Electrostatics

The Coulomb force acting on a particle of charge $q$ [C] at position $\vec{r}$ [m] due to a particle of charge $q_0$ [C] at position $\vec{r}_0$ [m] is,

$$ \vec{F}_{\text{Coulomb}} = \frac{qq_0}{4\pi\epsilon_0 |\vec{r}-\vec{r}_0 |^2}\hat{r} \hspace{1cm}\text{[N]}.$$

Here the unit vector $\hat{r}$ points from $\vec{r}_0$ to $\vec{r}$, and $\epsilon_0$ = 8.854187817×10-12 F/m is the permittivity constant. Since there is a force between the charges, work is required to change the separation between them and since this is a conservative force, a potential energy can be defined,

$$E_{pot} = \frac{qq_0}{4\pi\epsilon_0 |\vec{r}-\vec{r}_0 |} \hspace{1cm}\text{[J]}.$$

It is convenient to define an electric field and an electrostatic potential associated to each charge. The electric field of charge $q_0$ is proportional to the Coulomb force,

$$ \vec{E}(\vec{r})= \frac{q_0}{4\pi \epsilon_0}\frac{\vec{r}-\vec{r}_0}{|\vec{r}-\vec{r}_0|^3} \hspace{1cm}\text{[V/m]}\qquad \vec{F}_{\text{Coulomb}}=q\vec{E}\hspace{1cm}\text{[N]},$$

and the electrostatic potential is proportional to the potential energy,

$$\varphi(\vec{r})= \frac{q_0}{4\pi \epsilon_0 |\vec{r}-\vec{r}_0|} \hspace{1cm}\text{[V]}\qquad E_{pot}(\vec{r}) = q\varphi (\vec{r})\hspace{1cm}\text{[J]}.$$

The electric field is a vector field. There is a vector at every point in space. The electric field points away from positive charges and towards negative charges. The electric field is minus the gradient of the electrostatic potential,

$$\vec{E}(\vec{r}) = - \nabla \varphi (\vec{r}).$$

The difference in electrostatic potential between two points can be determined by integrating the electric field from one point to the other,

$$\varphi(\vec{r}_2) - \varphi(\vec{r}_1)=-\int\limits_{\vec{r}_1}^{\vec{r}_2}\vec{E}\cdot d\vec{r}.$$

The work necessary to move a charge $q$ from $\vec{r}_1$ to $\vec{r}_2$ is minus the integral of the force $q\vec{E}$ over the distance,

$$W = q(\varphi(\vec{r}_2) - \varphi(\vec{r}_1))=-\int\limits_{\vec{r}_1}^{\vec{r}_2}q\vec{E}\cdot d\vec{r}.$$

Electrostatics describes how the charge density of a system is related to the electric field and to the electrostatic potential.

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