# Work done in moving a charge

So, recently while I was studying electric fields and charges for a test, I came across the definition of electric potential as -

The amount of work done in moving a unit positive test charge from infinity to a point, without changing its kinetic energy, i.e, without acceleration against the force due to an electrostatic field.

But this contradicts to what the work-energy theorem says, that the work done equals the change in kinetic energy. What am I missing here?

Also, what exactly is electric potential? Why can we only define a potential for a conservative field?

• – Styg
Commented Feb 21, 2018 at 13:56

If you were using the work energy theorem then you would need to consider both the forces on the charge which are the external force and the force due to the electric field.
These two forces are equal in magnitude and opposite in direction so the net force on the charge is zero.
So the net work done on the charge is zero which leads to the conclusion that the change in the kinetic energy of the charge is zero.

What you are missing from your definition are the words done by an external force.

The amount of work done by an external force in moving a unit positive test charge from infinity to a point, without changing its kinetic energy, i.e, without acceleration against the force due to an electrostatic field.

ie that definition only includes one of the two forces acting on the charge.

• I don't like this definition of potential: it's correct but it seems to me unnecessarily complicated. Why bring in another force as well as the force from the electric field? I'd go for this: "The potential of a point P is the work done per unit charge by the electric field on a (small) charge going from P to infinity." Commented Feb 21, 2018 at 14:31
• @PhilipWood The two definitions are exactly equivalent or you could use minus the work done by the electric field in going from infinity to point $P$. Commented Feb 21, 2018 at 14:36
• Yes indeed. The definitions are equivalent, but my point is that bringing in an external force complicates things unnecessarily. Commented Feb 21, 2018 at 16:19
• @PhilipWood I prefer to think of a force exerted by me (external force) doing work on the charge rather than the more abstract field doing the work? :-) Commented Feb 21, 2018 at 16:48
• Fair enough. I can't argue with anyone else's preferences! Commented Feb 21, 2018 at 17:12

The work energy theorem says that the net work done on a point mass equals the change in kinetic energy. $$\int \vec F_{net} \cdot d\vec r = \frac{1}{2} m \Delta (v^2)$$

In the hypothetical scenario that is one way to define the electrostatic potential, the net force on the test charge is 0 (unaccelerated), but there are two distinct forces acting on it,

$$\vec F_{electrostatic} = q \vec E$$

and

$$\vec F_{external} = -q \vec E$$

The electric potential is defined as

$$\Phi = \frac{1}{q} \int_{r_o}^r \vec F_{external} \cdot d\vec r$$

However, the net work done is $0$ as $\vec F_{net} = \vec F_{external} + \vec F_{electrostatic} = 0$, and hence the kinetic energy of the charge does not change.