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Take the electric potential to vanish at infinity. The distance from the centre to each point charge is half the diagonal of a polygon of the square, namely $a/\sqrt{2}$. The superposition principle for the electric potential therefore gives, for four charges $+q$,Strictly, specifying only the magnitude of each electric charge leaves its sign undetermined. For signed charges $q_j$, the general answer is $V=\sqrt{2}{\sum }_{j=1}^4{q}_j/(4\pi {ϵ}_{0}a)$. For example, two positive and two negative charges of equal magnitude give zero electric potential at the centre. The boxed result uses the usual interpretation that all four charges have the same positive sign.

Fermat principle states that the optical path length $\int nds$, equivalently the travel time in the nondispersive stationary isotropic ray model, is stationary for the actual light ray between fixed endpoints. It need not be a minimum in every optical system.

For two homogeneous media separated by $y=0$, place the fixed endpoints at $(0,h_1)$ and $(D,-h_2)$, with positive heights. If the ray crosses at $(x,0)$, its optical path isStationarity with respect to the crossing point givesThe two fractions are $\sin {\theta }_1$ and $\sin {\theta }_2$, with angles measured from the interface normal. Thuswhich is Snell's law. Dividing $\mathcal{L}$ by the vacuum light speed gives the corresponding travel time; minimizing geometric distance without refractive-index weighting would incorrectly predict a straight ray through unlike media.

For an ideal inductor of constant self-inductance $L$, the passive-sign electric potential difference is $V_L=LdI/dt$. The supplied power is $V_{L}I$, so increasing the electric current quasistatically from zero storesAny electrical resistance would additionally dissipate energy as heat; it is not included in this stored magnetic energy.

For a long vacuum solenoid with $N$ turns, axial length $\ell$ and cross-sectional area $A$, Ampère's law gives $B={\mu }_0(N/\ell )I$. Its flux linkage is $N\Phi =NBA$, soThe nearly uniform field occupies volume $A\ell$, with negligible external-field energy in the long-solenoid approximation. Dividing its stored energy by this volume yieldsThe vacuum and linear-inductance assumptions matter; a nonlinear magnetic material requires the appropriate field-energy integral rather than simply substituting a varying inductance into $L{I}^2/2$.

Let $R_1<R_2$ and put charge per unit length $+{\lambda }_q$ on the inner shell and $-{\lambda }_q$ on the facing outer conductor. For a long coaxial Gaussian surface of radius $r$ in the gap, Gauss's law givesThe electric potential difference between the shells isWith total inner charge $Q={\lambda }_{q}L$, the capacitance isThis assumes a vacuum or approximately air-filled gap and negligible fringing at the ends. A uniform linear dielectric replaces ${ϵ}_0$ by its permittivity. The expression depends on the shell-to-shell potential difference, not on the arbitrary common potential reference.

The magnetostatic Ampère's law is $\oint \mathbf{B}\cdot d\mathbf{\ell }={\mu }_0{I}_{\mathrm{enc}}$, equivalently $\nabla \times \mathbf{B}={\mu }_0\mathbf{J}$ in vacuum. Let the wire radius be $a$ and take positive electric current along its axis. Cylindrical symmetry makes $\mathbf{B}$ azimuthal and constant in magnitude on a radius-$r$ circular path, so $2\pi rB={\mu }_0{I}_{\mathrm{enc}}$.

Uniform electric current density gives $J=I/(\pi a^2)$. Inside the wire, $I_{\mathrm{enc}}=J\pi r^2=I{r}^2/a^2$; outside, the full electric current is enclosed. ThereforeThe direction is set by the right-hand rule around the electric current. The magnitude starts at zero on the axis, rises linearly to ${\mu }_{0}I/(2\pi a)$ at the surface and then falls as $1/r$. The two expressions are continuous at $r=a$. The long-wire approximation neglects end effects and any high-frequency redistribution of electric current.

For a fixed number $N$ of identical bulbs, first approximate each as a resistor of electrical resistance $R$ connected to a fixed ideal battery electric potential difference $V$. A series circuit has electrical resistance $NR$, while a parallel circuit has electrical resistance $R/N$. ThusParallel connection therefore maximizes dissipation in this model: every bulb receives the full source electric potential difference instead of sharing it. For $N=1$ the two arrangements coincide.

An actual incandescent light bulb changes electrical resistance as its filament heats, so a single constant $R$ need not describe both operating points. The same qualitative conclusion holds for identical stable bulbs with increasing current-voltage function $i(U)$: parallel operation gives $P_p=NVi(V)$, whereas symmetric series operation gives $P_s=Vi(V/N)$. The former is larger for $N>1$. This comparison assumes a fixed collection and an ideal source as specified; an unlimited number of parallel bulbs on a zero-internal-resistance ideal battery would have no finite total-power maximum.

Let $\psi =kx-\omega t$ and use bold $\mathbf{B}$ for the magnetic flux density, to distinguish it from the scalar amplitude $B$ in $E_z$. For a plane electromagnetic wave travelling along $+x$, Faraday's law gives the wave-associated fieldHenceFor an explicit component check, $\nabla \times \mathbf{E}=(0,kB\sin \psi ,kA\cos \psi )$, while differentiating the displayed field gives $-\partial \mathbf{B}/\partial t$ equal to that vector. The field is transverse and also satisfies $\nabla \cdot \mathbf{B}=0$.

In vacuum, Ampère-Maxwell equation compatibility requires $\omega /k=c$, so each factor $k/\omega$ is $1/c$. More generally it is the inverse phase velocity on a suitable homogeneous-medium plane-wave branch. A time-independent background field could be added without changing the electric wave's Faraday equation; the requested propagating-wave field takes that independent background to be zero. Finally, $\mathbf{E}\times \mathbf{B}=(k/\omega )(E_y^2+E_z^2)\hat{\mathbf{x}}$ confirms the propagation direction and the relative signs.

Take the electric potential to vanish at infinity and work at $r>0$. For a positive point charge $q$ in vacuum,The field is radial and outward. Since a radial scalar function has gradient $\nabla V=(dV/dr)\hat{\mathbf{r}}$,Thus the two expressions have the required vector relationship. Adding an arbitrary constant to $V$ would leave the electric field unchanged; the chosen reference at infinity fixes that constant. The point-charge singularity itself is excluded from the ordinary derivative calculation.