Charge and Coulomb's law

Corresponding Wikipedia articles: Electric charge, Coulomb's law

Charge

The charge is an electric flux through a closed surface of limited space volume. The following equation is known as Gauss's law for electric field:

 Domain SI CGS Simplified $\tag{1}$ Vacuum (aether) $Q=\varepsilon_0\oint{\frac{c}{|\mathbf{\overrightarrow{v}}|}\mathbf{\overrightarrow{E}}\mathrm{d}\mathbf{\overrightarrow{S}}}$ $q=\varepsilon_0\frac{c}{|\mathbf{\overrightarrow{v}}|}div\;\mathbf{\overrightarrow{E}}+\varepsilon_0\mathbf{\overrightarrow{E}}\;grad\frac{c}{|\mathbf{\overrightarrow{v}}|}$ $Q=\frac{1}{4\pi}\oint{\frac{c}{|\mathbf{\overrightarrow{v}}|}\mathbf{\overrightarrow{E}}\mathrm{d}\mathbf{\overrightarrow{S}}}$ $q=\frac{c}{4\pi|\mathbf{\overrightarrow{v}}|}div\;\mathbf{\overrightarrow{E}}+\frac{1}{4\pi}\mathbf{\overrightarrow{E}}\;grad\frac{c}{|\mathbf{\overrightarrow{v}}|}$ $Q=\oint{\frac{c}{|\mathbf{\overrightarrow{v}}|}\mathbf{\overrightarrow{E}}\mathrm{d}\mathbf{\overrightarrow{S}}}$ $q=\frac{c}{|\mathbf{\overrightarrow{v}}|}div\;\mathbf{\overrightarrow{E}}+\mathbf{\overrightarrow{E}}\;grad\frac{c}{|\mathbf{\overrightarrow{v}}|}$ Matter $Q=\oint{\mathbf{\overrightarrow{D}}\mathrm{d}\mathbf{\overrightarrow{S}}}$ $q=div\;\mathbf{\overrightarrow{D}}$ $Q=\frac{1}{4\pi}\oint{\mathbf{\overrightarrow{D}}\mathrm{d}\mathbf{\overrightarrow{S}}}$ $q=\frac{1}{4\pi}div\;\mathbf{\overrightarrow{D}}$ $Q=\oint{\mathbf{\overrightarrow{D}}\mathrm{d}\mathbf{\overrightarrow{S}}}$ $q=div\;\mathbf{\overrightarrow{D}}$ $\tag{2}$

The actual value of the particle charge does not depend on the environmental permittivity, because the interaction force is inversely proportional to the permittivity by its definition ("Electric field and Lorentz force", 1).

Coulomb's law

The charge interaction force is determined by the potential energy $$U$$ of a point charge $$q$$ with an electric displacement $$\mathbf{\overrightarrow{D_q}}$$, positioned in the field $$\mathbf{\overrightarrow{E}}$$ of a charge $$Q$$ at a distance $$R$$ from it. The equation of the energy density ("Electric field and Lorentz force", 6) in the spherical coordinates centered at the position of charge $$Q$$, with substitutions $$\mathrm{d}V=\mathrm{d}S\mathrm{d}r$$, $$\mathbf{\overrightarrow{E}}\mathrm{d}S=E\mathrm{d}\mathbf{\overrightarrow{S}}$$, is: $U=\oint{\mathbf{\overrightarrow{E}}\mathbf{\overrightarrow{D_q}}\mathrm{d}V}=\int_0^\infty{E\mathbf{\overrightarrow{D_q}}\mathrm{d}\mathbf{\overrightarrow{S}}\mathrm{d}r}\tag{3 – SI, Sim.}$ $U=\frac{1}{4\pi}\oint{\mathbf{\overrightarrow{E}}\mathbf{\overrightarrow{D_q}}\mathrm{d}V}=\frac{1}{4\pi}\int_0^\infty{E\mathbf{\overrightarrow{D_q}}\mathrm{d}\mathbf{\overrightarrow{S}}\mathrm{d}r}\tag{3 – CGS}$ The field produced by the charge $$Q$$ follows from (2), and for a homogeneous dielectric medium is defined by the equation: $Q=\oint{\mathbf{\overrightarrow{D}}\mathrm{d}\mathbf{\overrightarrow{S}}}=\varepsilon_0\varepsilon E4\pi r^2\tag{4 – SI}$ $Q=\frac{1}{4\pi}\oint{\mathbf{\overrightarrow{D}}\mathrm{d}\mathbf{\overrightarrow{S}}}=\varepsilon Er^2\tag{4 – CGS}$ $Q=\oint{\mathbf{\overrightarrow{D}}\mathrm{d}\mathbf{\overrightarrow{S}}}=\varepsilon E4\pi r^2\tag{4 – Sim.}$ The field flux of the charge $$q$$ through the spherical surface of rafius $$r$$ is non-zero only if the charge $$q$$ is inside the sphere, that is when $$r\geq R$$. The consequent modifying of the integration limits and substituting the expressions for $$E$$ gives: $U=\frac{Q}{4\pi\varepsilon_0\varepsilon}\int_R^\infty{\oint{\mathbf{\overrightarrow{D_q}}\mathrm{d}\mathbf{\overrightarrow{S}}\frac{\mathrm{d}r}{r^2}}}=\frac{Qq}{4\pi\varepsilon_0\varepsilon}\int_R^\infty{\frac{\mathrm{d}r}{r^2}}=\frac{1}{4\pi\varepsilon_0\varepsilon}\frac{Qq}{R}\tag{5 – SI}$ $U=\frac{Q}{4\pi\varepsilon}\int_R^\infty{\oint{\mathbf{\overrightarrow{D_q}}\mathrm{d}\mathbf{\overrightarrow{S}}\frac{\mathrm{d}r}{r^2}}}=\frac{Qq}{\varepsilon}\int_R^\infty{\frac{\mathrm{d}r}{r^2}}=\frac{1}{\varepsilon}\frac{Qq}{R}\tag{5 – CGS}$ $U=\frac{Q}{4\pi\varepsilon}\int_R^\infty{\oint{\mathbf{\overrightarrow{D_q}}\mathrm{d}\mathbf{\overrightarrow{S}}\frac{\mathrm{d}r}{r^2}}}=\frac{Qq}{4\pi\varepsilon}\int_R^\infty{\frac{\mathrm{d}r}{r^2}}=\frac{1}{4\pi\varepsilon}\frac{Qq}{R}\tag{5 – Sim.}$ The interaction force is equal to the potential energy gradient. In this case, it is equal to the derivative with respect to a distance $$R$$, representing the well-known Coulomb's law, which is valid for the point charges at the long distances: $\mathbf{\overrightarrow{F}}=-\nabla U=\frac{1}{4\pi\varepsilon_0\varepsilon}\frac{Qq}{R^2}\tag{6 – SI}$ $\mathbf{\overrightarrow{F}}=-\nabla U=\frac{1}{\varepsilon}\frac{Qq}{R^2}\tag{6 – CGS}$ $\mathbf{\overrightarrow{F}}=-\nabla U=\frac{1}{4\pi\varepsilon}\frac{Qq}{R^2}\tag{6 – Sim.}$