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For a flexible uniform string with linear mass density $\rho$ and tension $T$, dimensional analysis gives $[T/\rho ]=L^2{T}^{-2}$ and hence $c\propto \sqrt{T/\rho }$. The small-slope force balance $\rho {\psi }_{tt}=T{\psi }_{xx}$ fixes the factor to one:Here $\rho$ has units kg/m, not the volume-density units used for a bulk fluid. The phase moves along the string at $c$, while a material point moves transversely with ${\psi }_t=A\omega \sin (kx-\omega t)$. Its maximum speed is $A\omega$, which is not the phase velocity. Small slope $Ak\ll 1$ makes $A\omega \ll c$ in this model.

The kinetic energy and extra elastic energy of a segment areThe second expression follows from extra length $\sqrt{1+{\psi }_x^2}dx-dx\simeq {\psi }_x^{2}dx/2$ against the pre-existing constant tension. Time averaging a harmonic wave gives $⟨{\mathcal{E}}_K⟩=\rho A^2{\omega }^2/4$ and $⟨{\mathcal{E}}_U⟩=T{A}^2{k}^2/4=\rho A^2{\omega }^2/4$. Therefore

At the driven endpoint the string pulls along its local tangent into $x>0$, with vertical component $T{\psi }_x$ for small slope. The point exerts the opposite force on the string, so the driven-string endpoint power isFor the outgoing wave $\psi =A\cos (kx-\omega t)$, at $x=0$ the derivatives are ${\psi }_x=Ak\sin \omega t$ and ${\psi }_t=-A\omega \sin \omega t$. Thus $P=T{A}^{2}k\omega {\sin }^2\omega t$ andThe energy flux of a transverse string wave follows by combining the energy density $\mathcal{E}=(\rho {\psi }_t^2+T{\psi }_x^2)/2$ and flux $P=-T{\psi }_x{\psi }_t$, which satisfy ${\partial }_t\mathcal{E}+{\partial }_{x}P=0$. For a pure right-going nondispersive wave, $P=c\mathcal{E}$ even instantaneously. energy is transported with the wave; a material string element merely oscillates about its unperturbed position.

In the Lorentz law, $q$ is signed electric charge, $\mathbf{E}$ and $\mathbf{B}$ are the electric and magnetic fields at the bead, and the velocity in $q\mathbf{V}\times \mathbf{B}$ is its full laboratory velocity. Here $\mathbf{V}=v\hat{\mathbf{r}}+w\hat{\mathbf{\theta }}$. Since $\hat{r}\times \hat{z}=-\hat{\theta }$ and $\hat{\theta }\times \hat{z}=\hat{r}$,The tangential component integrated around an instantaneous circle of nearly fixed radius givesThis is the stated tangential circulation work expression in the $w\gg v$ limit. It is tangential circulation work. The tangential circulation work versus particle work distinction matters because the bead actually follows an expanding spiral. The radial displacement may be small, but its force is proportional to $w$, so its power is of the same order as the tangential power. In factThe cancellation expresses the fact that magnetic forces do no work. Ignoring radial work while retaining the first-order tangential effect would incorrectly conclude that a magnetic field supplies net particle energy.

The electric field giving the same tangential force would be $\boxed{{\mathbf{E}}_{\mathrm{eq}}=-vB\hat{\theta }}$. This is a force-equivalent field for the moving guide, not evidence that a real circulating electric field has appeared in the laboratory merely because the loop expands in a static magnetic field. An electromotive force is the tangential driving-force circulation per unit charge, with units of energy per charge. For a material circuit moving with local velocity ${\mathbf{v}}_{\mathrm{loop}}$,Orient the ring counterclockwise with positive normal $+\hat{z}$. Its flux is $\Phi =B\pi R^2$, so $\dot{\Phi }=2\pi BRv$. The tangential circulation therefore givesIt is the motional term that matters here. For a fixed loop, the circulation of the actual electric field instead follows the partial time derivative of the magnetic field. The distinction between moving-wire magnetic forces and induction in a stationary loop is essential to the flux rule.^[ref]

For the magnetic power balance on an expanding circular guide, the guide must supply an additional inward reaction $-qBw\hat{r}$ opposing the radial magnetic force. With radial guide speed $v$, this contribution to its power on the bead is $P_{\mathrm{guide}}^{(B)}=-qBvw$, equal to the tangential magnetic power. The radial magnetic power is positive by the same amount, so the total magnetic power still vanishes; the mechanical exchange is associated with the moving mechanical constraint.

If “the ring's work” is meant as its full work on a finite-mass bead, the source statement needs a further qualification. For a frictionless bead of mass $m_b$ with constant imposed radial speed, its radial acceleration is $-w^2/R$, so the complete radial guide reaction and power areThe first term is present even when $B=0$ and describes the familiar change of tangential kinetic energy on an expanding guide. The requested $-qBvw$ is exactly the additional magnetic-field-dependent term, not that full reaction power. No bead mass or extra agency fixing $w$ is supplied, so it would be unjustified to assign a numerical total mechanical power or silently discard its centripetal contribution.

A time-independent conservative force defines a conservative vector field on a domain when the work between two points is independent of path, equivalently when every closed-path work integral vanishes. Define a potential energy $U$ by $\boxed{\mathbf{F}=-\nabla U}$ and $U(B)-U(A)=-{\int }_A^B\mathbf{F}\cdot d\mathbf{r}$. The additive constant is arbitrary; only differences affect work. A vanishing curl is necessary, and is sufficient on a simply connected domain under the usual smoothness assumptions. Local zero curl by itself need not guarantee a single-valued global potential on a domain with holes.

For a test mass outside a point or spherical mass, $U=-GMm/r$ with zero at infinity reproduces the inward gravitational force. Near a surface with approximately uniform downward gravity, $U=mgz$ when $z$ increases upward. A linear spring has $U=k{x}^2/2$. In electrostatics the electric potential is energy per unit charge, so $U=qV$ and $\mathbf{E}=-\nabla V$; potential and potential energy have different units and roles.

For motion under a time-independent conservative force, $d(K+U)/dt=0$. nonconservative forces change the mechanical energy by their work, and an explicitly time-dependent potential can exchange energy with the external agency changing it. A time-varying magnetic field can give a circulating electric field, so not every electric field is electrostatic or conservative. Conversely, zero magnetic work on a charge does not turn a velocity-dependent magnetic force into an ordinary position-potential force.

For a test particle outside a spherical central mass, the gravitational force is central, so angular momentum is conserved and the orbit is planar. With specific angular momentum $h=r^2\dot{\theta }$ and specific orbital energy $ϵ={\dot{r}}^2/2+h^2/(2r^2)-GM/r$, the radial effective potential combines attraction with the centrifugal barrier. Its nonradial solutions are conic sections:For nonzero $h$, negative energy gives an ellipse with $0\le e<1$, zero energy a parabolic Kepler orbit and positive energy a hyperbolic Kepler orbit. The circle is the special case $e=0$. Exactly radial motion is a separate zero-angular-momentum case and may hit the central body; a mathematical point-mass trajectory is not automatically a physically noncolliding orbit.

Conservation of $h$ gives constant areal velocity $h/2$, so equal areas are swept in equal times. For a bound ellipse of semimajor axis $a$, $ϵ=-GM/(2a)$ and $T^2=4{\pi }^2{a}^3/(GM)$. A circular orbit has speed $\sqrt{GM/r}$, while the minimum local escape speed is $\sqrt{2GM/r}$, obtained by setting the total energy to zero. These are the test-mass two-body results: a comparable second mass requires the relative-coordinate gravitational parameter $G(M+m)$, and nonspherical structure or drag can perturb the conic and its conserved quantities.

The ground-state energies are $1.50\mathrm{eV}$ and $0.376\mathrm{eV}$, respectively. For a particle in a box of width $L$, the wave function must vanish at both infinite walls. The lowest nonzero standing wave therefore has wavelength $2L$, giving the kinetic energy eigenvalueUsing the electron mass and converting joules to electronvolts givesDoubling the confinement length divides this kinetic energy by four. A zero potential energy inside the box does not imply zero total energy: a constant, zero-momentum wave function cannot satisfy both wall conditions unless it vanishes everywhere. This is a concrete manifestation of quantum confinement and the Heisenberg uncertainty principle.

The threshold must conserve both energy and momentum. For head-on photons, the total energy is $E_1+E_2$, while the total momentum has magnitude $|E_2-E_1|/c$. Their combined invariant mass $M$ satisfiesAt the threshold for photon-photon pair production, the electron and Positron have no relative momentum in their centre-of-momentum frame, so $M=2m_e$. ThereforeSimply equating the laboratory energy to $2m_e{c}^2$ would miss the very large laboratory momentum: the newly produced pair must carry it away.

With zero gravitational potential energy at the wheel axis, the release energy and the energy at the apex of the projectile motion areOnly the vertical velocity vanishes at the apex: the horizontal kinetic energy remains. Equating the two energies gives

Conservation of energy states that the total energy of an isolated system is constant. In a mechanical problem with conservative forces and no dissipation, it gives $K+U=\text{constant}$. It often determines a speed or turning height directly, without solving the full time-dependent equation of motion. With external work or dissipation, use the corresponding energy balance instead.

After a drop leaves the wheel, neglecting air resistance, the conservative force of the Newtonian gravitational field acts. Put the origin at the wheel axis and choose positive horizontal displacement to the right. In the printed diagram $\theta$ is measured upward from the leftward horizontal radius, and rotation is clockwise. The release position and velocity are thereforeFor upward release, $\cos \theta >0$. These conventions fix the signs used below.

The capacitor stores electrostatic energy, the inductor stores magnetic energy, and the resistor converts energy into heat. Their instantaneous values areThe sign convention for electric current does not change these expressions. With the initial voltage, $E_C(0)=C{V}_0^2/2=90\mu \mathrm{J}$ and $E_I(0)=P_R(0)=0$.

The two stored energies follow directly by squaring the expressions for electric charge and electric current:Both are nonnegative, with alternating maxima. Initially all stored energy is in the capacitor; at a zero of $q$ it is entirely in the inductor. The total isThe total energy decreases monotonically, with a horizontal tangent whenever the electric current is zero. Its envelope decays on the $1/(2\gamma )=20\mathrm{ms}$ scale, but the exact total is not a featureless exponential: the resistor's instantaneous loss varies within each oscillation. The figure in part (a) shows this gradual modulation of the downward slope as well as the component energies.

In a fixed point-mass Newtonian gravitational field, a test mass $m$ experiencesBecause this is a central force, its torque about the attracting centre vanishes and angular momentum is conserved. The orbit lies in a fixed plane and sweeps out equal areas in equal times. Since the force is also conservative, the specific orbital energy is constant:The last two terms form an effective potential. They show how the angular-momentum barrier competes with attraction and determines the turning radii. For nonradial motion the orbit is a conic section:Negative energy gives an ellipse, with $e<1$; zero energy gives a parabola; positive energy gives a hyperbola. The circular limit has $e=0$, $v_c=\sqrt{GM/r}$ and $\epsilon =-GM/(2r)$. For a bound ellipse of semimajor axis $a$, $\epsilon =-GM/(2a)$ and the orbital period is $T=2\pi \sqrt{a^3/(GM)}$. The local escape speed is $\sqrt{2GM/r}$. These are the point-mass, test-particle limits; extended mass distributions or noncentral perturbations change the orbit laws.

An electric dipole consists of opposite electric charges separated by a short displacement. Its electric dipole moment is $\mathbf{p}=q\mathbf{d}$, directed from negative to positive electric charge. At distances large compared with the separation, its leading electric potential and electric field in vacuum areThe absence of net electric charge removes the monopole term; the dipole electric potential therefore falls as $r^{-2}$ rather than $r^{-1}$. On its axis the electric field is parallel to $\mathbf{p}$ with magnitude $2p/(4\pi {ϵ}_0{r}^3)$; on the equatorial plane it points oppositely with half that magnitude.

In a uniform applied electric field, the two forces cancel as a net translation but form a couple. The torque and potential energy areAlignment parallel to the electric field is stable. A nonuniform electric field can also exert a net force, $\mathbf{F}=\nabla (\mathbf{p}\cdot \mathbf{E})$ for a fixed permanent electric dipole moment. An induced electric dipole instead depends on the applied electric field, so its energy must account for the work needed to create it.

Electromagnetic induction is the production of an electromotive force by changing magnetic flux or by moving a conductor through a magnetic field. With the loop and surface orientations related by the right-hand rule, Faraday's law of induction isHere $\mathbf{v}$ is the local wire velocity. For a stationary loop, the induced electric field can have a nonzero closed-path integral; its differential form is $\nabla \times \mathbf{E}=-\partial \mathbf{B}/\partial t$. For a straight rod moving perpendicularly to itself and to a uniform magnetic field, the motional electromotive force has magnitude $B\ell v$.

The minus sign expresses Lenz's law: the induced electric current produces a magnetic field that opposes the change in linked magnetic flux. Pulling a conducting loop through a magnetic field therefore requires work against an electromagnetic resisting force; that work supplies the resulting heat or stored energy.

For an inductor with constant self-inductance, $\mathcal{E}=-LdI/dt$ and its stored magnetic energy is $L{I}^2/2$. A changing electric current in one coil can induce an electromotive force in another through mutual inductance, the operating principle of a electrical transformer. These examples connect the flux law, the direction of the response and conservation of energy.

The Lorentz force determines the motion of a charged particle:At nonrelativistic speeds, $\mathbf{p}=m\mathbf{v}$. The electric field does work at rate $q\mathbf{E}\cdot \mathbf{v}$, while the magnetic part of the Lorentz force is perpendicular to velocity and does no work. A uniform electric field gives constant acceleration $q\mathbf{E}/m$; with a transverse initial velocity, the trajectory is a parabola.

In a uniform magnetic field, the parallel velocity is constant, while the perpendicular component undergoes circular motion. The angular cyclotron frequency, orbit radius and helical pitch areThe sign of $q$ determines the sense of rotation. For perpendicular uniform electric fields and magnetic fields, the guiding centre has the electric drift velocityIndeed, $\mathbf{E}+{\mathbf{v}}_D\times \mathbf{B}=0$, so the remaining motion is cyclotron rotation about that drifting centre, with possible motion along the magnetic field. The drift is independent of electric charge sign and mass. At relativistic energies, retain $\mathbf{p}=\Gamma m\mathbf{v}$ and the corresponding relativistic orbit radius rather than assuming constant nonrelativistic acceleration.

The original diagram has a horizontal bottle neck, so weight does not enter the axial force balance. Gauge pressure acts on the cork's end area and supplies outward force $F_g=\pi a^2{P}_g$. The radial contact force is the contact pressure times the cylindrical contact area, $N=2\pi a(l-x)P_c$. The limiting friction is $\mu N$, so outward sliding occurs whenThe assumption of equal static and kinetic friction coefficients removes a separate force drop when slipping starts.

Initially the pressure is marginal at $x=0$, so $\pi a^2{P}_g=2\pi al\mu P_c$. During extraction the contact length decreases, while the pressure force stays constant. The net force is therefore $F(x)=\pi a^2{P}_{g}x/l$. The cork's mass is $m_c=\rho \pi a^{2}l$, giving the pressure-driven cork ejection equationFor $x(0)=0$ and $\dot{x}(0)=v_0>0$, solving the constant-coefficient equation givesThe equilibrium is unstable: as the cork moves out, friction decreases and the unbalanced force grows.

Multiplying the equation of motion by $\dot{x}$ and integrating gives ${\dot{x}}^2=v_0^2+b^2{x}^2$. The contact ends at $x=l$, soEquivalently, the net work is ${\int }_0^{l}F(x)dx=\pi a^2{P}_{g}l/2$. Exactly zero initial velocity would leave the ideal marginal configuration stationary. The exit time $b^{-1}\mathrm{arsinh}(bl/v_0)$ diverges as $v_0\to 0^{+}$, even though the limiting exit speed is finite.

The density is $500\mathrm{kg}{\mathrm{m}}^{-3}$, so $v_{\mathrm{exit}}\simeq \sqrt{400000/500}=28.3\mathrm{m}{\mathrm{s}}^{-1}$. For an ideal upward launch at this speed, conservation of mechanical energy gives height above the exitThis neglects air resistance and the small gravitational work during extraction compared with the launch energy. Actual gas pressure need not stay constant, so it is an idealized estimate.

Adding a stopper that doubles the moving mass, while preserving the same pressured area and contact friction, leaves the extraction work unchanged. The squared exit speed and the ballistic height are therefore halved:

Use the diagram's downward displacement $z$, measured from the unloaded plate position, and take $z(0)=0$. The mechanical energy, with gravitational potential zero at that reference, isThe loaded equilibrium follows from $k{z}_{\mathrm{eq}}=mg$, so its downward shift is $\boxed{z_{\mathrm{eq}}=mg/k}$.

The shock absorber's force is $-b\dot{z}$, hence its power is force times velocity: $dE/dt=-b{\dot{z}}^2\le 0$. Differentiating the energy gives $\dot{z}(m\ddot{z}+kz-mg)=-b{\dot{z}}^2$, which yields, with continuity through instants of zero velocity,Set ${\omega }_0=\sqrt{k/m}$ and $\gamma =b/(2m)$.

For release from rest, the three step response of a damped oscillator regimes are easy to compare after shifting to $z-z_{\mathrm{eq}}$. With light damping $0<\gamma <{\omega }_0$, write ${\omega }_d=\sqrt{{\omega }_0^2-{\gamma }^2}$:It overshoots and oscillates around the loaded position with a decaying envelope. At critical damping, $\gamma ={\omega }_0$, it becomes $z/z_{\mathrm{eq}}=1-(1+\gamma t)e^{-\gamma t}$, approaching monotonically from below. For heavy damping $\gamma >{\omega }_0$, set $\eta =\sqrt{{\gamma }^2-{\omega }_0^2}$ and replace the trigonometric bracket by $\cosh \eta t+(\gamma /\eta )\sinh \eta t$. This is also monotonic, but its slow exponential delays settling.

Critical damping gives the fastest nonoscillatory settling from the unloaded rest condition, avoiding a repeatedly swinging reading without the sluggishness of overdamping. Its required coefficient isAt critical damping the repeated characteristic root is $-\gamma$. Thus the general displacement under constant load is $(A+Bt)e^{-\gamma t}+C$. The equilibrium and initial conditions give $C=z_{\mathrm{eq}}$, $A=-z_{\mathrm{eq}}$ and $B-\gamma A=v_0$, henceSubtracting the final position gives $z-z_{\mathrm{eq}}=[-z_{\mathrm{eq}}+(v_0-\gamma z_{\mathrm{eq}})t]e^{-\gamma t}$. The exponential is positive and the bracket starts negative. It becomes positive at some $t>0$ if and only if its slope is positive. Therefore critical-damping overshoot from initial velocity requireswith $v_0$ positive downwards. At equality the scale still approaches from below. Critical damping prevents oscillatory modes, but does not guarantee a non-overshooting response to an arbitrarily large impact velocity.

Here $V(x,y)$ is the potential energy and $E$ is the stationary energy eigenvalue. The Schrödinger equation is the eigenvalue equation for kinetic plus potential energy. Its boundary conditions select the allowed energies. For a normalized wavefunction, the Born rule assigns probability $|\psi (x,y){|}^{2}dxdy$ to a small area, and $\int |\psi {|}^{2}dxdy=1$. A stationary state's full time dependence is $\psi (x,y)e^{-iEt/\hslash }$; its probability density is time-independent.

The infinite walls impose $\psi =0$ on all four boundaries. For a separated solution $\psi =X(x)Y(y)$ in the zero-potential interior,The first term depends only on $x$ and the second only on $y$, so each is constant, $-k_x^2$ and $-k_y^2$. Vanishing at $x=0$ and $y=0$ removes the cosine terms. Vanishing at the other boundaries givesThe normalized two-dimensional infinite square well eigenbasis and energies areNormalization follows from the two sine-squared integrals $a/2$ and $b/2$. Zero quantum numbers would give an identically zero function. These separated products form a basis, not every possible wavefunction: a general state is their superposition, and a degenerate energy eigenspace can have nonseparable stationary combinations of products with that same energy.

For $a=1\mathrm{nm}$ and $b=2\mathrm{nm}$, the ground state is $(1,1)$ and the nearest excitation is $(1,2)$, along the longer side. The next quantum along the shorter side costs more energy. ThusThis transition has nonzero dipole matrix element for an electric-field component along $y$. Light polarized strictly along $x$ would not drive it, so the minimum assumes an allowed polarization rather than an arbitrary fixed one.

For the requested degeneracy, keep $b>a$. The first two states are necessarily $(1,1)$ and $(1,2)$, since $E_{21}-E_{12}=3{\pi }^2{\hslash }^2(a^{-2}-b^{-2})/(2m)>0$. The third level is the smaller of $E_{21}$ and $E_{13}$: states with both indices at least two lie above $E_{21}$, and higher states on the $n_x=1$ sequence lie above $E_{13}$. Make these two candidates equal:The third-level degeneracy in a rectangular quantum box then has lowest distinct energies $11/8$, $5/2$ and $35/8$ in units ${\pi }^2{\hslash }^2/(2m_e{a}^2)$, with multiplicities one, one and two. The third-level pair is exactly $(2,1)$ and $(1,3)$.

At the first two specified times, the displacement profiles areThe second curve is the first shifted right by $\pi /(2k)=\lambda /4$. More generally a point of fixed wave phase satisfies $\omega t-kx=\text{constant}$, so it moves along $x=(\omega /k)t+\text{constant}$. Hence the wave travels in the positive $x$ direction withThis phase velocity is distinct from the local transverse velocity $\partial y/\partial t$ of a string element.

For the force derivation take a short segment between $x$ and $x+\Delta x$. Let the tangent angles to the positive $x$ axis be ${\theta }_L$ and ${\theta }_R$. The left tension pulls along the negative tangent and the right tension along the positive tangent. With small slopes and approximately constant tension magnitude $T$, the horizontal components cancel to leading order andThe segment mass is $\rho \Delta x$, where $\rho$ is mass per unit length, not volume density. Newton's second law therefore givesThe approximation neglects bending stiffness, damping and changes in tension caused by finite-amplitude stretching.

Converting the supplied density gives $\rho =0.020\mathrm{kg}{\mathrm{m}}^{-1}$, soFor the pulse, define $g(x)=A{e}^{-a{x}^2/2}$ with $a>0$. The given initial velocity is $-c{g}^{\prime }(x)$, exactly the velocity associated with a right-moving displacement $g(x-ct)$. ThereforeIt moves right without changing shape in this nondispersive model. With the same initial displacement but zero initial transverse velocity, D'Alembert's formula instead gives the Gaussian pulse splitting from zero initial velocityThe two half-amplitude pulses travel in opposite directions. Their initial velocities cancel, while their initial displacements add to the prescribed Gaussian. The two initial-value problems have different kinetic energies; splitting into half-height pulses does not contradict energy conservation. These forms assume a string long enough that reflections from its ends have not returned.

In electrostatics, Gauss's law isThe surface normal points outward. Spherical symmetry makes the field radial and constant in magnitude on a concentric Gaussian surface.

For a uniformly charged sphere, enclosed charge is $4\pi {\rho }_0{r}^3/3$ inside and $Q=4\pi {\rho }_0{R}^3/3$ outside. Thus the radial field isSet potential zero at infinity. Outside, $V=Q/(4\pi {ϵ}_{0}r)$. Inside, integrate $dV/dr=-E_r$ from the known boundary value, obtainingThe field rises linearly from zero to its maximum at $R$, then falls as $r^{-2}$. The potential falls quadratically from ${\rho }_0{R}^2/(2{ϵ}_0)$ to its surface value and then as $r^{-1}$. Both are continuous at the surface; there is no separate sheet of surface charge.

For the electric field of a hydrogen ground-state electron cloud, its enclosed charge isWriting $u=2s/a_0$ and using $\int u^2{e}^{-u}du=-e^{-u}(u^2+2u+2)$ givesThe total electron-cloud charge tends to $-e$, as required. Dividing by $4\pi {ϵ}_0{r}^2$ yieldsThe expression is negative and therefore points inward. Its apparently singular terms cancel at small radius, leaving $E_{e,r}\simeq -er/(3\pi {ϵ}_0{a}_0^3)$.

A distant external charge polarizes the neutral atom. Its electric field displaces the positive and negative charge distributions relative to one another, giving an induced dipole $\mathbf{p}=\alpha {\mathbf{E}}_{\mathrm{ext}}$ with static electric polarizability $\alpha >0$. The field is stronger on the nearer side, so the opposite-charge attraction exceeds the like-charge repulsion. This is charge-induced dipole attraction, not a net monopole charge on the atom.

Quantitatively, for external charge $Q=\pm e$ at separation $r$, the polarization energy is $U=-\alpha E_{\mathrm{ext}}^2/2$, henceHere negative $F_r$ means decreasing separation. Reversing the external charge reverses the induced dipole but leaves this leading long-range force attractive and of the same magnitude. The factor one half accounts for the energy cost of inducing the dipole.

The supplied charge density describes the unperturbed atom. Adding its electron field to the nucleus field leaves $E_{\mathrm{atom}}=e{e}^{-2r/a_0}(r^{-2}+2/(a_{0}r)+2/a_0^2)/(4\pi {ϵ}_0)$, an exponentially small residual because the cloud has an infinite tail. A completely frozen density would give only that tiny sign-dependent force. Allowing polarization is essential to obtain the dominant attractive power-law force for either sign of external charge when $r\gg a_0$.

A charge $q$ moving nonrelativistically in prescribed fields obeys the Lorentz force lawThe magnetic part is perpendicular to velocity, so $d(m{v}^2/2)/dt=q\mathbf{E}\cdot \mathbf{v}$: the electric field can change kinetic energy, while the magnetic field alone changes direction without doing work.

In a uniform electric field with no magnetic field, acceleration is constant: $\mathbf{v}={\mathbf{v}}_0+(q\mathbf{E}/m)t$ and $\mathbf{r}={\mathbf{r}}_0+{\mathbf{v}}_{0}t+q\mathbf{E}{t}^2/(2m)$. The trajectory is a parabola when there is initial motion perpendicular to the field, and a line when motion is purely along it. Reversing the charge reverses the electric acceleration.

In a uniform magnetic field with no electric field, resolve velocity into parallel and perpendicular components. The parallel component is constant; the perpendicular component rotates at the cyclotron frequencyThe resulting path is a helix, or a circle if $v_{\parallel }=0$, with pitch $2\pi v_{\parallel }/{\omega }_c$. Charge sign determines its sense of rotation. These relations underlie magnetic bending and mass/charge analysis. At relativistic energies, use momentum $\mathbf{p}=\gamma m\mathbf{v}$; at fixed energy the cyclotron frequency becomes $|q|B/(\gamma m)$.

For uniform crossed fields $\mathbf{E}\perp \mathbf{B}$, the velocity ${\mathbf{v}}_D=\mathbf{E}\times \mathbf{B}/B^2$ satisfies $\mathbf{E}+{\mathbf{v}}_D\times \mathbf{B}=0$. The orbit is a gyromotion about a guiding center drifting withindependent of mass and charge sign. A velocity selector transmits particles with the appropriate direction and speed $E/B$ without deflection. If the electric field has a component along the magnetic field, that component accelerates the parallel motion instead. Nonuniform fields can add other drifts; the simple helix and uniform crossed-field expressions require the stated uniform-field assumptions.