A small metal sphere, carrying a net charge of q1 = -2.50 ?C , is held in a stat

Question

A small metal sphere, carrying a net charge of q1 = -2.50 ?C , is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2 = -7.50 ?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.

A small metal sphere, carrying a net charge of q1- 2.50 C , is held in a stationary position by insulating supports. A second small metal sphere with a net charge of q,--7.50 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 0.430 m apart? U E m/s Figure 1 of 1 Submit My Answers Give Up Part B How close does q2 get to q1? 42 u= 22.0 m/s 91 r= 0.800 m Submit My Answers Give Up

Explanation / Answer

when d = 0.8 m

PE = kq1q2/d = (9 x 10^9 x 2.50 x 10^-6 x 7.50 x 10^-6) / 0.8

PE = 0.221 J

KE = m v^2 /2 = (1.8 x 10^-3 x 22^2 / 2) = 0.436 J

when d = 0.43 m

PE = (9 x 10^9 x 2.50 x 10^-6 x 7.50 x 10^-6) / 0.8

PE = 0.392 J

KE = mv^2 /2

Using energy conservation,

0.221 + 0.436 = 0.392 + (0.0018 v^2 /2 )

v = 17.14 m/s

B) q2 travel towards q1 until its KE becomes zero.

now using energy conservation,

initial PE + KE = final PE + KE

0.221 + 0.436 = (kq1q2/d) + 0

(9 x 10^9 x 2.50 x 10^-6 x 7.50 x 10^-6) / d = 0.657

d = 0.257 m

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