Electrostatic equilibrium: Difference between revisions

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A conductor is said to have reached electrostatic equilibrium when its charge distribution becomes stable.


== Properties ==
Consider a solid conductor $(S)$ at electrostatic equilibrium.
By combination of [[Classical_mechanics#Newton's_2nd_law|Newton's 2nd law]] and [[Electric_charge#Electrostatic_force|Coulomb's law]], for any macroscopic charge $A$:
$$\vec a = \frac{Q_A\vec E}{m}$$
By definition of electrostatic equilibrium,
$$
\forall A \in (S), \vec a=0 \\
\therefore \vec E =0 \text{ for every point inside (S)  (1)}
$$
As a result, if conductor $(S)$ is charged, any charge must be destributed on the surface of $(S)$, $S(S)$, which creates an electric field on $S(S)$.
This field must satisfy $\vec E \bot S(S)$ at all points. (3) Otherwise, the macroscopic charges would move along the surface.
=== Summary ===
To summarise, the three properties for a conductor at electrostatic equilibrium are:
# $\vec E_{internal}=0$
# charges may exist only on the surface of the conductor
# on the surface, $\vec E =0 \lor \vec E \bot \text {surface}$

Revision as of 21:08, 1 October 2026

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A conductor is said to have reached electrostatic equilibrium when its charge distribution becomes stable.

Properties

Consider a solid conductor $(S)$ at electrostatic equilibrium.

By combination of Newton's 2nd law and Coulomb's law, for any macroscopic charge $A$:

$$\vec a = \frac{Q_A\vec E}{m}$$

By definition of electrostatic equilibrium,

$$\begin{gathered} \forall A \in (S), \vec a=0 \\ \therefore \vec E =0 \text{ for every point inside (S) (1)} \end{gathered}$$

As a result, if conductor $(S)$ is charged, any charge must be destributed on the surface of $(S)$, $S(S)$, which creates an electric field on $S(S)$.

This field must satisfy $\vec E \bot S(S)$ at all points. (3) Otherwise, the macroscopic charges would move along the surface.

Summary

To summarise, the three properties for a conductor at electrostatic equilibrium are:

  1. $\vec E_{internal}=0$
  2. charges may exist only on the surface of the conductor
  3. on the surface, $\vec E =0 \lor \vec E \bot \text {surface}$