Problem: Every .4200 seconds, a charge of 1.6 Coulombs moves a distance .1500 m parallel to an electric field of 60000 N/C. How much power is involved?
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Solution: Power is work/time. The work in this situation results from the force moving the charge through the given distance. The force on 1.6 Coulombs in a 'c/ N/C field is ( 1.6 C)( 60000 N/C) = 96000 N. The work associated with this force acting over distance .1500 m is ( 96000 N)( .1500 m) = 14400 J. The power is thus 14400 J / ( .4200 seconds) = 34280 J/s = 34280 watts.
Generalized Response:
.
.
.
.
.
.
.
.
.
.
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.
