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Set 51 Problem number 10


Problem

A charge of 3.16 Coulombs moves through a displacement of .028 m, parallel to and in the direction of the field, in an electric field of 225000 N/C. How much work is done by the charge?

Solution

The magnitude of the force on a charge of 3.16 C in a field of strength 225000 N/C is

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. If we choose the direction of motion as positive, the force is therefore directed in the negative direction so

A field force parallel to and in the same direction as the field implies that positive work is done on the object by the field so that negative work is done by the object against the field.  If the electrostatic force is the net force on the charge, the KE of the charge would therefore increase accordingly.

Generalized Solution

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, as in the present case, 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, as in the present case, the force exerted by the field is parallel to the displacement and positive if the force is in the opposite direction.

Explanation in terms of Figure(s), Extension

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:

Figure(s)

work_on_charge_moving_in_field.gif (4782 bytes)

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