# Electric Potential, Work, Potential Energy, and Electric Field [closed]

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

• 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/… Commented Mar 20, 2022 at 12:38
• 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. Commented Mar 20, 2022 at 16:05

## Work

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. $$|F_E| = \frac{kq_1q_2}{r^2}$$
• I am assuming that your text book has derivations of above formulae. If any doubt in concept, tag me and ask. Commented Mar 20, 2022 at 13:46