The integral form of Kirchoff’s Voltage Law for electrostatics (KVL; Section [m0016_Kirchoffs_Voltage_Law_Electrostatics_Integral_Form]) states that an integral of the electric field along a closed path is equal to zero: $\oint_{\mathcal C}{ {\bf E} \cdot d{\bf l} } = 0$ where $${\bf E}$$ is electric field intensity and $${\mathcal C}$$ is the closed curve. In this section, we derive the differential form of this equation. In some applications, this differential equation, combined with boundary conditions imposed by structure and materials (Sections [m0020_Boundary_Conditions_on_E] and [m0021_Boundary_Conditions_on_D]), can be used to solve for the electric field in arbitrarily complicated scenarios. A more immediate reason for considering this differential equation is that we gain a little more insight into the behavior of the electric field, disclosed at the end of this section.
The equation we seek may be obtained using Stokes’ Theorem (Section [m0051_Stokes_Theorem]), which in the present case may be written: $\int_{\mathcal S} \left( \nabla \times {\bf E} \right) \cdot d{\bf s} = \oint_{\mathcal C} {\bf E}\cdot d{\bf l} \label{m0152_eKVL2}$ where $${\mathcal S}$$ is any surface bounded by $${\mathcal C}$$, and $$d{\bf s}$$ is the normal to that surface with direction determined by right-hand rule. The integral form of KVL tells us that the right hand side of the above equation is zero, so: $\int_{\mathcal S} \left( \nabla \times {\bf E} \right) \cdot d{\bf s} = 0$ The above relationship must hold regardless of the specific location or shape of $${\mathcal S}$$. The only way this is possible for all possible surfaces is if the integrand is zero at every point in space. Thus, we obtain the desired expression: $\boxed{ \nabla \times {\bf E} = 0 } \label{m0152_eKVL}$ Summarizing: