A charge of 13 `microC with an initial kinetic energy of 377 J and a mass of 1.38 kg is acted upon by an electric field against which it does -63 J of work. If no force other than the electrostatic forces act on the charge, what will be its final kinetic energy and speed?
The work done by an object, in the absence of forces other than the electrostatic force, is done against an opposing force and therefore decreases the object's KE. Thus the new KE will be the original 377 J, less the -63 J used in doing the work, or 440 J.
The speed of the mass is found by setting .5mv^2 equal to this new KE.
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 force in its direction of motion. This would be in response to a negative force acting ON the object in its direction of motion, which will decrease its velocity and hence its KE.
The work `dW done by the net force exerted by an object and the change `dKE in its KE are related by the Work-Energy Theorem
which formalized the commonsense notion that if work `dW is done by the net force exerted by the object, its KE will decrease by the same amount.
In this case we start with an initial kinetic energy KE0 and the object does work `dW, which will decrease the KE to KE0 - `dW. This KE will be equal to .5 m v^2, and we will solve the resulting equation
for v.
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.
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