I don't quite understand these concepts. What is the relationship of electric potential with work, potential energy, and electric field?

  • $\begingroup$ You can find the answer in Wikipedia searching each topic and you will find their physical relationships. Also, I have found this link with further explanations quora.com/… $\endgroup$ Commented Mar 20, 2022 at 12:38
  • $\begingroup$ This is too broad. You need to research the relationship between these terms and then, and only then, if there is specific concept you can't understand come back with that. $\endgroup$
    – Bob D
    Commented Mar 20, 2022 at 16:05

1 Answer 1



Work done in physics is defined as force times displacement (due to that force)
More precisely, $$dW=\vec{F}.d\vec{r}$$ or $$\text{Total work done}=\int_{r_1}^{r_2}\vec{F}.d\vec{r}$$

Potential energy

Potential energy is defined only for a system (combination of two or more interacting bodies).
We cannot calucalte absolute potetnial energy of a system, only difference of potetnial energy between initial and final configurations (of a system) can be calculated i.e. $\Delta U$

Definition of PE: It is defined as negative of work done by the internal conservative forces of a system $$dU = -W_c=-\vec{F_c}.d\vec{r}$$

For example, let's take gravitational force
Gravitational force is a conservative force and work done by this force (in a system) causes change in potential energy of the same.
Let's take a stone and the Earth as our system. Initially this stone was at Earth's surface, now we slowly move this stone in vertically upwards to a height of $H$.

Therefore change in PE: $$\Delta U = U_f - U_i = -(\vec{F_g}.\vec{r}) = -(-mgH) = mgh $$

Generally we assume that PE at ground level is zero, thus pure potential energy at height $H$ can be written as $mgH$

Electric Field

Every charge particle in universe exerts force on other charge particles (electrostatic force).

$$\vec{F_E} = \frac{1}{4\pi \epsilon _o} \frac{q_1 q_2}{r^2}\hat{r}$$

I hope you are familiar with this force.

So coming straight to the point, electric field strength of a charges particle (or a system) is the force experienced by a unit positive charge in that field.

$$\vec{E} = \frac{\vec{F_E}}{q} = \frac{kq}{r^2}$$

Electric Potential

In electronics, electrostatic force is responsible for electric potential energy.
Here, potential energy of system of charges kept at infinity distance is taken as zero (reference) - This is global reference.

Electric potential is the change in potential energy per unit positive charge.

Since, $$\Delta U=\frac{kq_1q_2}{r}$$ Electric potential, $$\Delta V = \frac{kq}{r}$$

We can write: $U = qV$ i.e. potential energy of the system = charge placed at that point times potential of that point. (Here potential energy and potential at infinity is considered to be zero)

Some important relations

  1. $$dV = -\vec{E}.d\vec{r}$$
  2. $$U = \frac{kq_1q_2}{r}$$
  3. $$|F_E| = \frac{kq_1q_2}{r^2}$$
  • $\begingroup$ I am assuming that your text book has derivations of above formulae. If any doubt in concept, tag me and ask. $\endgroup$
    – Spencer
    Commented Mar 20, 2022 at 13:46

Not the answer you're looking for? Browse other questions tagged or ask your own question.