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The integral and differential forms of Gauss's law in vacuum areFor an isolated spherical conductor, symmetry gives $E(r)=Q/(4\pi {ϵ}_0{r}^2)$ outside and zero inside. Taking electric potential zero at infinity, $V(R)={\int }_R^{\infty }E(r)dr=Q/(4\pi {ϵ}_{0}R)$. Thus the capacitance of an isolated conducting sphere isCharging from zero gives the electrostatic potential energyThis is also the integral of electric field energy density ${ϵ}_0{E}^2/2$ outside the sphere. The later insertion into a coaxial cylindrical capacitor conserves its electric charge, not its original isolated electric potential.

Use an ideal inverting summing amplifier, grounding the non-inverting input and connecting the two sources through resistances $R$ and $R/2$. A feedback resistance $R_f=2R$ gives the desired weights. Operational-amplifier negative feedback makes the inverting node a virtual ground, and negligible input current lets Kirchhoff's current law give

Replacing the feedback resistor by a capacitor $C$ makes a weighted op amp integrator:The initial capacitor charge determines the integration constant. A nonzero DC weighted input produces a ramp; a real operational amplifier eventually reaches its output-voltage limit. A parallel feedback resistance can limit drift while preserving integration over an appropriate frequency range.

A grating spectrometer places an illuminated narrow slit at the front focal plane of a collimator. The resulting parallel beam reaches the diffraction grating at known incidence, here taken normal. A rotatable telescope focuses each outgoing diffracted parallel beam onto its crosshair or detector. Align the slit with the grating rulings, focus the collimator at infinity, and illuminate many grooves across a suitable aperture. A narrow slit supplies sufficient spatial coherence and prevents slit width from blurring the angular separation; the telescope must also resolve the two images.

The groove spacing is $d=(2000\times 100{)}^{-1}\mathrm{m}=5.00\mu \mathrm{m}$. The grating equation at normal incidence is $d\sin {\theta }_m=m\lambda$. For diffraction order $m=2$,ThereforeThe ideal resolving power of a diffraction grating is $|m|N$, so the Rayleigh criterion for this doublet requires about $N\ge 145$ illuminated grooves, spanning roughly $0.73\mathrm{mm}$. This is an ideal lower bound; finite slit width, aberrations and detector resolution can require a larger usable aperture.

The intensity of an isotropic radiator at distance $r$ is the radiated power divided by spherical area: $\mathcal{I}=P/(4\pi r^2)$. For a locally plane far-field electromagnetic wave in air, its mean Poynting flux is $\mathcal{I}=E_{\mathrm{rms}}^2/Z_0=E_{\mathrm{peak}}^2/(2Z_0)$, where the vacuum wave impedance is $Z_0\simeq 376.7\Omega$. ThusandBoth conventions are given because “field strength” alone does not specify RMS or peak. This estimate treats the stated transmitter power as radiated power and ignores ground reflection and absorption.

The continuity equation for electric charge is $\nabla \cdot J=-{\partial }_t\rho$. Taking the divergence of the magnetostatic Ampère's law would instead require $\nabla \cdot J=0$. The displacement current repairs this conflict: using Gauss's law, $\nabla \cdot ({ϵ}_0{\partial }_{t}E)={\partial }_t\rho$, so the Ampère-Maxwell equation isIn the parallel-plate capacitor, neglect fringing fields and take $E_z=V_0{e}^{-i\omega t}/d$. A circular Ampère loop of radius $s<a$ encloses displacement current $-i\omega {ϵ}_0\pi s^2{V}_0{e}^{-i\omega t}/d$. Therefore the magnetic field between circular capacitor plates has phasor and amplitudeIt is zero on the axis, grows linearly with radius, and is azimuthal by the right-hand rule. This is a quasistatic approximation: in addition to $d\ll a$, the uniform-field model requires $\omega a/c\ll 1$ and excludes the edge region. The answer gives the oscillation amplitude, rather than a time-independent instantaneous magnitude.

For steady currents in vacuum, Ampère's circuital law isHere $C$ is an oriented closed curve, $d\ell$ its tangent line element, and $I_{\mathrm{enc}}$ the signed electric current through a spanning surface oriented by the right-hand rule. A circular contour about a straight infinite wire has constant tangential magnetic field, so $B(r)={\mu }_{0}I/(2\pi r)$. The Lorentz force on a length $L$ of the other wire is $IL\hat{\ell }\times B$. ThereforeThe right-hand rule makes the force on each parallel, same-direction current point toward the other wire.

The Huygens-Fresnel principle constructs a later wavefront as the envelope of secondary wavelets emitted from the earlier one. At a planar interface, take a time interval during which one point on the incoming front travels distance $v_1\Delta t$ before reaching the interface, while the point that reached it first generates a wavelet of radius $v_2\Delta t$. The two right triangles share the interface segment, giving $\sin i/\sin r=v_1/v_2$. Since the refractive index is $n=c/v$, this yields Snell's law:The angles are measured from the interface normal. This geometric construction assumes homogeneous media and uses the propagation speeds at the given frequency.

The electric field of a signed point electric charge in vacuum follows Coulomb's law:For the conducting plane, the method of images replaces each charge $Q_j$ at height $h_j$ by an image charge $-Q_j$ at height $-h_j$. The real and image electric fields have cancelling tangential components and equal normal components at the surface. Thus the electric field at a conducting plane from axial point charges, just above the point directly below them, isNumerically,The surface value means the air-side limit. Inside the ideal conductor the electrostatic electric field is zero; its discontinuity across the surface is supported by induced surface charge. Omitting the image charges would give half the external answer.