Revision Electrostatics

Electric Field E

Electrostatic Potential V

Electric Dipole p

Electric flux and the Gauss theorem in electrostatics

Current electricity

Electromotive force (EMF)

Current I

Current density J

Magnetic effects

Magnetic dipoles (m)

The magnetic B-field

Magnetic flux and the Gauss theorem in magnestatics

Magnetic fields set up by electrical currents

The Biôt-Savart law

The Ampère Law

Electromagnetic induction

The Faraday-Lenz law

Time-dependent effects

The displacement current

Maxwell's equations

The em wave equation (radio, light, ...)

Vector operators in Mathematics

grad scalar field ∇V

div vector field ∇.E

curl vector field ∇×E

∇² scalar ∇²V

Throughout this section, we will be assuming that everything is in a vacuum.

Electric field E ----o----> E

We define the E-field as the force on a unit charge at the point concerned.

Coulomb's Law r q o- - - - - - - - -o Q X

Force on a charge q at X is (Qq)/(4πε₀r²).

ε₀ = permittivity of free space.

Hence, the E-field at X due to Q is:

E = (Q)/(4πε

₀r²)

Lines of E Fields

The lines of an electric field E are given by:

dy/dx = E

_y/E_x

Electrostatic Potential V

The work done in moving test charge q in the presence of the charge Q from infinity to an arbitrary point is the potential V of this arbitary point.

A o __ F |'.. /| : / ' | ''. / o <---- q | / '. r_A / ,->\ ,''. | / | '..' '. / | : | / dl : PATH / '.. |/ ''... o- - - - - - - - - - - - - - -'-o Q r_B B _ F /| / q /)θ o----o / dl / L F_ext

F_ext is the force an external agent requires to hold charge q in position against force F.

When q moves by dl, the work done by the external agent force F_ext is:

dW

_ext = F_ext × (-dl cos θ)

But F_ext = F = qE, so:

dW

_ext = -q E dl cos θ

dW

_ext = -q E_l dl

Hence, in qoing from A to B, the work done by the external agent is (by adding together the work done for each little bit dl):

B W_ext = - ∫ q E cos θ dl A B = - ∫ q E dr A

But E = Q/(4πε₀r²), so:

B qQ.dr W_ext = - ∫ ----- A 4πε₀r² [ qQ.dr ]B qQ qQ = [ ----- ] = ------ - ------ [ 4πε₀r²]A 4πε₀r_B 4πε₀r_A

...which is independent of the route taken (and is thus a conservative force).

If q is a unit charge,

Q Q W_ext = ------ - ------ 4πε₀r_B 4πε₀r_A

...which we write as:

W_ext = v_B - v_A

...where

Q V(r) = ------ 4πε₀r_r

...which is the potential at the point concerned, and is the work required to bring a unit test charge (q=1) from ∞ (r_A=∞) to the point concerned (r_B).

The relation between E and V _A->B B B W_ext = - ∫ E_ldl = - ∫ E.dl A A

for unit charge

B V_B-V_A = - ∫ E.dl A

Suppose A and B coincide (close loop). Then,

V

_B = V_A

Hence,

∮ E.dl = 0 closed loop

This is only correct in electrostatics.

From earlier:

dW

_ext = -q E_l dl

If q=1, dW_ext = dV

Hence,

dV = -E

_ldl

i.e.

-∂V E_l = -- ∂l -∂V E_x = -- etc. ∂x

Since E is a vector, we have

E = -grad(v) = -∇V

∂V ∂V ∂V E = - i -- - j -- - k -- ∂x ∂y ∂z

Electric Dipole p

Two point charges +q and -q separated by a distance a constitute a dipole.

a o---------o -q +q --------> p

Usually we assume a is small.

The dipole moment p has magnitude

p = qa

Potential due to a dipole o X / / / r / / /) θ --->

The potential at X due to a dipole is given by:

p cos θ v = ------- 4πε₀r²

Derivation of the potential due to a dipole X ,o r_- .'/: ,' / :r₊ ,' r/)θ: o---+---o -q a +q

From this diagram one sees that:

q q V = ------ - ------ 4πε₀r₊ 4πε₀r_- q ( r_- - r₊ ) = ----- ( ------- ) 4πε₀ ( r₊r_- )

But, if you drop a perpendiculous from +q to r_-, and assume that r>>a, you can approximate that r_- - r₊ ≅ a cos θ and r_-r₊ ≅ r², so:

qa cos θ p cos θ V = -------- = ------- 4πε₀r² 4πε₀r²

The electric field caused by a dipole . /|\ E_θ |_ \ r | | \-----------------+------> E_r \'θ X dipole -∂V Radial component: E_r = -- ∂r -∂V -1 ∂V Tangential component: E_θ = -- = - -- ∂s r ∂θ

...where ds = rdθ is a deplacement in the tangential direction.

Hence:

2p cos θ E_R = -------- 4πε₀r³ p cos θ E_θ = ------- 4πε₀r³

To find the electric field E, we use geometry:

. _ E_θ /|\ ,| E | ,' \ r |,')) φ \-----------------+------> E_r p \'θ X

Hence:

E = (E_r² + E_θ²)^½ p = ----- ( 3 cos² θ + 1 )^½ 4πε₀r³ E_θ tan φ = -- = ½ tan θ E_r

The electric field caused by a dipole _____________________\ +q / E o--->qE /| ____________ /_|_____\ / | / a/ |a sin α / | ________ /_____|_____\ / _| / -q/)α | | qE<---o--------+ _____________________\ /

Couple (Torque) acting on a dipole is denoted by the greek letter Γ and is:

Γ = (Eq)×(a sin α)
Γ = pE sin α

In vector notation:

Γ = p × E

Electric flux and the Gauss theorem in electrostatics | _ normal-->| α /| E |,^./ | / ___|_/ / r|/\ \_____/<-- arbitrary area dS / / /

Electric Flux passing through dS is defined as: (normal Component of E) × (area) = E_n dS = E cos α dS = E.dS (for small dS — E same throughout dS)

For a large surface, the flux passing through a surface is:

∬ E_n dS surface

...were E may be a function of the position on dS.

For a closed surface we can show that:

charged enclosed ∯ E_n dS = ---------------- whole ε₀ surface

i.e. Electrical Flux leaving a closed surface is equal to the charge enclosed divided by ε₀

This is the Gauss Theorem.

Charge density

ρ(v) = charge per unit volume at the point concerned.

_____ ( ,. ) C []' ) (_ _) ,. (_____) []' = volume dv

Clearly, the charge enclosed is given by:

∭ ρ(v) dv volume

Gauss Theorem

The Gauss Theorem becomes:

1 ∯ E_n dS = --- ∭ ρ(v) dv surface ε₀ volume

Gauss Theorem in differential form

The divergence div(F) of a vector quantity F is the scalar which represents the net flux per unit volume of F leaving a small volume.

Let's consider the divergence of the electrical field E:

______________ ___ /| LHFACE /' /| /|\ : / | / | | : / / | |' / | / |.'|dz +-------------+ ..'| | . | | ...''|' | | ''|'....''' | | \|/ \ non uniform | + - - - -|- - + __...'''/ electric field E | | ../'' /\ | / ......''|' / / ..'|''''' | / /dy |/ |/ / +-------------+ \/_ RHFACE <-------------> x x+dx

Flux of E leaving LHFACE = -E_x dydz (negative because leaving in -x direction)

At RHFACE, the field is (E_x+dE_x), so the flux leaving RHFACE = (E_x+dE_x)dydz

Adding these two expressions, we get the net flux leaving in x-direction:

dE

_x dydz

...where dE_x is:

∂E_x dE_x = -- dx ∂x

Substituting:

∂E_x ∂E_x -- dxdydz = -- dv ∂x ∂x

Similarly for other directions, giving:

∂E_x ∂E_y ∂E_z ( -- + -- + -- ) dv = Net flux leaving whole volume ∂x ∂y ∂z

The flux leaving per unit volume is that divided by dv, thus:

∂E_x ∂E_y ∂E_z ( -- + -- + -- ) = div E ∂x ∂y ∂z

...which can be written ∇.E

Now the net flux leaving is equivalent to the charge enclosed divided by ε₀:

div E dv = ρ(v) dv / ε

₀

...giving:

div E = ρ/ε

₀

Electromotice Force (EMF)

Consider a charged capacitor:

E_Q B|- <- +|A |- <- +| |- <- +| +---------|- <- +|---------+ : |- <- +| : : |- <- +| : : |- <- +| : : : :..........................: mathematical circuit

Inside the capacitor there is an electric field E_Q due to the charges.

B V_B-V_A = - ∫ E.dl = E_Q x A ∫ E.dl = 0 closed loop

(This is all electrostatics -- no moving charges.)

Consider now an electrochemical cell.

| | | | B|- <- +|A |- →→ +| Key: |- <- +| +---------|- →→ +|---------+ <- E_Q due to charges : |- <- +| : : |- →→ +| : →→ Field set up by : |- <- +| : chemical interactions : ↑ | : : Electrolyte | : : | : : : :..........................: mathematical circuit

Owing to chemical interactions, - charges are driven to A, and + charges are driven to B.

This continues until the electric field E_Q due to charge separation balances the field set up by the chemical interaction.

Therei s then no net electric field inside the cell.

The A terminal is at a higher potential than terminal B (as evidenced by the fact that current flows).

The difference V_A - V_B is called the EMF ε of the cell (on an open circuit).

B ∫ E.dl (via circuit) = V_A-V_B = ε A

But:

B ∫ E.dl (via cell) = 0 A

...because the two fields cancel each other inside the cell. Hence:

∫ E.dl = ε + 0 = ε whole circuit

The general definition of the EMF in a circuit is therefore:

ε = ∫ E.dl whole circuit

Note that the circuit could be fictional, this is just any mathematical closed path.

Magnetism

There are only magnetic dipoles — e.g. small permanent magnets, small coils of current carrying wire. — no magnetic monopoles. (At least, not in the macroscopic world.)

For small coils carrying a circuit I the magnetic dipole moment is given by:

m = I × area

. /|\ m __|___ / | |\ I \_____/ | ↑ | area dS

m = Id S (for small areas)

The Magnetic B-field

Direction is that in which a freely suspended dipole points (a compass). Magnitude is obtained by measuring the torque needed to hold the compass needle at π/2 rads to the field.

_____________________\ __ / B /| / ' ________ /___________\ α(/ / / / _____________________\ /

Γ = m × B

If α = π/2, Γ = mB, ∴ B = Γ/m (where m is the magnetic dipole magnitude).

Magnetic Field set up by a Dipole . /|\ B_θ |_ \ r | | \-----------------+------> B_r \'θ \ _\| m (magnetic dipole)

In E-fields,

2p cos θ E_R = -------- 4πε₀r³ p cos θ E_θ = ------- 4πε₀r³

...and by analogy, the magnetic dipole case is similar:

μ₀ 2m cos θ B_R = -- . -------- 4π r³ μ₀ m cos θ B_θ = -- . ------- 4π r³

μ₀ = permeability of free space.

Magnetic Flux | _ normal-->| α /| B |,^./ | / ___|_/ / r|/\ \_____/<-- arbitrary area dS / / /

Magnetic flux passing through dS is B_ndS = B cos α dS = B.dS

Gauss Theorem in Magnetostatics

Magnetic flux leaving a closed surface is zero.

∫∫ B.dS = 0 closed surface ∴ div B = 0

Magnetic Fields set up by Current Carrying Circuits

Biôt-Savart Law (_ B |\ I / ,' \ / ,' | r/ ,' / / ,' \__ θ /,' \^/' dl ,\ ,'_/ ,' / ,' \__ \ (_

What is the field dB at the point marked B due to dl.

We find that:

μ₀ I dl sin θ dB = -- . ---------- 4π r²

...and it is into the diagram at the point B, so:

μ₀ I dl × r dB = -- . -------- 4π r³

We can integrate this law to give:

Field due to a long straight wire | /|\ I | | | ...''|''... .'' | ''. : |-. h : : +-+-------x : | __: '.. | ../ B '''.....''' | | | μ₀ I B = --- 2πh Field at the centre of a circular coil. ..'.. .' '. : : /:\ I : B : : ---<----:------+-+----+---<--- : |_| : : | : : |a : '. | .' ''|'' ↑ COIL μ₀ I B = --- 2 a Field inside a cylindrical coil. ,o8888888888888888888888888888888888888o. ,dP' `9b. `9b. `9b. `9b. `9b. `9b. ,8' `8. `8. `8. `8. `8. `8. 8' `8. `8. `8. `8. `8. `8. oP 'o 'o 'o 'o 'o 'o B 8 8 8. 8. 8. 8. 8 --<--O]------ - -O] /O]\ /O]\ /O]\ /O]\ O] 8. 8 / 8 |' / 8 |' / 8 |' / 8 |' 8 `o ,P ,P ,P ,P ,P ,P 8. ,8 ,8 ,8 ,8 ,8 ,8 `8. ,8' ,8' ,8' ,8' ,8' ,8' `9b. .dP' .dP' .dP' .dP' .dP' .dP' `98888888888888888888888888888888888888P' ↑ n turns per unit length
B = μ₀nI inside (uniform and independent of radius).

The Ampère Circuital Law

The line integral of B around a closed path is equal to μ₀ times the current enclosed.

∫ B .d

_l = μ₀ × (current enclosed)

This is still just for a vacuum.

Current passing through area ds is:

dI = J_n dS = J cos α dS = J.dS | _ normal→| α /| J |,-./ ,---|--/-. / _|_/_ \ / / |/ / \ ( /____/←dS ) \ / `--->------' path ∫ B . dl = μ₀ ∫∫ J . dS line surface integral integral around path Field due to a long straight wire. | /|\ I | | | ...''|''... .'' | ''. : |-. h : : +-+-------x : | __: '.. | ../ B '''.....''' | | |

The lines of B are circles, by symmetry. Around a circle of radius h, B is a constant. Therefore:

∫ B . dl = B ∫ dl closed closed = B 2πh

But ∫B.dl - μ₀ × (current enclosed) = μ₀I ∴ B 2πh = μ₀I and B = μ₀I/(2πh) as before.

Field inside a long solenoid. n turns per unit length ,o8888888888888888888888888888888888888o. ,dP' `9b.⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀`9b. ,8' `8. `8. 8' `8. -->-->-->-->-->-->-->-->-->-- `8. oP 'o 'o 8 8 -->-->-->-->-->-->-->-->-->-- `8 O O O 8. 8 -->-->-->-->-->-->-->-->-->-- 8 `o ,P 1 2 ,P 8. ,8 -->- +-->-->-->--+ ->-->-->-- ,8 `8. ,8' | | ,8' `9b. .dP'⨂⨂⨂⨂⨂⨂⨂⨂|⨂⨂⨂⨂⨂⨂⨂⨂⨂⨂⨂|⨂⨂⨂⨂⨂⨂⨂⨂⨂⨂,dP' `98888888888888|88888888888|88888888888P' | | 4 +--<--------+ 3 l path

We perform the line integral in steps.

2 1→2 ∱ B.dl = Bl 1 3 2→3 ∱ B.dl = 0 (B at right angles to dl) 2 4 3→4 ∱ B.dl = 0 (B = 0 outside) 3 1 4→1 ∱ B.dl = 0 (B at right angles to dl) 4

Using the Ampère theorem,

∲ B.dl = Bl

We perform the line integral in steps.

\begin{matrix} 1 \to 2 & \int_{1}^{2} B . ⅆ l = B l \\ 2 \to 3 & \int_{2}^{3} B . ⅆ l = 0 & B at \frac{π}{2} to ⅆ l \\ 3 \to 4 & \int_{3}^{4} B . ⅆ l = 0 & B = 0 outside \end{matrix}

grad	scalar field	∇V
div	vector field	∇.E
curl	vector field	∇×E
∇²	scalar	∇²V