Derive Maxwell’s third equation in integral and differential form

Maxwell's third equation is Faraday's law of electromagnetic induction: \(\oint \vec E\cdot d\vec l = -\dfrac{d}{dt}\int \vec B\cdot d\vec S\) (integral form) and \(\nabla\times\vec E = -\dfrac{\partial \vec B}{\partial t}\) (differential form). For the given \(\vec D\), the flux is 100 C.

(Numbering used: 1st \(\nabla\cdot\vec D = \rho\), 2nd \(\nabla\cdot\vec B = 0\), 3rd Faraday's law, 4th Ampere–Maxwell law.)

Integral form. Faraday's law: the emf induced in a closed loop equals the negative rate of change of magnetic flux through it:

\[e = -\frac{d\phi_B}{dt}\]

The emf is the work done per unit charge around the closed path, \(e = \oint_C \vec E\cdot d\vec l\), and the flux is \(\phi_B = \int_S \vec B\cdot d\vec S\), where \(S\) is any surface bounded by \(C\). So

\[\oint_C \vec E\cdot d\vec l = -\frac{d}{dt}\int_S \vec B\cdot d\vec S\]

For a fixed (stationary) surface, the time derivative can go inside:

\[\oint_C \vec E\cdot d\vec l = -\int_S \frac{\partial\vec B}{\partial t}\cdot d\vec S\]

Differential form. By Stokes' theorem, \(\oint_C \vec E\cdot d\vec l = \int_S (\nabla\times\vec E)\cdot d\vec S\). Hence

\[\int_S (\nabla\times\vec E)\cdot d\vec S = -\int_S \frac{\partial\vec B}{\partial t}\cdot d\vec S\]

This holds for every surface, so the integrands are equal:

\[\nabla\times\vec E = -\frac{\partial\vec B}{\partial t}\]

Meaning: a time-varying magnetic field produces a curling (non-conservative) electric field.

Numerical. \(\vec D = 20x\,\hat i + 10\,\hat j\) C/m². The area of 1 m² normal to the x-axis has \(d\vec S = \hat i\,dS\), so only the x-component contributes. At \(x = 5\) m:

\[D_x = 20\times5 = 100\ \mathrm{C/m^2}\]

\[\psi = \int \vec D\cdot d\vec S = 100\times 1 = 100\ \mathrm{C}\]

The y-component (10 \(\hat j\)) is parallel to the surface and contributes no flux.