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--- eqn-electromagnetism [February 05, 2026 at 23:13] – yanevskiv
+++ eqn-electromagnetism [May 14, 2026 at 11:38] (current) – external edit 127.0.0.1
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+# (WIP) Equations of Electromagnetism
+## Electrostatics
+### Coulomb's law
+Attractive force $Q_2$ feels thanks to $Q_1$'s electric field:
+$$\mathbf F_ = \frac{1}{4\pi\varepsilon_0}\frac{Q_1Q_2}{r^2}\mathbf r_{012}$$
+### Electric field
+Electric field created by a single point charge $Q$:
+$$\mathbf E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}\mathbf r_0$$
+Electric field created by multiple point charges $Q_1, Q_2, ..., Q_N$:
+$$\mathbf E = \frac{1}{4\pi\varepsilon_0}\sum_{i=1}^{N}\frac{Q_i}{r^2_i}\mathbf r_{0i}$$
+Electric field created by charge distributed along a 1D curve $L$, where $Q'$ $[C / m]$ is the charge density:
+$$\mathbf E = \frac{1}{4\pi\varepsilon_0}\int_C\frac{Q'\mathrm d\ell}{r^2}\mathbf r_0$$
+Electric field created by charge distributed across a 2D surface $S$ where $\rho_S$ $[C / m^2]$ is the charge density:
+$$\mathbf E = \frac{1}{4\pi\varepsilon_0}\int_S\frac{\rho_S\mathrm dS}{r^2}\mathbf r_0$$
+Electric field created by charge distributed across a 3D volume $V$ where $\rho$ $[C / m^3]$ is the charge density:
+$$\mathbf E = \frac{1}{4\pi\varepsilon_0}\int_V\frac{\rho\mathrm dV}{r^2}\mathbf r_0$$
+### Electric potential
+Electric field $\mathbf E$ created by the electric potential $V$:
+$$\mathbf E = -\nabla V$$
+Nabla operator $\nabla$ is defined as:
+$$\nabla = \frac{\partial}{\partial x}\mathbf i_x +  \frac{\partial}{\partial y}\mathbf i_y +  \frac{\partial}{\partial z}\mathbf i_z$$
+Electric potential $V$ created by a single point charge $Q$
+$$V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}$$
+Electric potential $V$ created by point charges $Q_1, Q_2, ..., Q_N$:
+$$V = \frac{1}{4\pi\varepsilon_0}\sum_{i=1}^{N}\frac{Q_i}{r_i}$$
+Electric potential $V$ created by charge distributed along a 1D curve $C$ with $Q'$ as the charge density:
+$$V = \frac{1}{4\pi\varepsilon_0}\int_C\frac{Q'\mathrm d\ell}{r}$$
+Electric potential $V$ created by charge distributed across a 2D surface $S$ with $\rho_S$ as the charge density:
+$$V = \frac{1}{4\pi\varepsilon_0}\int_S\frac{\rho_S\mathrm dS}{r}$$
+Electric potential $V$ created by charge distributed across a 3D volume $V$ with $\rho$ as the charge density:
+$$V = \frac{1}{4\pi\varepsilon_0}\int_V\frac{\rho\mathrm dV}{r}$$
+Electric potential $V$ as the line integral of an electric field $\mathbf E$ from arbitrary point $P$ to a reference point $R$:
+$$V = \int_P^R\mathbf E\cdot\mathrm d\mathbf\ell$$
+Electric potential $V'$ when moving the refernce point from $R$ to $R'$ in previous relation:
+$$V' = V + \int_{R}^{R'}\mathbf E\cdot\mathrm d\mathbf \ell$$
+Electric potential around a charge distributed along a line with charge density Q':
+$$V = \frac{Q'}{2\pi\varepsilon_0}\ln\frac{r_R}{r}$$
+Potential difference bewteen two points $A$ and $B$:
+$$U_{AB} = V_A - V_B = \int_A^B\mathbf E\cdot\mathrm d\mathbf\ell$$
+### Gauss's law
+Gauss's law (integral form):
+$$\oint_S\mathbf E \cdot\mathrm dS = \frac{1}{\varepsilon_0}\int_V\rho\mathrm dV$$
+Gauss's law in differential form:
+$$\nabla\mathbf E = \frac{\rho}{\varepsilon_0}$$
+### Maxwell's equations in Electrostatics
+Faraday's law in differential form in electrostatics:
+$$\nabla\cdot\mathbf E = 0$$
+Gauss's law in differential form in electrostatics:
+$$\nabla\cdot\mathbf E = \frac{\rho}{\varepsilon_0}$$
+Faraday's law in integral form in electrostatics
+$$\oint_C\mathbf E\cdot\mathrm d\mathbf\ell = 0$$
+Gauss's law in integral form in electrostatics
+$$\oint_S\mathbf E\cdot\mathrm d\mathbf S = \frac{1}{\varepsilon_0}\iiint_V\rho\cdot\mathrm dV$$
+### Poisson and Laplace equations
+Electric potential in vaccuum satisfies the Poisson equation:
+$$\nabla^2 V = -\frac{\rho}{\varepsilon_0}$$
+Without charge $\rho$, Poisson equation reduces to Laplace equation:
+$$\nabla^2 V = 0$$
+Nabla squared operator is defined as:
+$$\nabla = \frac{\partial^2}{\partial x^2}\mathbf i_x +  \frac{\partial^2}{\partial y^2}\mathbf i_y +  \frac{\partial^2}{\partial z^2}\mathbf i_z$$
+### Electrostatic dipole
+Electric potential of an electric dipole:
+$$V = \frac{1}{4\pi\varepsilon_0}\left(\frac{Q}{r^+} - \frac{Q}{r^-}\right)$$
+Electric field of an electric dipole ($\mathbf E = -\nabla\cdot V$):
+$$\mathbf E = -\nabla\cdot V$$