• Electrostatics - Chapters 22, 23 and 24
• Circuits - Chapters 25 and 26
• Magnetism - Chapters 28 and 29
• Induction and E.M. Waves - Chapters 30 and 32

• Introduction Lecture
• Problem Solving Lecture
• Workshop
• Follow-up Lecture
• Seminar

Full instructions are in the Question Booklet

• Gravity - matter always attracts
• Electromagnetic - holds atoms together
• Strong nuclear - binds atomic nucleus together
• Weak nuclear - allows nuclear reactions

Last two are very short-range (sub-atomic £ 10^-15 m)

• Charges exert a force proportional to separation
• Like charges repel
• Unlike charges attract

Ratio charge/mass » 1000× that of smallest ion
(now know m_p/m_e = 1836)

1889 - Measured charge of electron and hence mass

Name Electron due to Johnstone Stoney (1883)

• Coulomb's Law, Electric Field (Chapter 22)

  • Electric Charge, Insulators, Conductors
  • Coulomb's Law
  • Electric Field
  • Charged Particles in Uniform Electric Fields
  • Electric Dipole
• Gauss's Law and Electric Field (Chapter 23)

  • Electric Field from Coulomb's Law
  • Gauss's Law
  • Electric Field from Gauss's Law
  • Electrostatic Properties of a Conductor
• Electric Potential (Chapter 24)

  • Electric Potential, Potential Difference
  • Calculating Electric Potential
  • Equipotential Surfaces

Law named after Charles Augustin Coulomb (1736 - 1806)

Other work by Joseph Priestly, Henry Cavendish, and
James Clerk Maxwell

Describes the electric force between two point charges
Q₁ and Q₂:

The force exerted on Q₁ by Q₂ is:

F12 =  1 4pe0  Q1 Q2 r122   ^ r

where [^(r)] is a unit vector directed from Q₁ to Q₂ along the
line joining them

e₀ is the permittivity of free space
(8.85×10^-12 C² N^-1 m^-2)

Tipler uses k = [ 1/(4pe₀)] as the ``Coulomb Constant''.

N.B. Force repulsive if charges of same sign

The Coulomb is the charge carried past a point in circuit
by 1 Amp flowing for 1 second

Electron only carries charge of 1.6×10^-19 C.

For two 1 C charges 1 m apart, force = 9×10⁹ N

FE.M. FG =  1 G 4pe0  e2 mp me = 2 ×1039

m_e = 9.1 ×10^-31 kg;
m_p = 1.7 ×10^-27 kg;
G = 6.7 ×10^-11 N m² kg^-2

Why does gravity dominate the Universe?

Q) What is the force binding a crystal of Salt?

Typical distance d between ions ?

~ 1Å (10^-10 m)

Force between postive/negative ion » 2×10^-8 N

How many bonds in one square meter ?

About 1/d² » 10²⁰ bonds

Force to break them is 10²⁰ ×2 ×10^-8

= 2 ×10¹² N

Oversimplified, but clearly crystals can be very strong!

Field: a quantity that can be associated with a position
- vector (e.g. electric force, gravity)
- scalar (e.g. electric potential, temperature)

The electric field E at point P due to a charge Q is the
electric force exerted by that charge on a test particle
divided by the charge q₀ on the test particle:

E =  F q0 =  1 4pe0  Q r2   ^ r

Test charge q₀ assumed small - does not disturb E field.

For a distribution of charges Q₁, Q₂, ... Q_i use the
principle of superposition to get F and/or E

E =  1 4pe0 å i   Qi ri2   ^ ri

E points away from a positive charge

E points towards a negative charge

Use this principle when drawing electric field lines

The net electric flux F_net through any closed surface S is equal to the net charge enclosed by the surface (Q_inside) divided by e₀:

Fnet =  Qinside e0 = 4pk Qinside

A Gaussian surface is any closed surface over which the flux is evaluated. Use whatever surface is easiest.

Similar to Coulomb's Law, but can also be applied when electric field is due to a varying magnetic field

Fnet =  Qinside e0 = ó (ç) õ S  E · ^ n   dA = ó (ç) õ S  En dA

The net flux F_net of a uniform electric field E through surface area A is the dot product of E and the unit vector normal to the surface ([^n]) integrated over the surface.

The dot (or scalar) product of two vectors A and B is A.B = ABcosq, where q is the angle between them

Consider a spherical surface of radius r (the Gaussian surface) surrounding a point charge Q. The electric field E is uniform in all directions and directed radially (i.e. it is normal to the surface).

Using Gauss's Law:

Fnet = ó (ç) õ S  En dA = E ó (ç) õ S  dA = E(4pr2)

The total charge enclosed is just Q_inside, thus

E(4pr2) =  Qinside e0       or       E =  Qinside 4pe0 r2

Using Coulomb's Law:

The charge Q_inside exerts a force on a test-particle q₀, and hence

E =  F q0       thus       E =  Qinside 4pe0 r2   ^ r

For static charge distributions can use either law

Conductors are a special case - Section 23-5

• E is normal to the surface of a conductor
• E is zero inside a conductor

The electric potential energy U(r) of a test-particle of charge q₀ at distance r from a point charge is the work done against the electric force when moving the test-particle from infinity to distance r from the point charge

U(r) = - ó õ r ¥  F ·dl = -q0 ó õ r ¥  E ·dl

We define U(r) to be zero when r=¥

Example of a line integral

Electric force is a conservative force - work done is independent of path (zero for a closed path)

The electric potential V is the electric potential energy U of a test-particle at that point divided by its charge q₀

V =  U q0

Unit: Volt, after Count Alessandro Volta (1745-1827)

V(r) = - ó õ r ¥  E ·dl

You can also derive V from E using the gradient of the E field (read section 24-3).