Problem: A charge of 88 Coulombs moves a distance 25 m, parallel to and in the direction of the field, in an electric field of 18 N/C. How much work is required?
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Solution: The magnitude of the force on a charge of 88 C in a field of strength 18 N/C is 88 C ( 18 N/C) = 1584 N. To move the charge parallel to and in the direction of the field requires that this force be exerted opposite to the direction of motion. Therefore the work will be negative, and will be equal to negative the product of the magnitude of the force and the distance: Work = - ( 1584 N)( 25 m) = 39600 Joules.
Generalized Response: When a charge Q, in Coulombs, experiences an electric field E, in N/C, it experiences a force F = Q * E Newtons in or opposite to the direction of the field, depending on whether the product Q * E is positive or negative. If the charge moves a distance `ds, then the work done by this field force on the charge is
W(field on charge) = F `ds = Q * E * `ds.
The work required to move the charge is equal in magnitude but of opposite sign to the work done on the charge by the field. So the work required to move the charge is
W(opposed to field) = - W(field on charge) = - Q * E * `ds.
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Figure description: The figure below shows a straight path of length `ds from a starting point to an ending point, with the path parallel to an electric field . If a charge Q travels from start to finish, it experiences a force F = Q * E in its direction of motion. Thus the work done on the charge is
W(field on charge) = F `ds = Q * E * `ds.
The charge does work to oppose the action of the field, so
W(charge on field) = -W(field on charge) = - Q * E * `ds.
