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6. ElectrostaticsSavchenko Formulas, chapter 6 of 14, 40 formulas

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6.1Coulomb's law. Electric field

Coulomb's law Закон Кулона

law 6.1

F = \frac{1}{4\pi\varepsilon_0}\,\frac{q_1 q_2}{r^2} = k\,\frac{q_1 q_2}{r^2}

НмКл$\displaystyle k = \frac{1}{4\pi\varepsilon_0} = 9\cdot10^9\ \frac{\text{Н м}^2}{\text{Кл}^2}$

СГС$\displaystyle F = \frac{q_1 q_2}{r^2}\ \text{(СГС)}$

q₁, q₂: the charges
r: distance between them
$\varepsilon_0$: the electric constant, $8.85\cdot10^{-12}$ F/m

Point charges interact with a force inversely proportional to the square of the distance, like charges repelling. The force lies along the line joining the charges, and forces from several charges add as vectors. In CGS units the factor $$k$$ is one.

Field strength. Force on a charge Напряжённость поля. Сила на заряд

definition 6.1

\vec E = \frac{\vec F}{q}, \qquad \vec F = q\vec E

$\vec E$: electric field strength
q: test charge
$\vec F$: force on the charge in that field

The field strength is the force per unit positive charge placed at a point. Fields of different sources add as vectors, and the force on a charge is $$qE$$ , giving an acceleration $$qE/m$$ . A charge in gravity and an electric field hangs where $$qE$$ balances $$mg$$ .

Field of a point charge Поле точечного заряда

law 6.1

E = \frac{1}{4\pi\varepsilon_0}\,\frac{q}{r^2}, \qquad \vec E = \sum_i \vec E_i

наосикольца$\displaystyle E = \frac{1}{4\pi\varepsilon_0}\,\frac{Qh}{(R^2 + h^2)^{3/2}}\ \text{(на оси кольца)}$

q: the charge producing the field
r: distance from it

A point charge's field points along the radius and falls as $$1/r^2$$ . The field of a set of charges is the vector sum of each one's field, and that of an extended body is an integral over its elements $$dq$$ . On the axis of a charged ring only the axial component survives.

Equilibrium point between two charges Точка равновесия между двумя зарядами

method 6.1

\frac{q_1}{x^2} = \frac{q_2}{(l - x)^2}, \qquad x = \frac{\sqrt{q_1}}{\sqrt{q_1} + \sqrt{q_2}}\,l

q₁, q₂: two like charges
l: distance between them
x: distance from the first charge to the point of zero field

Between like charges the field vanishes where their fields are equal in size, and the distances are as the square roots of the charges. A third charge placed there is in equilibrium, but unstable against displacement along the line if all three charges have one sign.

Appears in problems (16) 2.3.8 2.3.9 2.3.31 2.3.40 3.1.7 5.1.4 6.1.8 6.5.1

Charged balls on strings Заряженные шарики на нитях

method 6.1

к$$\tan\alpha = \frac{F_{\text{к}}}{mg}, \qquad q = 2l\sin\alpha\sqrt{4\pi\varepsilon_0\, mg\tan\alpha}$$

$\alpha$: angle of the string to the vertical
l: length of the string
$к$F_{\text{к}}$$: Coulomb force between the balls

A ball hangs where the string's tension balances both gravity and the repulsion, so the ratio of the Coulomb force to $$mg$$ is the tangent of the deflection angle. The separation $2l\sin\alpha$ enters Coulomb's law and the charge follows from one equation. For small angles $\sin\alpha \approx \tan\alpha$ .

Appears in problems (5) 6.1.9 6.1.14 6.1.15 7.1.27 9.1.5

Dipole moment Дипольный момент

definition 6.1

\vec p = q\vec l, \qquad F_x = p\,\frac{\partial E}{\partial x}

q: charge at each end of the dipole
$\vec l$: vector from the negative charge to the positive
$\partial E/\partial x$: field gradient

Two equal and opposite charges a small distance apart are described by one vector, the dipole moment. In a uniform field a dipole feels only a torque turning it along the field, in a nonuniform one also the force $p\,\partial E/\partial x$ pulling it toward the stronger field. A polarized ball's dipole moment is an integral over its surface charge.

Appears in problems (4) 6.6.15 6.6.28 6.6.29 7.2.9

6.2The flux of the electric field. Gauss's theorem

Field of a charged plane Поле заряженной плоскости

law 6.2

уповерхностипроводника$$E = \frac{\sigma}{2\varepsilon_0}, \qquad E = \frac{\sigma}{\varepsilon_0}\ \text{(у поверхности проводника)}$$

снаружислоятолщины$\displaystyle E = \frac{\rho h}{2\varepsilon_0}\ \text{(снаружи слоя толщины }h)$

заряднаповерхностипроводника$\displaystyle \sigma = \varepsilon_0 E\ \text{(заряд на поверхности проводника)}$

$\sigma$: surface charge density
$\rho$: volume charge density in a slab

An infinite plane produces on both sides a uniform field $\sigma/2\varepsilon_0$ , independent of distance. At a conductor's surface the fields of all its charges add to $\sigma/\varepsilon_0$ outside and zero inside. Two oppositely charged plates give $\sigma/\varepsilon_0$ between them and zero outside.

Gauss's law Теорема Гаусса

law 6.2

внутр$$\oint \vec E\cdot d\vec S = \frac{q_{\text{внутр}}}{\varepsilon_0}$$

потокчерезплощадку$\displaystyle \Phi = ES\cos\alpha\ \text{(поток через площадку)}$

потокточечногозарядачерезтелесныйугол$\displaystyle \Phi = \frac{q\,\Omega}{4\pi\varepsilon_0}\ \text{(поток точечного заряда через телесный угол)}$

$\Phi$: flux of $\vec E$ through a closed surface
$внутр$q_{\text{внутр}}$$: charge enclosed by the surface
$\varepsilon_0$: the electric constant

The flux of the field through any closed surface equals the enclosed charge over $\varepsilon_0$ , whatever charges sit outside. For symmetric distributions one picks a surface on which $$E$$ is constant and normal, a sphere, a cylinder or a box, and the field comes out in one line. The flux of a point charge through a patch is $q\Omega/4\pi\varepsilon_0$ with $\Omega$ the solid angle.

Field of a charged ball and shell Поле заряженного шара и сферы

law 6.2

E = \frac{Q}{4\pi\varepsilon_0 r^2}\ (r > R), \qquad E = \frac{\rho r}{3\varepsilon_0}\ (r < R)

всферическойполостиотцентрашаракцентруполости$\displaystyle \vec E = \frac{\rho\,\vec l}{3\varepsilon_0}\ \text{(в сферической полости, }\vec l\text{ от центра шара к центру полости)}$

\displaystyle E = \frac{\rho R^3}{3\varepsilon_0 r^2}\ (r > R)

Q: total charge
R: radius of the ball or shell
$\rho$: volume charge density

Outside, a uniformly charged ball or shell acts as a point charge at the centre. Inside a shell the field is zero, inside a uniformly charged ball it grows linearly with radius, since only the charge within $$r$$ counts. The field in a spherical cavity of a ball is uniform, $\rho\vec l/3\varepsilon_0$ , the difference of two balls' fields.

Field of a charged line and cylinder Поле заряженной нити и цилиндра

law 6.2

E = \frac{\lambda}{2\pi\varepsilon_0 r}, \qquad E = \frac{\rho r}{2\varepsilon_0}\ (r < R)

$\lambda$: linear charge density
r: distance from the axis
$\rho$: volume charge density of the cylinder

A long line's field is radial and falls as $$1/r$$ , which follows from Gauss's law for a cylinder of length $$l$$ holding $\lambda l$ . Inside a uniformly charged cylinder the field grows linearly with $$r$$ , outside it equals that of a line with the same charge per length.

Appears in problems (7) 6.2.6 6.2.7 6.3.17 6.4.7 7.4.34 8.1.11 9.2.3

Flux through a solid angle Поток через телесный угол

law 6.2

\Phi = \frac{q\,\Omega}{4\pi\varepsilon_0}, \qquad E_\perp = \frac{\sigma\,\Omega}{4\pi\varepsilon_0}

$\Omega$: solid angle the surface subtends at the charge
$\sigma$: surface charge density
$E_\perp$: field component normal to the charged surface

A point charge's flux through a surface is proportional to the solid angle it subtends, since the whole sphere of $4\pi$ carries $q/\varepsilon_0$ . Hence the flux through a cube face with the charge at the centre is $q/6\varepsilon_0$ , and through a face with the charge at a corner zero. Conversely the normal field of a charged surface at a point is $\sigma\Omega/4\pi\varepsilon_0$ , $\Omega$ being the solid angle of the surface from that point, which gives the field of a disc, a cone or a hemisphere with no integrals.

Appears in problems (7) 6.1.19 6.2.1 6.2.3 6.2.5 9.3.7 9.3.10 9.3.18

Field of a charged slab Поле заряженного слоя

law 6.2

внутриотсерединыснаружи$$E = \frac{\rho x}{\varepsilon_0}\ \text{(внутри, от середины)}, \qquad E = \frac{\rho h}{2\varepsilon_0}\ \text{(снаружи)}$$

$\rho$: volume charge density
h: thickness of the slab
x: distance from the mid-plane

Inside a uniformly charged flat slab the field grows linearly from the middle, since by Gauss's law only the charge between $$-x$$ and $$x$$ counts, outside the slab acts as a plane with $\sigma = \rho h$ . The field in a cavity of a slab or a ball is found by subtraction, as for the missing charge. Two slabs of opposite sign give a field only in between.

Appears in problems (6) 6.2.6 6.2.11 6.2.12 6.2.14 6.3.26 6.5.4

6.3The potential of an electric field. Conductors in a constant electric field

Field and potential Связь поля и потенциала

law 6.3

\vec E = -\nabla\varphi, \qquad \varphi_1 - \varphi_2 = \int_1^2 \vec E\cdot d\vec l, \qquad E = \frac{U}{d}

\displaystyle \varphi(r) = \varphi(R) + \int_r^R E\,dr

однородноеполе$\displaystyle U = Ed\ \text{(однородное поле)}$

$\varphi$: potential
U: potential difference, voltage
d: distance along the field

The field is minus the gradient of the potential, and the potential difference between two points is the integral of $$E$$ along any path between them. In a uniform field that is just $$Ed$$ . The potential inside a charged ball or cylinder comes from integrating the known field inward from the surface.

Accelerating voltage Ускоряющее напряжение

law 6.3

\frac{mv^2}{2} = eU, \qquad v = \sqrt{\frac{2eU}{m}}

U: potential difference traversed
e: charge of the particle
m: its mass

A charge that freely crosses a voltage $$U$$ gains kinetic energy $$eU$$ , whatever the path and the field shape. An electron accelerated by a kilovolt moves at about $2\cdot10^7$ m/s. The same relation gives the stopping voltage that halts a particle.

Energy of a charge in a field Энергия заряда в поле

law 6.3

\frac{mv_1^2}{2} + q\varphi_1 = \frac{mv_2^2}{2} + q\varphi_2, \qquad \frac{mv^2}{2} = eU

энергиядвухзарядов$\displaystyle W = \frac{q_1 q_2}{4\pi\varepsilon_0 r}\ \text{(энергия двух зарядов)}$

\displaystyle v = \sqrt{\frac{2eU}{m}}

q: charge of the particle
$\varphi$: potential at the point
U: potential difference traversed

The electrostatic field is conservative, and a charge in it has potential energy $q\varphi$ , conserved together with the kinetic energy. Crossing a voltage $$U$$ an electron gains $$eU$$ , hence the electron volt. Two charges have energy $q_1q_2/4\pi\varepsilon_0 r$ , positive for like charges.

Appears in problems (15) 2.3.38 2.3.39 6.3.1 6.3.2 6.3.35 6.5.25 7.1.12 7.1.13

Potential of a point charge Потенциал точечного заряда

law 6.3

\varphi = \frac{1}{4\pi\varepsilon_0}\,\frac{q}{r}, \qquad \varphi = \sum_i \varphi_i

q: the charge
r: distance from it
$\varphi$: potential, zero at infinity

A point charge's potential falls as $$1/r$$ and is zero at infinity. Potentials are scalars, so a set of charges has the plain sum and an extended body the integral $\int dq/4\pi\varepsilon_0 r$ . Hence the potential at the centre of a ring or a shell is $Q/4\pi\varepsilon_0 R$ , all elements being at one distance.

Appears in problems (14) 6.3.3 6.3.4 6.3.5 6.3.6 6.3.7 6.3.21 6.3.24 6.3.25

Conductor in an electrostatic field Проводник в электростатическом поле

law 6.3

внутри$$E_{\text{внутри}} = 0, \qquad \varphi = \text{const}, \qquad E_n = \frac{\sigma}{\varepsilon_0}$$

зарядназаземлённойсфереотзаряданарасстоянии$\displaystyle q' = -q\frac{R}{L}\ \text{(заряд на заземлённой сфере от заряда }q\text{ на расстоянии }L)$

внешниегранидвухпластин$\displaystyle \sigma_1' = \frac{\sigma_1 + \sigma_2}{2}\ \text{(внешние грани двух пластин)}$

$внутри$E_{\text{внутри}}$$: field inside the metal
$\sigma$: surface charge density
$\varphi$: potential of the conductor, the same at all its points

In equilibrium a conductor's charges rearrange so that the field inside is zero and the whole conductor, cavities included, sits at one potential. Charge lives on the surface, where the field is normal and equals $\sigma/\varepsilon_0$ . On a stack of plates the outer faces carry half the total charge each and the inner faces are opposite. A grounded sphere has zero potential, giving an induced charge $$-qR/L$$ .

Appears in problems (13) 6.3.9 6.3.11 6.3.12 6.3.15 6.3.16 6.3.18 6.3.24 6.3.31

Capacitance of an isolated conductor Ёмкость уединённого проводника

definition 6.3

шара$$C = \frac{q}{\varphi}, \qquad C_{\text{шара}} = 4\pi\varepsilon_0 R$$

сферическийконденсатор$\displaystyle C = 4\pi\varepsilon_0\frac{R_1R_2}{R_2 - R_1}\ \text{(сферический конденсатор)}$

q: charge of the conductor
$\varphi$: its potential
R: radius of the ball

An isolated conductor's potential is proportional to its charge, with a coefficient set only by shape and size. A ball has capacitance $4\pi\varepsilon_0 R$ , and scaling a body up $$k$$ times multiplies its capacitance by $$k$$ . A spherical capacitor's follows from the potential difference of its plates.

Appears in problems (10) 6.3.39 6.3.40 6.4.1 6.4.5 6.4.6 6.4.17 6.5.6 6.5.15

Potential of a charged shell and ball Потенциал заряженной сферы и шара

law 6.3

внутришара$$\varphi = \frac{Q}{4\pi\varepsilon_0 R}\ (r \le R), \qquad \varphi(r) = \frac{\rho}{6\varepsilon_0}(3R^2 - r^2)\ \text{(внутри шара)}$$

\displaystyle \varphi = \frac{Q}{4\pi\varepsilon_0 r}\ (r \ge R)

центроднородныйшар$\displaystyle \varphi_{\text{центр}} = \frac{3Q}{8\pi\varepsilon_0 R}\ \text{(однородный шар)}$

Q: charge of the shell or ball
R: radius
$\rho$: volume charge density of the ball

Outside, a shell's potential is that of a point charge, inside the field is zero and the potential everywhere equals its surface value. Inside a uniform ball the potential rises toward the centre by $\rho R^2/6\varepsilon_0$ , from integrating the linear field. The potential at the centre of a ring or a shell can also be computed as $\int dq/4\pi\varepsilon_0 R$ , all charges being at one distance.

Appears in problems (10) 6.3.4 6.3.5 6.3.7 6.3.24 6.3.25 6.3.26 6.3.27 6.3.31

Method of images Метод изображений

method 6.3

F = \frac{1}{4\pi\varepsilon_0}\,\frac{q^2}{(2h)^2}, \qquad q' = -q\frac{R}{L}

q: charge above a conducting plane
h: its distance from the plane
$$q'$$: image charge in a grounded sphere of radius $$R$$
L: distance of the charge from the sphere's centre

Outside the metal the field of a charge above a grounded plane equals that of the charge and its mirror image $$-q$$ at depth $$h$$ , since that pair gives zero potential on the plane. The charge is drawn to the plane as to the image $$2h$$ away. For a grounded sphere the image $$-qR/L$$ sits at $$R^2/L$$ from the centre.

Appears in problems (8) 6.3.24 6.3.28 6.3.29 6.3.30 6.3.31 6.3.32 6.3.33 6.6.16

Spherical capacitor. Concentric shells Сферический конденсатор. Концентрические сферы

law 6.3

C = 4\pi\varepsilon_0\frac{R_1R_2}{R_2 - R_1}, \qquad \varphi_1 - \varphi_2 = \frac{q}{4\pi\varepsilon_0}\left(\frac{1}{R_1} - \frac{1}{R_2}\right)

R₁, R₂: radii of the inner and outer shells
q: charge of the inner shell

Between concentric shells only the inner charge makes the field, and the potential difference is the integral of $q/4\pi\varepsilon_0 r^2$ . As $R_2 \to \infty$ this is the isolated ball's capacitance, for a narrow gap the flat formula. Each shell's potential in a system is the sum of all shells' contributions, inner ones giving $q/4\pi\varepsilon_0 r$ and outer ones $q/4\pi\varepsilon_0 R$ .

Appears in problems (8) 6.3.21 6.3.24 6.3.35 6.4.5 6.4.6 6.4.17 6.5.6 8.3.45

6.4Capacitors

Parallel-plate capacitor Плоский конденсатор

law 6.4

C = \frac{\varepsilon\varepsilon_0 S}{d}, \qquad E = \frac{U}{d} = \frac{q}{\varepsilon_0 S}

пластинадиэлектрикатолщинывзазоре$\displaystyle C = \frac{\varepsilon_0 S}{d - h + h/\varepsilon}\ \text{(пластина диэлектрика толщины }h\text{ в зазоре)}$

двадиэлектрикарядом$\displaystyle C = \frac{\varepsilon_0 S}{2d}\,(\varepsilon_1 + \varepsilon_2)\ \text{(два диэлектрика рядом)}$

S: plate area
d: gap
$\varepsilon$: permittivity of the filling

The field between close plates is uniform, $\sigma/\varepsilon_0$ , the voltage is $$Ed$$ , hence the capacitance. A dielectric weakens the field $\varepsilon$ times and raises the capacitance by the same factor. A slab in part of the gap and a filling over half the area are treated as capacitors in series and in parallel.

Capacitance of a capacitor Ёмкость конденсатора

definition 6.4

C = \frac{q}{U}, \qquad q = CU

q: charge of one plate
U: voltage between the plates

A capacitor's charge is proportional to the voltage, and the capacitance depends only on geometry and the dielectric. The plates carry equal and opposite charges. In a circuit with a source, the voltage across a charged capacitor equals the emf and the charge is $C\mathcal{E}$ .

Energy of a capacitor Энергия конденсатора

law 6.4

W = \frac{CU^2}{2} = \frac{q^2}{2C} = \frac{qU}{2}

заряженнаясфера$\displaystyle W = \frac{Q^2}{8\pi\varepsilon_0 R}\ \text{(заряженная сфера)}$

работаприраздвиженииобкладок$\displaystyle A = W_2 - W_1\ \text{(работа при раздвижении обкладок)}$

C: capacitance
U: voltage
q: charge

Charging a capacitor moves charge against a rising voltage, and the work is the triangle's area $$qU/2$$ . The energy sits in the field between the plates. Pulling the plates apart slowly at constant charge, the outside force's work equals the gain in $$q^2/2C$$ , at constant voltage the source joins the balance. Half the source's work when charging through a resistor becomes heat.

Appears in problems (18) 6.4.15 6.4.16 6.4.17 6.4.18 6.5.18 6.5.19 6.5.20 6.5.28

Capacitors in parallel and in series Соединение конденсаторов

law 6.4

парпосл$$C_{\text{пар}} = C_1 + C_2, \qquad \frac{1}{C_{\text{посл}}} = \frac{1}{C_1} + \frac{1}{C_2}$$

$пар$C_{\text{пар}}$$: parallel combination, a common voltage
$посл$C_{\text{посл}}$$: series combination, a common charge

In parallel the voltages agree and the charges add, in series the charges agree and the voltages add. A grounded plate between two others splits a capacitor into two in parallel. In a complicated network one writes charge conservation for each isolated node and the sum of voltages around each loop.

Appears in problems (16) 6.4.4 6.4.7 6.4.8 6.4.9 6.4.13 6.5.27 6.5.28 6.6.9

Charge redistribution on plates Перераспределение зарядов на пластинах

method 6.4

контур$$q_1 + q_2 = \text{const}, \qquad \sum_{\text{контур}} U_i = 0$$

соединениедвухзаряженныхконденсаторов$\displaystyle U = \frac{C_1U_1 + C_2U_2}{C_1 + C_2}\ \text{(соединение двух заряженных конденсаторов)}$

q₁, q₂: charges on plates joined by a wire or isolated together
U_i: voltages across the elements of a loop

The charge of an isolated group of plates is conserved, and the sum of voltages around a closed loop is zero, since the field is conservative. These two rules find the charges after any switching. A plate between two grounded ones splits its charge between them inversely to the distances. Joining two capacitors gives a weighted mean voltage.

Appears in problems (7) 6.1.5 6.3.31 6.4.9 6.4.11 6.4.13 6.4.14 6.6.8

Cylindrical capacitor Цилиндрический конденсатор

law 6.4

C = \frac{2\pi\varepsilon\varepsilon_0 l}{\ln(R_2/R_1)}

R₁, R₂: radii of the inner and outer cylinders
l: length of the capacitor

The field between coaxial cylinders is $q/2\pi\varepsilon_0 l r$ , and integrating over $$r$$ gives the logarithm of the radius ratio. For a narrow gap the logarithm is near $$d/R$$ and the formula becomes the flat one. A thin wire near a plane or a cylinder inside a cylinder of small radius has a capacitance per length of order $2\pi\varepsilon_0/\ln$ .

Appears in problems (3) 6.4.7 6.4.8 6.4.17

6.5Electrical pressure. The energy of an electric field

Electric pressure on a charged surface Электрическое давление на заряженную поверхность

law 6.5

p = \frac{\sigma^2}{2\varepsilon_0} = \frac{\varepsilon_0 E^2}{2} = \frac{\sigma E}{2}

притяжениеобкладок$\displaystyle F = \frac{q^2}{2\varepsilon_0 S} = \frac{\varepsilon_0 S U^2}{2d^2}\ \text{(притяжение обкладок)}$

черезэнергию$\displaystyle F = -\frac{dW}{dx}\ \text{(через энергию)}$

$\sigma$: surface charge density
E: field just outside the surface
p: force per unit area, outward

A patch's charge feels not the full field $$E$$ but the field of all other charges, $$E/2$$ , since its own field does not push on it. Hence the pressure $\sigma E/2$ , always outward, like a negative pressure. This gives the attraction of plates, the stretching of a charged soap bubble and the rise of liquid in a capacitor. The same follows from $$F = -dW/dx$$ .

Appears in problems (15) 6.1.1 6.4.15 6.5.1 6.5.2 6.5.3 6.5.5 6.5.6 6.5.7

Energy of the electric field Энергия электрического поля

law 6.5

w = \frac{\varepsilon\varepsilon_0 E^2}{2}, \qquad W = \int \frac{\varepsilon_0 E^2}{2}\,dV

заряженнаясфера$\displaystyle W = \frac{Q^2}{8\pi\varepsilon_0 R}\ \text{(заряженная сфера)}$

равномернозаряженныйшар$\displaystyle W = \frac{3Q^2}{20\pi\varepsilon_0 R}\ \text{(равномерно заряженный шар)}$

w: energy density of the field
E: field strength
$\varepsilon$: permittivity of the medium

The capacitor's energy $$CU^2/2$$ written through the field is $\varepsilon_0E^2/2$ per unit volume, and this holds for any field. A charged sphere's energy comes from integrating $$E^2$$ outside, a ball's with the inner part added. Field energy is not additive in the fields, so bringing charges together and cutting a ball apart cost work.

Appears in problems (9) 6.5.14 6.5.16 6.5.18 6.5.19 6.5.20 6.5.21 7.4.34 12.1.28

Energy of a system of charges Энергия системы зарядов

law 6.5

W = \frac{1}{2}\sum_i q_i\varphi_i = \sum_{i<j}\frac{q_iq_j}{4\pi\varepsilon_0 r_{ij}}

$\varphi_i$: potential at charge $$i$$ due to all the others
r_ij: distance between the charges

The interaction energy is a sum over all pairs, and the half in the first formula keeps each pair from being counted twice. Assembling the system charge by charge costs work against those already placed, so $$n$$ equal charges at one spot need $$n(n-1)/2$$ pair energies. The work of outside forces in a rearrangement is the energy difference.

Appears in problems (6) 6.5.19 6.5.22 6.5.23 6.5.25 6.5.26 7.4.10

Energy of a charged shell and ball Энергия заряженной сферы и шара

law 6.5

сферашар$$W = \frac{Q^2}{8\pi\varepsilon_0 R}\ \text{(сфера)}, \qquad W = \frac{3Q^2}{20\pi\varepsilon_0 R}\ \text{(шар)}$$

Q: charge
R: radius

A charged shell's energy is $Q\varphi/2$ , half the charge times the potential, or $$Q^2/2C$$ with capacitance $4\pi\varepsilon_0R$ . The same comes from integrating the field's energy density outside. A uniform ball adds the field inside, and its energy is $$6/5$$ as large. Splitting a shell in two or halving its radius costs the difference of these energies, the work being done against the electric pressure.

Appears in problems (5) 6.5.16 6.5.18 6.5.19 6.5.20 6.5.25

Classical electron radius Классический радиус электрона

value 6.5

м$$m_ec^2 = \frac{e^2}{8\pi\varepsilon_0 R}, \qquad R \approx 1{,}4\cdot10^{-15}\ \text{м}$$

m_e: electron mass
e: elementary charge
R: radius at which the field energy equals $$m_ec^2$$

If all of the electron's rest energy is ascribed to its electric field, the field of a shell of radius $$R$$ gives $e^2/8\pi\varepsilon_0R$ , hence $R \sim 10^{-15}$ m. It is an estimate of the scale where electrostatics stops working, not the electron's actual size.

Appears in problems (2) 6.5.16 6.5.17

6.6The electric field in the presence of a dielectric

Permittivity Диэлектрическая проницаемость

definition 6.6

пол$$\varepsilon = \frac{E_0}{E}, \qquad \vec E = \vec E_0 + \vec E_{\text{пол}}$$

своб$\displaystyle \vec D = \varepsilon\varepsilon_0\vec E, \qquad \oint \vec D\cdot d\vec S = q_{\text{своб}}$

\displaystyle \varepsilon_1 E_{1n} = \varepsilon_2 E_{2n}, \qquad E_{1\tau} = E_{2\tau}

$\varepsilon$: relative permittivity
E₀: field of the free charges, without the dielectric
$пол$\vec E_{\text{пол}}$$: field of the bound charges of the polarized dielectric

A dielectric filling the whole field weakens it $\varepsilon$ times, since the bound charges on its surface produce an opposing field. The capacitance grows $\varepsilon$ times, and a capacitor on a source takes an extra charge $(\varepsilon-1)CU$ . At a boundary of two dielectrics the normal component of $\varepsilon E$ and the tangential $$E$$ are continuous, hence the refraction of field lines.

Appears in problems (18) 6.5.3 6.6.1 6.6.4 6.6.5 6.6.6 6.6.7 6.6.8 6.6.9

Polarization and bound charge Поляризация и связанные заряды

law 6.6

\vec P = \varepsilon_0(\varepsilon - 1)\vec E = n\vec p, \qquad \sigma' = P_n

$\vec P$: polarization, dipole moment per unit volume
n: number density of molecules
$\vec p$: dipole moment of one molecule
$\sigma'$: bound surface charge density

In a field the molecules of a dielectric acquire dipole moments, and on the surface the dipoles face a bound charge appears with density equal to the normal component of $$P$$ . For a slab in a parallel-plate capacitor the bound charge $\sigma(\varepsilon-1)/\varepsilon$ opposes the free one, which is what weakens the field. A jump of $$P_n$$ at the boundary of two dielectrics gives bound charge too.

Appears in problems (13) 6.6.2 6.6.3 6.6.4 6.6.12 6.6.14 6.6.15 6.6.18 6.6.19

Dielectric drawn into a capacitor Втягивание диэлектрика в конденсатор

law 6.6

F = \frac{\varepsilon_0(\varepsilon - 1)\,b\,U^2}{2d}, \qquad h = \frac{\varepsilon_0(\varepsilon - 1)U^2}{2\rho g d^2}

припостоянномнапряжении$\displaystyle F = \frac{U^2}{2}\frac{dC}{dx}\ \text{(при постоянном напряжении)}$

припостоянномзаряде$\displaystyle F = -\frac{q^2}{2C^2}\frac{dC}{dx}\ \text{(при постоянном заряде)}$

b: plate width across the motion
d: gap
U: voltage
h: rise of a liquid dielectric
$\rho$: density of the liquid

Capacitance grows as a dielectric enters the gap and the energy of the system with its source falls, so a slab or a liquid is drawn in. The force comes from the derivative of the capacitance, $F = U^2\,dC/dx/2$ at constant voltage, from $$q^2/2C$$ with the opposite sign at constant charge. A liquid rises until the pressure $\varepsilon_0(\varepsilon-1)E^2/2$ is balanced by the column $\rho g h$ .

Appears in problems (7) 6.5.27 6.6.16 6.6.17 6.6.19 6.6.20 6.6.21 6.6.22

Molecular polarizability. Clausius Mossotti Поляризуемость молекулы. Формула Клаузиуса Моссотти

law 6.6

лок$$p = \alpha E_{\text{лок}}, \qquad \frac{\varepsilon - 1}{\varepsilon + 2} = \frac{n\alpha}{3\varepsilon_0}, \qquad p = 4\pi\varepsilon_0 r^3 E$$

лок$\displaystyle \vec E_{\text{лок}} = \vec E + \frac{\vec P}{3\varepsilon_0} = \frac{\varepsilon + 2}{3}\vec E$

силанадипольвнеоднородномполе$\displaystyle F_x = p\,\frac{\partial E}{\partial x}\ \text{(сила на диполь в неоднородном поле)}$

$\alpha$: polarizability of a molecule
$лок$E_{\text{лок}}$$: local field at the molecule, larger than the mean field in a dense medium
r: radius of a conducting ball, the atom model
n: number density of molecules

A conducting ball in a field $$E$$ gets a dipole moment $4\pi\varepsilon_0 r^3E$ , the order of magnitude of an atom's polarizability. In a dilute gas $\varepsilon - 1 = n\alpha/\varepsilon_0$ , in a dense medium a molecule feels the local field with the surroundings' $P/3\varepsilon_0$ added, which gives the Clausius Mossotti formula. A dipole is drawn into the stronger field with force $p\,\partial E/\partial x$ , for an induced dipole proportional to $$E^2$$ .

Appears in problems (6) 6.6.2 6.6.3 6.6.15 6.6.28 6.6.29 6.6.30

Capacitor with dielectric layers Конденсатор со слоями диэлектрика

law 6.6

C = \frac{\varepsilon_0 S}{d - h + h/\varepsilon}, \qquad \frac{1}{C} = \frac{1}{\varepsilon_0 S}\left(\frac{d_1}{\varepsilon_1} + \frac{d_2}{\varepsilon_2}\right)

h: thickness of the dielectric slab in the gap $$d$$
d₁, d₂: thicknesses of the layers with permittivities $\varepsilon_1, \varepsilon_2$

Layers across the field act as capacitors in series, since $$D$$ is the same in all and the field in each is $D/\varepsilon\varepsilon_0$ . A slab of thickness $$h$$ shortens the effective gap by $h(1 - 1/\varepsilon)$ , a metal plate by its full thickness. Layers along the field act as capacitors in parallel.

Appears in problems (5) 6.6.4 6.6.11 6.6.12 6.6.22 11.6.9

Heat when a capacitor is filled Тепло при заполнении конденсатора

law 6.6

Q = W_1 - W_2 = \frac{q^2}{2C}\,\frac{\varepsilon - 1}{\varepsilon}\ (q = \text{const})

припостоянномнапряжениисучётомработыисточника$\displaystyle Q = \frac{CU^2}{2}(\varepsilon - 1)\ \text{(при постоянном напряжении, с учётом работы источника)}$

зарядкачерезсопротивление$\displaystyle Q = \frac{C\mathcal{E}^2}{2}\ \text{(зарядка через сопротивление)}$

q: charge of the capacitor
C: capacitance before filling
$\varepsilon$: permittivity of the dielectric

When a dielectric flies into the gap of a disconnected capacitor, the energy $$q^2/2C$$ drops $\varepsilon$ times and the difference goes to heat and the slab's oscillation. With the source attached the capacitor takes more charge, the source does work $\mathcal{E}\Delta q$ , twice the energy gain, and half of it heats the circuit. The same holds when any capacitor is charged through a resistor.

Appears in problems (3) 6.6.23 6.6.24 6.6.25

Boundary conditions at a dielectric surface Граничные условия на поверхности диэлектрика

law 6.6

\varepsilon_1 E_{1n} = \varepsilon_2 E_{2n}, \qquad E_{1\tau} = E_{2\tau}, \qquad \frac{\tan\alpha_1}{\tan\alpha_2} = \frac{\varepsilon_1}{\varepsilon_2}

E_n: field component normal to the boundary
$E_{\tau}$: tangential component
$\alpha$: angle between $\vec E$ and the normal

At a boundary free of charge the normal component of $\varepsilon E$ is continuous, since the flux of $$D$$ through a thin box is zero, and the tangential component of $$E$$ is continuous, since the work around a small loop is zero. Field lines refract so that the tangents of the angles are as the permittivities. In a layered capacitor it follows that the field is inversely proportional to the layer's $\varepsilon$ .

Appears in problems (3) 6.5.3 6.6.5 6.6.12

Electric displacement Электрическое смещение

law 6.6

своб$$\vec D = \varepsilon\varepsilon_0\vec E = \varepsilon_0\vec E + \vec P, \qquad \oint \vec D\cdot d\vec S = q_{\text{своб}}$$

$\vec D$: electric displacement
$\vec P$: polarization
$своб$q_{\text{своб}}$$: free charge inside the surface, bound charge excluded

Gauss's law for $$D$$ involves only free charges, so in problems with a dielectric one first finds $$D$$ from symmetry and then $E = D/\varepsilon\varepsilon_0$ in each layer. The jump of the normal component of $$D$$ equals the free surface charge, and the jump of $\varepsilon_0E_n$ the total, free and bound together.

Appears in problems (3) 6.5.3 6.6.14 6.6.18