The picture below shows a metal sphere, which is held stationary by a support (w

Question

The picture below shows a metal sphere, which is held stationary by a support (which cannot conductor charge). It has a net charge of q1 = -2.70 C . Another metal sphere is launched towards it; you may ignore the effect of gravity in this problem. The second metal sphere has a net charge of q2 = -7.70 C and a mass of 1.80 g . When the spheres are 0.800 m apart, the smaller sphere is moving toward the larger one with a speed of 22.0 m/s . Both spheres are small enough, compared to the distances between them, that you can treat them as point charges if you need to.

(Figure 1)

Part A

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

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Part B

How close does q2 get to q1?

Figure 1 of 1

The picture below shows a metal sphere, which is held stationary by a support (which cannot conductor charge). It has a net charge of q1 = -2.70 C . Another metal sphere is launched towards it; you may ignore the effect of gravity in this problem. The second metal sphere has a net charge of q2 = -7.70 C and a mass of 1.80 g . When the spheres are 0.800 m apart, the smaller sphere is moving toward the larger one with a speed of 22.0 m/s . Both spheres are small enough, compared to the distances between them, that you can treat them as point charges if you need to.

(Figure 1)

Part A

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

v = m/s

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Part B

How close does q2 get to q1?

r = m

Figure 1 of 1

Explanation / Answer

E = PE + KE

E = kq1q2/d + 1/2mv^2

when they both 0.800 m apart then velocity of q2 is 22 m/s

E = 9*10^9 * 2.7*10^-6 * 7.7*10^-6 /0.800 + 1/2 * 1.8 * 10^-3 * 22^2

E = 0.669 J this total energy of system

part d )

when d = 0.410 m

PE = 0.456 J

KE = E-PE

KE = 0.213 J

0.213 = 1/2 * 1.8*10^-3 * v^2

v = 15.38 m/s

part b )

maximum close point when KE = 0

E = PE

0.669 = kq1q2/d

d = 0.28m

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