)

Ef = 9*10-4 Vf2 + 0.5092

from conservation of energy , Ei = Ef

0.6966 = 9*10-4 Vf2 + 0.5092

Vf= 14.42 m/s

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B) let d be the distance of closet approach.At this point q2 will be momentarily stop.So from energy conservation:

( 1/2)*m2*Vi2 + ( k*q1*q2 / ri) = k*q1*q2 / d

(1/2)*1.80*10-3*222 + ( 9*109 *-2.90*10-6*-8*10-6 / 0.800) =  9*109 *-2.90*10-6*-8*10-6 / d

0.6966 = 0.2088 /d

d= 0.299 m

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