why is the work done while calculating the potential due to a point charge negative?

Asked by  | 17th May, 2011, 04:56: PM

Expert Answer:

When a particle with charge q is placed in an external electric field E, (i.e. an electric field produced by other charges), then an electric force = qE will act on it.  If this force is not balanced by other forces, then the particle will accelerate, and its kinetic energy will change.  The electric force will do positive work on the particle.  If the electric force is balanced by another external force Fext = -qE, and this external force moves the particle against the electric force, than the external force Fext does positive work.  The work done is

W = òFext·d= -qòE·dr.

The work done by the external force is equal to the change in the electrostatic potential energy of the particle in the external field.  The change in the potential energy of a charge q when being moved from point A to point B, is the work done by an external force in moving the charge.

The work done while calculating the potential due to a point charge is negative because work has been done against the field.


The integral is taken along a particular path.  But the electrostatic force is a conservative force.  The integral is independent of the path.  DU therefore depends only on the endpoints A and B of the path, not on the actual path itself.

Answered by  | 18th May, 2011, 03:03: AM

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