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Problem: A charge of 81 `microC is located at the origin and fixed at that point. A charge of 54 `microC is located at the point ( 27 m,0,0). What is the x component of the acceleration of the second charge, assuming that it is attached to a mass of 100 kg?
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Solution: The charges are separated by 27 m. The magnitude of the force between the charges is therefore given by Coulomb's Law as (9 x 10^9 N m^2/C^2) ( 81 x 10^-6 C)( 54 x 10^-6 C)/( 27 m)^2 = .05340 N. This force will act on the 100 kg mass to which the charge is attached, giving it an acceleration of magnitude .05340 N / 100 kg = .0005340 m/s^2. Since the particles repel, this acceleration will be away from the origin; the x component of the acceleration will therefore be .0005340
Generalized Response: Coulomb's Law generalizes our observations of forces between charges into the statement F = k q1 q2 / r^2, where q1 and q2 are two charges and r the distance between them. If q1 is fixed at the origin and q2, of mass m, is located at (x,0,0), then we see that r = x so the magnitude of F is k q1 q2 / x^2. The acceleration of mass m attached to this charge is therefore a = F / m = k q1 q2 / (x^2 m).
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Figure description: The figure below shows how the distance r between charges is r = (x - 0) = x. The repulsive nature of the force is indicated by the direction of the force vector. The acceleration of the free charge q2 will be in the direction of the force vector.