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The Lorentz force isFor $\mathbf{E}=0$, the magnetic part is perpendicular to the velocity, so its instantaneous power $\mathbf{F}\cdot \mathbf{v}$ vanishes. The speed and kinetic energy therefore remain constant. Decompose the initial velocity into $v_{\parallel }$ along the uniform magnetic field and $v_{\perp }$ perpendicular to it. The parallel component is unchanged, while the perpendicular component rotates uniformly.

For a nonrelativistic particle with $qB\ne 0$, the resulting helical motion hasHere $r_L$ is the Larmor radius, ${\omega }_c$ the cyclotron frequency in angular units, and $h$ the axial advance per turn. Opposite signs of electric charge give opposite senses of rotation. Pure perpendicular motion gives a circle; pure parallel motion gives a straight line. If $q=0$ or $B=0$, there is no magnetic deflection. At relativistic speeds, replace $m$ in the radius and period by $\gamma m$; $\gamma$ stays constant because the magnetic force does no work.

Define $\mathbf{d}$ to point from the negative charge to the positive charge. The electric dipole moment is $\mathbf{p}=q\mathbf{d}$. In a uniform electric field, the two forces cancel, so the net force is zero. They nevertheless form a couple with torqueFor a positive angular displacement increasing $\theta$, the restoring torque component is $-pE\sin \theta$. Thus $dU/d\theta =pE\sin \theta$ andwith the additive constant chosen to make $U=0$ at $\theta =\pi /2$. Alignment is a stable minimum and antialignment an unstable maximum. The zero-force result uses a uniform field; a field gradient can exert a net force on a dipole.

The Lorentz force is $\mathbf{F}=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$. With no electric field, the exact nonrelativistic component equations for the given quadrupole areThe magnetic force does no work, so the total speed stays constant. In the paraxial approximation, $\dot{z}\simeq v$ and its small change is second order in the transverse motion. For $q>0$, put $\kappa =\sqrt{qA/(mv)}$. Initial travel parallel to the $z$ axis means zero entrance slopes, yieldingThus $x$ is focused while $y$ is defocused. The first focus in the focusing plane occurs atFor negative $q$, interchange the focusing and defocusing planes and use $|q|$ in $\kappa$. A single quadrupole magnet gives a line focus of a general beam, not simultaneous focusing of both transverse coordinates onto one point. The magnet must extend at least to the computed focus, and the slopes must remain small. For a relativistic beam, the corresponding paraxial factor is ${\kappa }^2=|q|A/p$, with $p$ the longitudinal relativistic momentum.

Choose $\mathbf{B}=B\hat{\mathbf{z}}$ and a signed tangential velocity $\mathbf{v}=u\hat{\mathbf{\varphi }}$, with $u>0$ counterclockwise viewed from $+z$. The magnetic part of the Lorentz force is $euB\hat{\mathbf{r}}$, while the electric part is $-eV\hat{\mathbf{r}}/(\eta r)$. The radial acceleration for a circular orbit is $-u^2/r$, regardless of the sign of $u$. ConsequentlyOne valid signed-velocity convention therefore has $a=m$, $b=eBr$, $c=-eV/\ln (r_2/r_1)$. If $v$ is required to be a nonnegative speed, the two senses of rotation instead satisfywith the plus sign for counterclockwise motion. For positive $V$, the electric force is inward, so it can supply circular motion in either direction; in one direction the magnetic force opposes it, while in the other it helps. A signed negative root must not be discarded as an unphysical speed; it denotes the opposite orbit direction.

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.

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 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.

Use the electric dipole above a grounded conducting plane approximation, valid for $d\ll a$. Let $\hat{\mathbf{z}}$ point upward, and choose $\mathbf{p}=p(\sin \theta \hat{\mathbf{x}}+\cos \theta \hat{\mathbf{z}})$. Reflecting the two constituent point charges and reversing their signs produces an image electric dipole momentat height $-a$. Thus the parallel electric dipole moment reverses and the normal one is unchanged. At the real electric dipole, the separation from its image is $2a\hat{\mathbf{z}}$, giving the image electric fieldThe physical torque is $\mathbf{\tau }=\mathbf{p}\times {\mathbf{E}}_{\mathrm{im}}$, soFor $0<\theta <\pi /2$, its direction reduces $\theta$, tending to align the electric dipole with the upward normal. In general it favours either normal orientation; a perfectly parallel orientation has zero torque but is unstable to tilting.

The conductor-response potential energy includes one half of the real-image interaction:This factor accounts for the induced surface charge changing as the physical source is assembled. Differentiation with respect to height at fixed orientation givesThere is no horizontal force, since translating the entire arrangement parallel to the plane leaves $U$ unchanged. For example, the normal and parallel orientations give respective downward force magnitudes $3p^2/(32\pi {ϵ}_0{a}^4)$ and $3p^2/(64\pi {ϵ}_0{a}^4)$.

A single physical point charge induces total plane electric charge $-q$. One can verify this by integrating its induced surface charge density, $-qh/[2\pi ({\varrho }^2+h^2{)}^{3/2}]$, over the plane. The two physical electric charges therefore induce opposite totals, givingThe local surface charge density is generally nonzero. The zero total holds for the finite neutral pair as well as its ideal electric dipole limit.

The method of images replaces electrical conductors by fictitious image charges outside the physical region. Choose the image charges so that the physical electric charges and their images produce the prescribed constant electric potential on every conducting boundary. The uniqueness theorem for the electrostatic potential then identifies this constructed electric potential with the physical solution in the charge-containing region. Several conducting planes can require several, or infinitely many, image charges; a single reflected image is sufficient for one infinite plane.

Put the plane at $z=0$ and the point charge at $(0,0,d)$. For $z>0$, use an image charge $-q$ at $(0,0,-d)$:This vanishes on the plane. An arbitrary constant can be added if a different reference electric potential is chosen. Excluding the point charge's own electric field, Coulomb's law givesThe separation from the image charge is $2d$, and the force is attractive for either sign of $q$. No imposed external electric field is included.

The electric dipole moment is $\mathbf{p}=qa\hat{\mathbf{y}}$. Write $\theta$ for the angle measured from $+y$, with $\hat{\mathbf{\theta }}$ pointing toward increasing $\theta$. The exact electric potential isFor $r\gg a$, expand the reciprocals to first order in $a/r$. Their common $1/r$ terms cancel, giving the electric dipole approximationThe radial electric field follows from $E_r=-\partial \varphi /\partial r$, and the tangential electric field from $E_{\theta }=-(1/r)\partial \varphi /\partial \theta$. The omitted terms are of relative order $a^2/r^2$ away from zeros of a component. Multiplying by the test electric charge gives the force componentsThe electric field lines emerge from the positive electric charge and terminate on the negative electric charge. They are mirror-symmetric about the electric dipole axis and bow out across its equatorial plane. In the point-electric dipole limit their meridional curves obey $dr/(rd\theta )=E_r/E_{\theta }=2\cot \theta$, hence $r=C{\sin }^2\theta$.

In a Cartesian coordinate system the leading electric dipole field givesThereforeThe force is downwards on the equator and upwards on the negative axial side. A free positive electric charge released from rest at the left-hand point initially accelerates downwards. As it moves into $x<0,y<0$, the force also acquires a positive $x$ component. Its path bends toward the right; the force continually changes as its position changes. An initial velocity was not specified, so the following more precise trajectory assumes release from rest and neglects radiation reaction.

There is a useful exact result within the point-electric dipole approximation. Put $A=k_{e}pQ>0$; the potential energy is $U=Ay/r^3$, a homogeneous function of degree $-2$. Thus $\mathbf{r}\cdot \mathbf{F}=2U$. For a particle of mass $m$, Newton's second law and conservation of energy implyAt release, both kinetic energy and potential energy vanish, so $E=0$; also $d(r^2)/dt=0$. Consequently $r=d$ throughout the motion. Parametrize the lower semicircle by $x=-d\cos \alpha$, $y=-d\sin \alpha$. ThenThe electric charge gains speed until the bottom point, where the force is inward and the speed is $\sqrt{2A/(m{d}^2)}$. It then slows, reaches $(d,0)$ at rest, and reverses. Thus, under these assumptions,Its trajectory is not an electric field line: the force determines acceleration, rather than velocity direction. For the finite two-charge electric dipole, corrections of order $a^2/d^2$ perturb the circular trajectory; an arbitrary nonzero initial velocity also changes it.

Faraday's law of induction states that the oriented electromotive force around a closed material circuit is minus the rate of change of its linked magnetic flux:The surface orientation and circuit direction obey the right-hand rule. The minus sign expresses Lenz's law: the induced electric current, when allowed to flow, produces a magnetic field opposing the change of linked magnetic flux. For a stationary circuit this reduces to $\oint \mathbf{E}\cdot d\mathbf{\ell }=-d{\Phi }_B/dt$.

The source's “Lorenz force” is the Lorentz force, $q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$. A wire element of cross-section $A_c$ contains $n{A}_{c}d\ell$ carriers of charge $q$ with drift velocity ${\mathbf{v}}_d$. Adding their magnetic forces and using $I=nq{A}_c{v}_d$ givesfor a straight wire in a uniform magnetic field. The length vector follows conventional electric current, irrespective of the carrier sign.

For the unlabelled upward-motion request, define $\hat{\mathbf{t}}=(\cos \varphi ,0,-\sin \varphi )$ down the slope, $\hat{\mathbf{b}}=(0,1,0)$ across the rails and $\hat{\mathbf{n}}=(\sin \varphi ,0,\cos \varphi )$ out of their plane. Choose the vertical magnetic field as $\mathbf{B}=B\hat{\mathbf{z}}$. An electric current $I$ across the bar in direction $-\hat{\mathbf{b}}$ gives a horizontal magnetic force $-IBs\hat{\mathbf{x}}$. The other forces are weight $-mg\hat{\mathbf{z}}$ and the rail normal force $N\hat{\mathbf{n}}$. Uniform upward velocity requiresConsequentlyEach symmetric rail supplies half of $N$. The electric current must make $I\mathbf{L}\times \mathbf{B}$ point horizontally toward the uphill side; for the coordinates just defined it flows along $-\hat{\mathbf{b}}$. Reversing the vertical magnetic field reverses this current direction. The external supply must maintain that electric current against the motional electromotive force during ascent. For $\cos \varphi =0$, no finite horizontal magnetic force can support uniform motion along the vertical rails.

In SI units, Maxwell's equations in vacuum, allowing charge density $\rho$ and electric current density $\mathbf{J}$, areThe source-free specialization sets $\rho =0$ and $\mathbf{J}=0$; it cannot describe the charged objects at their locations. Applying the divergence theorem to the first equation gives Gauss's law:For an isolated point charge $q_1$, spherical symmetry makes the electric field radial and constant in magnitude on a concentric sphere. Therefore $4\pi r^2{E}_r=q_1/{ϵ}_0$. Multiplying by a second point charge gives Coulomb's law,The force points away from the first charge for like signs and toward it for opposite signs.

For the spherical capacitor, a Gaussian sphere between the shells encloses $q$, so $E_r=q/(4\pi {ϵ}_0{r}^2)$ for $a<r<b$. Integrating the electric field inward from the outer shell yieldsAn arbitrary constant $V(b)$ remains unless a reference is specified. The capacitance is determined by the potential difference:For the subsequent charging operation, take the initial outer shell to be earthed, or interpret the stated $V$ as the inner-to-outer voltage with the outer reference zero. Isolating the inner shell fixes its electric charge at $q=CV$. Removing the outer shell leaves an isolated conducting sphere. Its electric field and electric potential, now with zero at infinity, areConsequentlyThese changes of electric potential are compatible with constant electric charge. Without the initial reference convention the absolute stated $V$ does not determine $q=CV$: for example, if the outer shell were initially isolated and neutral, the absolute inner electric potential would instead be $q/(4\pi {ϵ}_{0}a)$.

Finally take the connected spheres to be sufficiently far apart that the long wire's capacitance and their mutual influence can be neglected. Their common electric potential requiresso $q_a=q/3$ and $q_{2a}=2q/3$. The magnitudes of the surface electric fields areThis is the usual widely separated-sphere approximation. A wire enforces an equipotential, but alone does not make this ratio exact for arbitrary arrangements. As a concrete counterexample, if the second shell enclosed the first concentrically and they were connected, all net electric charge would reside on the exterior and the inner sphere's surface electric field would be zero.

Write $k=1/(4\pi {ϵ}_0)$. The separation of the two point charges is $2l\sin \varphi$, so Coulomb's law gives the horizontal repulsive force $F=k{Q}^2/(4l^2{\sin }^2\varphi )$. Resolving the rod's tension vertically and horizontally gives $T\cos \varphi =mg$ and $T\sin \varphi =F$. HenceThe left side increases from zero to infinity for $0<\varphi <\pi /2$, so this determines a unique outward equilibrium. For a small angle, ${\varphi }^3\simeq Q^2/(16\pi {ϵ}_{0}mg{l}^2)$. With $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, direct numerical solution gives

A large electrical conductor between the point charges provides electrostatic shielding: one must not retain their original mutual Coulomb force through an ideal infinite plate. In the usual broad-plate approximation, let $\sigma$ denote the unperturbed surface charge density on each face. The electrostatic boundary condition gives an outward electric field $E_0=\sigma /{ϵ}_0$. Neglecting local electrostatic induction, the rod's tension then satisfies $mg\tan \varphi =Q{E}_0$, givingThere are two physical qualifications. First, if $\sigma$ means the total charge per projected plate area, shared equally between the two faces, then $E_0=\sigma /(2{ϵ}_0)$ and the same approximation gives $\varphi \simeq 0.989^{\circ }$. The factor of two is a convention about the specified surface charge density, not a different Gauss's law.

Second, a point charge near an electrical conductor makes its surface charge density nonuniform. For an infinite plane with prescribed background electric field $E_0$, the method of images adds an attractive force $k{Q}^2/(4x^2)$, where $x=l\sin \varphi$. The more accurate equilibrium equation isIts larger, stable root is approximately $1.96^{\circ }$ for the per-face convention, or $0.877^{\circ }$ for the total-area convention. This ideal-plane refinement is not an exact answer for an unspecified finite plate: its dimensions and the meaning of the quoted surface charge density would be needed. In particular, adding a uniform-sheet electric field to the original mutual repulsion would describe a different, nonshielding model.

Let ${\hat{\mathbf{e}}}_r$ point from the intersection to the real point charge. Reflection symmetry cancels the component of electrostatic force perpendicular to the wedge bisector. An image charge $q_j$ at angular separation $\Delta$ from the real point charge gives radial forceThe two adjacent negative image charges have $\Delta =\pm 60^{\circ }$, distance $d$, and combined radial force $-k{q}^2/d^2$. The two positive image charges have $\Delta =\pm 120^{\circ }$, distance $\sqrt{3}d$, and combined radial force $+k{q}^2/(\sqrt{3}{d}^2)$. The opposite negative image charge is at distance $2d$ and contributes $-k{q}^2/(4d^2)$. ThusThe electrostatic force is towards the intersection, independently of the sign of $q$. One sums the five image charge electric fields and multiplies by $q$; the factor $1/2$ used in induced-charge electrostatic energy must not be inserted into this direct force calculation.

Add a uniform electric field satisfying $\mathbf{E}=-{\mathbf{v}}_0\times \mathbf{B}$. Then the Lorentz force $q(\mathbf{E}+{\mathbf{v}}_0\times \mathbf{B})$ vanishes for the desired velocity. With the axes used above, $\mathbf{E}=E\hat{\mathbf{y}}$, perpendicular to both ${\mathbf{v}}_0$ and $\mathbf{B}$, withThe crossed fields form a velocity selector. All particles must first have traversed the same electric potential difference from rest: that is what converts selection of speed into selection of charge-to-mass ratio. Equal charge-to-mass ratios yield equal speeds even if the separate masses and charges differ. For other ratios, the initial transverse force is $q(E-vB)$ and is nonzero.

Use the source orientation, anticlockwise viewed from $+z$. The position and oriented line element areThe Lorentz force on the electric current element is $d\mathbf{F}=Id\mathbf{l}\times \mathbf{B}$. Evaluating the cross product givesEach component integrates to zero around $0\le \theta <2\pi$, so ${\mathbf{F}}_{\mathrm{net}}=0$. Equivalently, uniform magnetic flux density permits $I\oint d\mathbf{l}\times \mathbf{B}=I(\oint d\mathbf{l})\times \mathbf{B}=0$. Uniformity is essential; a nonuniform magnetic field can exert a net force on a magnetic dipole.

time average of harmonic power. Expand the two real parts:The terms proportional to $e^{\pm 2i\omega t}$ average to zero over complete periods, while $F{u}^{\ast }$ is time-independent. For real peak amplitudes this is $F_0{u}_0\cos \varphi /2$.

String dynamics. A short element of length $dx$ experiences transverse force $T[{\psi }_x(x+dx)-{\psi }_x(x)]$. Newton's second law therefore gives $\rho dx{\psi }_{tt}=Tdx{\psi }_{xx}$, orThe approximation is a flexible, lossless string with small slope and constant tension; $\rho$ is its linear mass density.

energy transport. Across a cut toward increasing $x$, the transverse force is $F_y=-T{\psi }_x$ and the transverse velocity is $u={\psi }_t$. For a rightgoing wave these have complex amplitudes $ikTA$ and $i\omega A$. HenceThis real mechanical impedance is force divided by transverse velocity for a travelling wave, with units $\mathrm{kg}{\mathrm{s}}^{-1}$.

Matching and power ratios. At an ordinary massless junction without an external transverse force, displacement is continuous and transverse tension is balanced: ${\psi }_1={\psi }_2$, $T_1{\psi }_{1x}=T_2{\psi }_{2x}$. For the same tension, the slope is therefore continuous even if density changes. With real positive impedances,Their sum equals $P_i$; the transmitted power includes the impedance ratio and is not simply $t_{\psi }^2{P}_i$.

Response and resonance. The terms $m\ddot{x}$, $b\dot{x}$ and $kx$ describe inertia, linear viscous resistance and the spring's restoring force; the right side is the applied harmonic force. Substituting the complex trial motion and taking real parts givesThe steady response excludes the decaying homogeneous transient. Its magnitude is $[(k-m{\omega }^2{)}^2+b^2{\omega }^2{]}^{-1/2}$. Differentiating the denominator gives $2\omega [-2m(k-m{\omega }^2)+b^2]=0$, soA nonzero-frequency amplitude maximum exists only for $b^2<2km$; otherwise the maximum is at zero frequency. Increasing damping lowers and broadens the peak and shifts it downward. Every curve starts at $|R(0)|=1/k$ and tends to $1/(m{\omega }^2)$ at high frequency.

Series circuit. The PDF circuit has the resistor, inductor and capacitor in one loop. With current $I=\dot{q}$, Kirchhoff's voltage law gives $V=L\dot{I}+RI+q/C$, henceThus $m,b,k$ correspond to $L,R,1/C$. The resistor's rms voltage isIt is maximized exactly at ${\omega }_{\max }={\omega }_0=1/\sqrt{LC}$ for $R>0$. The charge-amplitude resonance is lower. Expanding its square root for $R^{2}C/L\ll 1$ givesThe two maxima refer to different observables: resistor voltage follows current, whereas charge follows displacement.