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Problem: A charge of 96 `microC is located at the origin, and a charge of 39 `microC is located at the point ( 90 m,0,0). The first charge isattached to a large mass, so that it can be considered stationary.The second charge is attached to a much smaller mass of 90 kg. What velocity,perpendicular to the x direction, should the second charge be givenif it is to orbit the first charge in a circular orbit?
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Solution: The key here is that the Coulomb force of attraction must provide the centripetal force which results in the centripetal acceleration v^2/r of the orbiting mass. The force of attraction has magnitude |(9 x 10^9 N m^2/C^2)( 96 x 10^-6 C)( 39 `micro C)/( 90 m)^2| = .004114 N; this is the centripetal force. The centripetal acceleration is thus F / m = .004114 N / ( 90 kg) = .00004571 m/s^2. This is equal to v^2/r, where r is the radius of the orbit--in this case the 90 m separation of the charges. Thus we have v^2/r = v^2/( 90 m) = .00004571 m/s^2. Solving, we get v = `sqrt[ .00004571 m/s^2 ( 90 m)] = .06413 m/s.
Generalized Response: An orbit of radius r and velocity v requires a centripetal acceleration of a = v^2 / r. If the orbiting mass is m, then the corresponding centripetal force required to maintain the circular path is F = ma = m v^2 / r. In this case the centripetal force is supplied by the Coulomb attraction | F | = k | q1 | | q2 | / r^2 . Thus we have
m v^2 / r = k | q1 | | q2 | / r^2.
We easily solve this equation for v to obtain
v = `sqrt( k | q1 q2 | / (m r) ).
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Figure description: The figure below shows the Coulomb force F on the orbiting charge. The circular orbit bounds the shaded circle. The centripetal acceleration a is directed toward the center of the circle, where the fixed charge q1 resides.