For a resistor, Ohm's law givesThe electric current and voltage are in phase.

For explicit sketches take a symmetric, 50% duty cycle square wave with levels $\pm V_0$. The source does not specify levels or duty cycle; a unipolar 50% wave follows by adding its mean and rescaling. Let $\tau =RC$ and $M$ be the positive output extremum in the periodic steady state. The square-wave response of an RC low-pass filter hasextended periodically. The extremum formula follows by imposing $V_{\mathrm{out}}(0)=-M$ and $V_{\mathrm{out}}(T/2)=M$. It makes the three regimes below quantitative, including their finite amplitudes.

The RC low-pass filter satisfies $RC{\dot{V}}_{\mathrm{out}}+V_{\mathrm{out}}=V$ for $t>0$. With zero initial capacitor voltage, the first-order linear differential equation givesThe output is continuous at switching and approaches $V$ with time constant $RC$.

The half-power amplitude condition for the RC low-pass filter gives $1+(\omega RC{)}^2=2$. Since angular frequency is positive,At this cutoff angular frequency, the output amplitude is smaller by $1/\sqrt{2}$ and the phase shift is $-\pi /4$.

The output is measured across the capacitor. The voltage divider gives the RC low-pass filter transfer functionThereforeThe amplitude ratio tends to $1$ and the phase shift to zero as $\omega \to 0$. For $\omega \to \infty$, the amplitude ratio is asymptotic to $1/(\omega RC)$ and the phase shift tends to $-\pi /2$.

The grounded outer cylinder fixes $\Phi (a)=0$. Integrating the annular electric field inward givesEquivalently, the coaxial cylindrical capacitor has $C_{\mathrm{cyl}}=2\pi {ϵ}_{0}l/\log (a/b)$ and carries charge $Q=4\pi {ϵ}_{0}RV$, so $\Phi (b)=Q/C_{\mathrm{cyl}}$. The original sphere potential $V$ is used to infer its conserved charge; it is not the final potential of the sphere or of the inner cylinder.

For $b<r<a$, a coaxial Gauss surface of length $l$ encloses net charge $Q$: the sphere and the cavity-wall charge cancel, leaving $+Q$ on the outside of the inner conductor. In the long-cylinder approximation,The electric field points radially outward for $V>0$. End effects and axial nonuniformity near the openings are excluded by the ideal coaxial cylindrical capacitor approximation.

Initially the isolated sphere has $Q=4\pi {ϵ}_{0}RV$. After insertion without contact, its charge remains $Q$. In the long-cylinder approximation, with the sphere far from the openings, neglect electric flux escaping through the ends. A Gauss surface running through the shielding metal and closed by end sections of negligible flux then encloses zero net charge. Its cavity wall therefore carries $-Q$. Since that cylinder is isolated and initially neutral, its outside wall carries $+Q$. The grounded outer cylinder supplies charge $-Q$ on its inside wall, terminating the annular electric field lines.

For positive $V$, the cavity electric field lines run from the sphere to the inner wall, while annular electric field lines run from the outside of the inner conductor to the grounded outer conductor. Negative $V$ reverses their directions. The transverse picture is schematic: exact cavity fields depend on the sphere's position and the ends. The following annular formulas neglect end effects and place the sphere well inside the long cylindrical assembly.

Assume normal dispersion, so shorter visible wavelengths have larger refractive indices. At fixed external angle $\theta$, the relevant criticality parameter iswhich increases with $n$ on this route. Thus shorter-wavelength blue light is totally reflected at B and reaches C, while longer-wavelength red light can transmit through B; green lies at the critical threshold. The prism separates the routes by wavelength near the green cutoff.

The totally reflected rays emerging from C have the same external deviation $180^{\circ }-2\alpha +2\theta$ despite their different refractive indices; their exit positions can differ. Red light also has a partially reflected component by the Fresnel equations, so C is not a mathematically perfect colour cutoff. The observation concerns total versus partial reflection, with the transmitted B rays spread by wavelength. Anomalous dispersion, if present, would require the actual $n(\lambda )$ instead of the normal ordering assumed here.

Take the positive horizontal direction along B. The initial direction is $(\alpha -90^{\circ })-\theta$. Reflection at the horizontal face B reverses the vertical direction. At C, symmetry of the equilateral geometry gives internal incidence $r$, so Snell's law makes the emerging angle to C's normal equal to $\theta$. Its direction is therefore $(90^{\circ }-\alpha )+\theta$. Subtracting the incoming direction givesThis is the smaller angle between the incident and emergent propagation directions in the requested reflection interval. The cancellation of $n$ occurs because the entry and exit refraction angles have equal magnitudes; refractive index still determines whether the reflected route exists and where it exits.

For $n=1.5$, the critical angle is $i_c=\arcsin (1/1.5)\simeq 41.8103^{\circ }$. Here $0\le r<i_c<\alpha =60^{\circ }$, so the condition at B is $\alpha -r>i_c$. At the threshold,Thus the requested positive-angle interval for strict total internal reflection isAt the upper equality the transmitted ray is grazing; normal incidence $\theta =0$ also gives total internal reflection at B if the endpoint is included. The distinction follows the question's positive-angle convention.

Let $r=\arcsin (\sin \theta /n)$ be the angle inside the prism relative to the inward normal of face A. For the entry position shown, below the midpoint of A, and $0<\theta <90^{\circ }$, the transmitted ray first reaches face B. Its incidence angle there is $i_B=|\alpha -r|$. If $n\sin i_B<1$, a transmitted ray leaves B according to Snell's law. If $n\sin i_B>1$, it undergoes total internal reflection at B and then reaches C. At the equality, transmission is grazing.

On the reflected route through this equilateral prism, the incidence angle at C is $r$, so $n\sin r=\sin \theta <1$ permits emergence from C. The one-reflection path through an equilateral prism is drawn alongside the direct-transmission route.

Real interfaces also produce partial Fresnel reflection below the critical angle; these additional weaker rays are omitted from the principal-path sketch. Which face a ray first reaches also depends on its entrance position, so the route statement uses the entrance geometry illustrated in the source.

For a displacement $(0,y)$ of the middle wire, the horizontal attractions towards the two fixed outer wires cancel. Each attraction has magnitude ${\mu }_0{I}^2/(2\pi \sqrt{d^2+y^2})$ per unit length. Projecting them in the $y$ direction givesNeglecting dissipative effects and other transverse forces, this is simple harmonic motion. Release from rest at $y_0$ givesThe inverse-angular-frequency timescale is ${\Omega }^{-1}=0.500\mathrm{s}$; it differs from the oscillation period by $2\pi$. The assumed displacement plane matters: small displacements towards one outer wire along $x$ are unstable, rather than oscillatory.

This result concerns the specified magnetic motion with gravity absent from, or compensated in, the $y$ direction. The source diagram fixes coordinate directions, not their orientation relative to gravity. If $y$ is physically vertical with upward positive and the wire is unsupported, add $-{\mu }_{m}g$ to the force per unit length. The maximum restoring magnetic force is only ${\mu }_0{I}^2/(2\pi d)=2.0\times 10^{-5}\mathrm{N}{\mathrm{m}}^{-1}$, compared with weight $9.8\times 10^{-3}\mathrm{N}{\mathrm{m}}^{-1}$; that different setup would fall and could not oscillate about the original plane.

Put the three wires at $(x,y)=(-d,0),(0,0),(d,0)$ with $d=1\mathrm{cm}$ and each current along $+z$, out of the page. Adding their Biot-Savart law magnetic fields givesFor $y\ne 0$, all three $x$-components have the same sign, so they cannot cancel. On $y=0$, a zero of the remaining component satisfiesExcluding the wire positions themselves, the two and only two finite zeros areThe magnetic field lines circulate counterclockwise about each current near that wire. Far away they circle the combined total current $3I$. The two nulls separate the nearby circulation regions in the original magnetic field line plot below.

Each long wire lies in the other's magnetic field, $B={\mu }_{0}I/(2\pi d)$. The magnetic Lorentz force per unit length has magnitudeFor the two-wire circuit's self-inductance, a nonzero wire radius is essential. Introduce a radius $a_w\ll d$. Between the wires their magnetic fields add. The external magnetic flux linkage per unit length through the connecting ribbon is, to leading order in $a_w/d$,It follows that the leading thin-wire loop self-inductance per unit length isIt is the loop value for the pair, not the contribution of just one conductor. Corrections from the wire interior depend on the current distribution: for nonmagnetic round wires with uniform direct electric current, the pair contributes another ${\mu }_0/(4\pi )$; for a surface current the internal contribution is zero. The leading logarithm is common to both models.

TODO (source data): the source gives no wire radius or current-distribution convention. There is no finite radius-independent value of the requested self-inductance: taking ideal zero-width wires sends $L_{\mathrm{ext}}^{\prime }$ to infinity. The force and the parameterized leading self-inductance above are determined, but a unique finite answer requires that additional specification. For example, changing $a_w$ to $a_w/2$ changes the leading result by ${\mu }_0\ln 2/\pi$ even at fixed separation.

For a current element $Id\mathbf{\ell }$, the Biot-Savart law isTake a straight wire along $z$, extending from $-a$ to $a$, and observe its magnetic field at $(s,0,0)$. All contributions point in $+\hat{\mathbf{\varphi }}$ for positive $I$ along $+z$, by the right-hand rule. Their magnitude integrates toFor $a/s\to \infty$ this becomes ${\mu }_{0}I/(2\pi s)$, so $\oint \mathbf{B}\cdot d\mathbf{\ell }=2\pi s{B}_{\varphi }={\mu }_{0}I$, exactly as Ampère's law requires for an infinite straight wire.

With $V=+20\mathrm{V}$ in the outer-minus-inner convention, $\eta =\ln 1.5=0.405465$ and the radial electric field at the orbit is $E_r=-986.52\mathrm{V}{\mathrm{m}}^{-1}$. The signed roots areUsing the proton mass and charge givesThe speeds are therefore $48.8$ and $96.7\mathrm{km}{\mathrm{s}}^{-1}$, in opposite senses of rotation. Both are nonrelativistic. The slower counterclockwise orbit has an outward magnetic force; the faster clockwise orbit has an inward magnetic force. Reversing $\mathbf{B}$ interchanges the senses, while reversing the electrode polarity changes the quadratic and may eliminate real orbits; its sign must therefore be specified.