Electric potential and capacitance

The electrostatic field is a potential field, which is equal to a gradient of the electric field potential. The potential $$U$$ is the potential energy $$W$$ of a charge, on which the field acts, related to the charge value: $\mathbf{\overrightarrow{F}}=-\nabla W=-grad\;W\tag{1}$ $\mathbf{\overrightarrow{E}}=-\nabla U=-grad\;U\tag{2}$ The electrical conductor is a medium for the charge carriers. The perfect conductor has no the resistance against this motion. Because of tending of the potential energy of the charge carriers to a minimum, an ideal conductor has the same potential over the entire volume, at least in case of the direct currents.
The capacitance is a quantity, which links the charge and the potential of an electrical conductor, depending on the conductor geometry and the environmental permittivity$C=\frac{Q}{U}=\frac{\oint{\mathbf{\overrightarrow{D}}\mathrm{d}\mathbf{\overrightarrow{S}}}}{\int{\mathbf{\overrightarrow{E}}\mathrm{d}\mathbf{\overrightarrow{l}}}}\tag{3 – SI, Sim.}$ $$C=\frac{Q}{U}=\frac{\oint{\mathbf{\overrightarrow{D}}\mathrm{d}\mathbf{\overrightarrow{S}}}}{4\pi\int{\mathbf{\overrightarrow{E}}\mathrm{d}\mathbf{\overrightarrow{l}}}}\tag{3 – CGS}$$ The potential in (3) is zero if the conductor is uncharged.
The electric energy of the conductor is determined by the energy volumetric density of the electric field ("Electric field and Lorentz force", 5) in the 3-dimensional space: $\frac{1}{2}\int{\mathbf{\overrightarrow{E}}\mathbf{\overrightarrow{D}}\mathrm{d}V}=\frac{1}{2}\int{\oint{\mathbf{\overrightarrow{E}}\mathbf{\overrightarrow{D}}\mathrm{d}\mathbf{\overrightarrow{S}}\mathrm{d}\mathbf{\overrightarrow{l}}}}\tag{4 – SI, Sim.}$ $\frac{1}{8\pi}\int{\mathbf{\overrightarrow{E}}\mathbf{\overrightarrow{D}}\mathrm{d}V}=\frac{1}{8\pi}\int{\oint{\mathbf{\overrightarrow{E}}\mathbf{\overrightarrow{D}}\mathrm{d}\mathbf{\overrightarrow{S}}\mathrm{d}\mathbf{\overrightarrow{l}}}}\tag{4 – CGS}$ This integral (4) contains the constant charge value ("Charge and Coulomb's law", 2) over entire volume, so: $\frac{1}{2}\oint{\mathbf{\overrightarrow{D}}\mathrm{d}\mathbf{\overrightarrow{S}}}\int{\mathbf{\overrightarrow{E}}\mathrm{d}\mathbf{\overrightarrow{l}}}=\frac{QU}{2}=\frac{CU^2}{2}\tag{5 – SI, Sim.}$ $\frac{1}{8\pi}\oint{\mathbf{\overrightarrow{D}}\mathrm{d}\mathbf{\overrightarrow{S}}}\int{\mathbf{\overrightarrow{E}}\mathrm{d}\mathbf{\overrightarrow{l}}}=\frac{QU}{2}=\frac{CU^2}{2}\tag{5 – CGS}$ The voltage is a difference of the electric potentials between two spatial points. The capacitance, defined by the voltage, characterizes two points (two capacitor plates etc.) and is determined by the electric displacement flux between them$C=\frac{\int{\mathbf{\overrightarrow{D}}\mathrm{d}\mathbf{\overrightarrow{S}}}}{\int{\mathbf{\overrightarrow{E}}\mathrm{d}\mathbf{\overrightarrow{l}}}}\tag{6 – SI, Sim.}$ $$C=\frac{\int{\mathbf{\overrightarrow{D}}\mathrm{d}\mathbf{\overrightarrow{S}}}}{4\pi\int{\mathbf{\overrightarrow{E}}\mathrm{d}\mathbf{\overrightarrow{l}}}}\tag{6 – CGS}$$ The electromotive force (EMF) is a circulation of the electric field in a closed loop (electric circuit), and its measurement units are the same as for the voltage.