A charge of 15.01 `microC is located at the origin and fixed at that point. A charge of -58.06 `microC is located at the point ( .005 m,0,0). What is the x component of the acceleration of the second charge, assuming that it is attached to a mass of .0016 kg on which there is no force except the electrostatic force between the charges?
The charges are separated by .005 m. The magnitude of the force between the charges is therefore given by Coulomb's Law as
This force will act on the .0016 kg mass to which the charge is attached, giving it an acceleration of magnitude
Since the particles repel, this acceleration will be away from the origin; the x component of the acceleration will therefore be 1.96E+08 m/s^2
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).
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.
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