Problem: A charge of 8 `microC with an initial kinetic energy of 319 J and a mass of 2.02 kg is acted upon by an electric field against which it does 69 J of work. In the absence of dissipative forces or potential energy changes, what will be its final kinetic energy and speed?
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Solution: The work done by an object, in the absence of dissipative forces or PE changes, is done against an opposing force and therefore decreases the object's KE. Thus the new KE will be the difference of the original 319 J and the 69 J used in doing the work, or 250 J. The speed is found by setting .5mv^2 equal to this new KE. We obtain the equation .5( 2.02 kg) v^2 = 250 J, which is solved to obtain v = 15.7 m/s. In this case the work done BY the object is positive, which implies that in response to the force on it the object exerts a positive force in its direction of motion, thereby decreasing its KE. This would be in response to a negative force acting ON the object in its direction of motion, which would tend to decrease its KE.
Generalized Response:
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Figure description: The figure below shows a charge q moving through a displacement `ds in the direction of a potential gradient dV / dx. The potential difference will be the product of the potential gradient and the displacement: `dV = dV / dx * `ds. The work done on the charge is therefore `dW = q `dV; this work is also the product `dW = Fave * `ds of the average force on the charge and the displacement. Setting these two expressions for work `dW equal we arrive at the conclusion that Fave = q dV/dx = q E.
E = dV / dx is called the electric field. Since E = Fave / q, we see that the electric field E is the force per unit charge experienced by a charge.
