The electric field is force per unit positive stationary test charge, in the limit that the test charge does not disturb the source:The magnetic flux density is defined by the velocity-dependent part of the Lorentz force, ${\mathbf{F}}_B=q\mathbf{v}\times \mathbf{B}$. Equivalently, a straight current element in a uniform field experienceswhich fixes its magnitude and right-hand-rule direction by force measurements with different current orientations.

Isolation fixes charge, not voltage: $Q=C_0{V}_0$. The electrostatic energy is therefore $U(t)=Q^2/[2C(\delta (t))]$. Inverting the small-tilt capacitance givesThe leading variation has frequency $2{\omega }_0$, since tilts of opposite sign give the same capacitance. With $U_0=C_0{V}_0^2/2$, the result is $U_0[1-{\delta }_0^2/(6a^2)+({\delta }_0^2/(6a^2))\cos 2{\omega }_{0}t]$ to this order. Energy decreases as the gap becomes less uniform at fixed charge; the difference is exchanged mechanically with the oscillator.

Take $x\in [-s/2,s/2]$ along the tilt, with local separation $h(x)=a+2\delta x/s$. Since $s\gg a$ and the tilt is small, each narrow strip is a local parallel-plate capacitor at the same voltage $V$. Its surface charge densities are approximatelyParallel strip capacitances add, so for $|\delta |<a$,Using $\log [(1+z)/(1-z)]=2(z+z^3/3+z^5/5+\cdots )$ givesThe denser charge at the smaller gap explains why tilting increases the capacitance. The local-strip approximation neglects fringing and slope corrections of order $(\delta /s{)}^2$, smaller than the retained $(\delta /a{)}^2$ term.

Ignoring edge fields, Gauss's law gives $E=\sigma /{ϵ}_0$ between the parallel plates. Their voltage is $V=Ea=\sigma a/{ϵ}_0$, and their charge magnitude is $Q=\sigma s^2$. Therefore

Let the column radius be $b=\sqrt{A/\pi }$, and $J=I_0/A$ be its uniform axial current density. Treating it as concentric shells, only the current at radii below $r$ contributes to $B(r)$: a surrounding shell has zero interior field. ThusThe magnetic force density is radially inward, $\mathbf{J}\times \mathbf{B}=-({\mu }_0{J}^{2}r/2)\hat{\mathbf{r}}$. Radial mechanical equilibrium at a fixed height requiresConsequentlyThe pressure is higher at the centre: its outward force balances the inward magnetic pinch. Vertical hydrostatic variation is separate from this radial pressure difference.

For a thin current-carrying cylindrical shell, a coaxial Amperian loop encloses zero current inside and the entire current outside. HenceMore generally $B(r)=0$ for $r<R$ and $B(r)={\mu }_0{I}_0/(2\pi r)$ for $r>R$. The tangential field discontinuity is supported by the surface current.

Let the wire lie along $z$, and observe at perpendicular distance $r$. In the Biot-Savart law, the azimuthal contributions from line elements add with magnitudeIndependently, cylindrical symmetry makes the field constant and azimuthal on a circular Amperian loop, so Ampère's law gives $2\pi rB={\mu }_0{I}_0$. Both yieldwith direction fixed by the right-hand rule.

Use north-minus-south terminal voltages for both oppositely wound coils, with coil A measured at high input impedance so its AC current is zero. Then $v_B=L_B{\dot{I}}_B$ and $v_A=M{\dot{I}}_B$, with $M<0$. HenceThe magnitude is the turns ratio times the drive, and these terminal references give a phase reversal. Reversing one voltmeter lead reverses the displayed sign. A fixed DC bias in A does not change this incremental result, while short-circuiting A would require solving both coupled current equations.

Let $B_E>0$ be the northward component of the ambient magnetic field. Define $\eta =+1$ if positive north-to-south current in A produces a northward field, and $\eta =-1$ otherwise. With opposite windings and the same signed current in both,A compass reverses only when this northward component becomes negative. For current chosen to make the coil field southward,At equality the ideal axial field is zero and the needle has no preferred axial direction. If $n_1=n_2$, the opposing coil fields cancel and no finite current reverses it in this model. A numerical threshold cannot be given without $B_E$ and the coil parameters.

Reversing one winding reverses its signed coupling to the other, while its self-inductance is unchanged. With the same north-to-south current references,Equal currents then produce opposing magnetic fields: the signed axial field is proportional to $(n_1-n_2)I$, rather than $(n_1+n_2)I$. Which direction is positive must be fixed by the right-hand rule and the actual winding orientation.

For chosen positive terminal currents, define self-inductance by the coil's own flux linkage ${\Lambda }_j=L_j{I}_j$ and mutual inductance by ${\Lambda }_2=M{I}_1$ when only coil A carries current. With matching winding directions and terminal references,The reciprocal flux-linkage relation gives the same $M$ from B to A, and $M^2=L_A{L}_B$ in this ideal fully coupled long-solenoid limit. The induced emf is $-d\Lambda /dt$; passive terminal voltage drops use the opposite sign, $v_A=L_A{\dot{I}}_A+M{\dot{I}}_B$ and similarly for B.

Let $A_t=\pi d^2/4$ be the tube cross-section. A long uniformly wound solenoid with $n_1$ total turns has field $B_A={\mu }_0{n}_{1}I/s$, approximately uniform inside and negligible outside. Therefore the flux through one cross-section, or one turn of either coil, isThe linked flux of coil A is $n_1\Phi$; that of coil B is $n_2\Phi$. Distinguishing flux from flux linkage prevents an extra factor of the turn count.

For normal incidence, adjacent grating lines have optical path difference $s\sin \theta$. Constructive interference at order $m$ requires $s\sin {\theta }_m=m\lambda$. Differentiating at fixed $m$ yields the angular dispersionIt is zero in the zero order and increases in magnitude toward grazing diffraction. An order exists only if $|m\lambda |\le s$. The number of illuminated lines controls peak width and resolving power, but does not alter this angle-versus-wavelength relation.

The first single-slit minimum is at $\sin \theta =\lambda /w$, so the characteristic angular resolution of an aperture of width $D$ is of order $\lambda /D$. With $D=1$ m, $\lambda =550$ nm, and distance $\mathcal{D}=10^8$ m, this slit-based estimate isFor the actual circular lens, the Rayleigh criterion uses the first zero of the Airy disk, $\Delta \theta \simeq 1.22\lambda /D$. It gives $6.71\times 10^{-7}$ rad and a resolved separation of about $67$ m. The factor $1.22$ belongs to the circular aperture, not to the one-dimensional slit derivation.

In the Fraunhofer approximation, contributions from positions $\xi$ across the slit have relative phase $k\xi \sin \theta$. With uniform illumination, the field relative to its axial value isThus the single-slit diffraction intensity isOn the detector $\sin \theta =y/\sqrt{d^2+y^2}$; near the axis, $\beta \simeq \pi wy/(\lambda d)$. The central maximum takes the continuous limit $\sin \beta /\beta \to 1$, and the minima satisfy $w\sin \theta =m\lambda$ for nonzero integers $m$. The stated far-field condition justifies adding the aperture contributions with a linear phase.

Here $q_1=3.8q_3$ and $q_2=-1.9q_3$. The part of the electrostatic potential energy that changes isAt the two endpoints, $U(-0.9)=37k_e{q}_3^2/L$ and $U(0.9)=-17k_e{q}_3^2/L$. The fixed-source interaction energy is unchanged, and conservation of energy givesThe force points toward increasing $f$ throughout this interval: the left charge repels $q_3$ and the right charge attracts it, so release from rest reaches the stated endpoint without a turning point.

With $q_2=-q_1$, both field contributions between the charges point the same way. Their signed force on $q_3$ isSince $f$ is dimensionless, the small-displacement expansion is $|f|\ll 1$, givingThe force at the midpoint is nonzero and the first variation is quadratic, not linear. In particular, the midpoint is not an equilibrium supporting a linear restoring-force approximation. The printed comparison of $f$ with $L$ is dimensionally inconsistent; the expansion above keeps their distinct roles.

Put $q_1$ at $x=-L$, $q_2$ at $x=+L$, and $q_3$ at $x=fL$, with $-1<f<1$. For a nonzero test charge, the force vanishes when the electric field vanishes:For equal-sign nonzero $q_1,q_2$, take the positive square root $s=\sqrt{q_2/q_1}$. The distances are positive, so $1-f=s(1+f)$ andFor $s>1$ the zero moves toward the weaker charge $q_1$; equal charges give $f=0$.

For an initially uncharged capacitor and zero initial current, Kirchhoff's voltage law is $L\dot{I}+RI+Q/C=V_0$, with $I=\dot{Q}$. LetDifferentiating the circuit equation at $t>0$ gives $\ddot{I}+2\alpha \dot{I}+{\omega }_n^{2}I=0$, while the original equation fixes $I(0)=0$ and $\dot{I}(0)=V_0/L$. The underdamped response isFor critical damping, $\alpha ={\omega }_n$, its limit is $I=(V_0/L)t{e}^{-\alpha t}$. For overdamping, put $\beta =\sqrt{{\alpha }^2-{\omega }_n^2}$ and obtain $I=V_0{e}^{-\alpha t}\sinh (\beta t)/(L\beta )$. These forms cover every positive $R,L,C$.

The source does not explicitly specify the capacitor's initial charge. If its initial voltage is $V_{C0}$, replace $V_0$ in the numerators by $V_0-V_{C0}$; an open switch alone does not discharge a capacitor. The boxed expression uses the usual uncharged initial condition.

Use positive current in the charging direction and passive voltage drops across $R,L,C$. The five ideal elements obeyThe battery maintains its emf independently of current. An ideal open switch has $I=0$ with its voltage unconstrained; an ideal closed switch has $v_{\mathrm{switch}}=0$ with its current unconstrained. These statements are the switch's piecewise current-voltage relation. Inductor current continuity and capacitor-voltage continuity hold when no impulse voltage or current is imposed.