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The original diagram has an ideal $5\mathrm{V}$ electric battery feeding series resistors of $8\Omega$ above the tapping point and $2\Omega$ below it. The long battery bar is positive, so terminal $A$ is above terminal $B$ in electric potential. With an open circuit at the output, the internal electric current is $5/(8+2)=0.5\mathrm{A}$. The voltage divider therefore givesFor the Thévenin resistance, suppress the ideal voltage source by replacing it with a short circuit. The two resistors are then in parallel as seen from $A,B$:By Thévenin's theorem, the terminal voltage under an outward load electric current $I$ is $V_A-V_B={\mathcal{E}}_0-I{R}_{\mathrm{Th}}$. Consequentlyflowing through an external short circuit from $A$ to $B$. Directly shorting the original terminals bypasses the $2\Omega$ resistor and leaves $5\mathrm{V}$ across $8\Omega$, independently confirming the result.

For normal incidence on a diffraction grating, adjacent grooves contribute a path difference $d\sin \theta$. Constructive interference occurs when the grating equation $d\sin \theta =m\lambda$ holds. Thus the groove spacing isAn allowed propagating diffraction order must have $|\sin \theta |\le 1$. ThereforeThe fifth diffraction order is the highest geometrically permitted order for this wavelength. Its actual brightness depends on groove shape and the single-slit diffraction envelope, which the source does not specify; geometric permission alone does not guarantee a bright observable maximum.

The current density is the moving charge density times its velocity:For a cylindrical Gaussian surface of radius $r<R$ and length $\ell$, cylindrical symmetry makes the electric field radial. Gauss's law gives $E(2\pi r\ell )=\rho \pi r^2\ell /{ϵ}_0$. A circular Ampère's circuital law contour encloses electric current $I(r)=\rho v_0\pi r^2$, giving $B(2\pi r)={\mu }_{0}I(r)$. HenceThe right-hand rule fixes the positive azimuthal direction of the magnetic field for positive electric charge moving along $+z$. Both the electric field and the magnetic field vanish on the axis. These are the steady long-cylinder electric field and magnetic field; end effects and changes of the prescribed beam distribution are excluded.

Use the parallel-plate capacitor approximation, neglecting the fringing field of a capacitor. With plate area $A=\pi (0.10\mathrm{m}{)}^2$, the initial capacitance and opposite plate electric charges areAfter the electric battery is disconnected, the isolated plate electric charges stay fixed. Doubling the separation halves the capacitance. The quasistatic pulling work is the increase of electrostatic energy:The final electric potential difference is thereforeAn electron released at rest gains kinetic energy $e\Delta V_f$, where $e$ is the positive elementary charge. Neglecting radiation and the negligible perturbation caused by this one electron, conservation of energy givesThis is $0.0280c$, so the nonrelativistic kinetic energy is adequate. The circular plates have finite radii; the numerical results are leading parallel-plate capacitor values rather than exact finite-plate capacitances.

In a source-free vacuum, the Maxwell equations in SI arewhere ${ϵ}_0$ and ${\mu }_0$ are the vacuum permittivity and vacuum permeability. Take the curl of the last equation, substitute Faraday's law of induction, and use the curl of the curl identity:The zero divergence therefore gives the wave equationFor a propagating plane wave, write $\mathbf{B}=\mathrm{Re}[{\mathbf{B}}_0{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}]$. The wave equation gives ${\omega }^2=c^2|\mathbf{k}{|}^2$, and Gauss's law for magnetism gives $\mathbf{k}\cdot {\mathbf{B}}_0=0$. Thus the oscillating magnetic field is a transverse wave: it is perpendicular to the propagation wavevector. A static background magnetic field, if present, is separate from this propagating component.