A charge of .3 Coulombs requires .24 seconds to move a distance .065 m parallel to and in the direction of an electric field of 70000 N/C. How much power is required?
Power is work/time. The work in this situation results from the force moving the charge through the given distance. The force on .3 Coulombs in a 'c N/C field is ( .3 C)( 70000 N/C) = 21000 N. The work associated with this force acting over distance .065 m is ( 21000 N)( .065 m) = 1365 J. The power is thus 1365 J / ( .24 seconds) = 5687 J/s = 5687 watts.
If we know the charge Q, the displacement of the charge and the electric field parallel to the displacement we can use the charge and field to find the force on the charge, and then use the displacement and the force to find the work:
If we also know the time required for the charge to move, we easily find the power:
We note that since `ds / `dt = v, the velocity of the charge, P = Q * E * v.
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