The region between two capacitor plates contains a uniform electric field of 425000 N/C. How much work is required to move a charge of 2 Coulombs through a displacement of .3 m within this region, with the motion parallel to and opposite to the direction of the field?
The magnitude of the force on a charge of 2 C in a field of strength
To move the charge parallel to and oppsite to the direction of the field requires that this force be exerted in the direction of motion. Therefore the work will be positive, and will be equal to the product of the magnitude of the force and the distance
When a charge Q, in Coulombs, is subject to an electric field E, in N/C, it experiences a force of magnitude | F | = | Q * E | Newtons. The direction of the force is in or opposite to the direction of the field, depending on whether the product Q * E is positive or negative. If the charge moves through a distance `ds parallel to the force, then the work done by this field force on the charge has magnitude
This work is positive if the force is parallel to the displacement and negative if the force is in the opposite direction.
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 has magnitude
This work is negative if the force exerted by the field is parallel to the displacement and positive if the force is in the opposite direction.
The figure below shows a straight path of length `ds from a starting point to an ending point, with the displacement from 'start' to 'end' parallel to and in the direction of the electric field. As the charge Q travels from start to finish, it experiences a force F = Q * E in its direction of motion; Q * E will be positive if Q is positive and negative if Q is negative.
Thus the work done on the charge is
The force exerted by the charge on the field is equal and opposite to the force exerted by the field on the charge. Ths charge therefore does work which is equal and opposite to the work done by the field on the charge:
We note that this work will be positive if Q is negative, and negative if Q is positive.
If the either the field or the direction of the path is reversed, then the force F = Q * E exerted by the field will be opposite to the direction of motion and the work done by the field on the charge will be the negative of that previously obtained:
Similarly, the work done by the charge against the field will be the negative of the previous result:
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