Definition from Wikipedia:

An electric potential (also called the electric field potential, potential drop or the electrostatic potential) is the amount of work needed to move a unit of positive charge from a reference point to a specific point inside the field without producing an acceleration.

If that is the definition of a voltage. Shouldn't the formula to calculate it be $V = Fe \cdot d$. Why do we divide it by $q$. Is it because it is a positive test charge as explained above?


2 Answers 2


The force exerted on a charge $q$ in a uniform electric field $E$ is


The work done on moving the charge a distance $d$ in the field is


The potential difference, or voltage, between two points is defined as the work per unit charge to move the charge between the two points, or


So your formula is for work, not voltage.


To answer your follow up comment:

Why do a need this idea of work per charge, we didn't have that when we explaind gravity, right?

In classical mechanics, the "gravitational potential" at a location is equal to the work per unit mass that would be done by the force of gravity if an object were moved from a specific location to a fixed reference location. It is therefore analogous to the "electrical potential" with mass fulfilling the same role as charge.

On the other hand, "Electrical potential energy" = $qEd$ is analogous to "gravitational potential energy" = $mgh$.

Or to put it another way, electrical potential is not the same thing as electrical potential energy and gravitational potential is not the same thing as gravitational potential energy.

Hope this helps.

  • $\begingroup$ Why do a need this idea of work per charge, we didn't have that when we explaind gravity, right? $\endgroup$
    – EHM
    Apr 15, 2019 at 12:32
  • $\begingroup$ See the ADDENDUM to my answer. Hope that helps. $\endgroup$
    – Bob D
    Apr 15, 2019 at 12:58
  • $\begingroup$ ... the gravitational potential near the surface of the Earth is $V_g=gh$, meaning that height is analogous to electric potential. The analogy is useful for understanding electric potential. $\endgroup$
    – garyp
    Apr 15, 2019 at 13:03

This is because, the definition is in terms of unit charge.


That is the formula. Voltage is the work done per unit charge.

So if you write down a fraction - say the speed of an object, you would say distance divided by time. So, let us say it is something like $3m/s = \frac{6}{2}$. That is, $6m$ in $2s$. But that is $3m$ in one second.

That's the analogy.


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