A small metal sphere, carrying a net charge of q1 = -2.70 mu C , is held In a st

Question

A small metal sphere, carrying a net charge of q1 = -2.70 mu C , is held In a stationary position by Insulating supports. A second small metal sphere, with a net charge of q2 = -7.60 muC and mass 1.50 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 (Fjgure 1). Assume that the two spheres can be treated as point charges. You can ignore the force of gravity. What is the speed of when the spheres are 0.450 m apart? How close does q2 get to q1?

Explanation / Answer

1.
When the two spheres are at a distance of r = 0.8m, the potential eergy of the system is U = kq1q2/r

Given q1 = -2.70C

              = -2.7 * 10^-6 C

and q2 = -7.6C

          = -7.6 * 10^-6 C

we have k = 8.99 * 10^9 Nm^2/C^2

therefore

U = (8.99 * 10^9 Nm^2/C^2)(-2.7 * 10^-6 C)(-7.6 * 10^-6 C)/(0.8m)

    = 0.2305935 J

At this position the q2 is moving with a speed of v = 22 m/s

then the kinetic energy of teh system is K = (1/2)mv^2

given the mass of q2 m = 1.5 g

                                    = 1.5*10^-3 kg

then K = (1/2)(1.5*10^-3 kg)(22m/s)^2

          = 0.363 J

then the total energy of the system is E = 0.2305935 J + 0.363 J

                                                            = 0.5935935 J

Now at r = 0.45 m the potential energy of the system is

U = (8.99 * 10^9 Nm^2/C^2)(-2.7 * 10^-6 C)(-7.6 * 10^-6 C)/(0.45 m)

   = 0.409944 J

If v is the speed of q2 at this point then the kinetic energy of teh system will be

              K = (1/2)(1.6*10^-3 kg)v^2

then the total energy of the system will be

           E =0.409944 J + (1/2)(1.6*10^-3 kg)v^2

Then from conservation of energy we should have

  0.409944 J + (1/2)(1.5*10^-3 kg)v^2 = 0.5935935 J

or (1/2)(1.5*10^-3 kg)v^2 = 0.5935935 J - 0.409944 J

or v^2 = 244.865 (m/s)^2

or v = 15.65 m/s

Therefore the speed of q2 at r = 0.45 is v = 15.65 m/s

2.

let r is the closest distance between the charges q1 and q2, at this point charge q2 will come to rest.

therefore the system will have only the potential energy.

then from conservation of energy we should have E = kq1q2/r

or

r = kq1q2/E

= (8.99 * 10^9 Nm^2/C^2)(-2.7 * 10^-6 C)(-7.6 * 10^-6 C)/(0.5935935 J)

= 0.31 m

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