o Aim of the lecture Gauss’ Law:

Flux More mathematical descriptions Generalised field distributions

o  Main learning outcomes familiarity with

Electric fields, potentials Coulomb’s and Gauss’ Laws Calculation of fields and forces

Lecture 3

+ -

+ -F F

Reminder:

o Objects have a property called ‘charge’Basic property like mass

o Charge is quantised – comes in units of 1.6 x 10-19CCharge comes in two types, which are called

PositiveNegative

o The total charge in any closed system cannot be changedo Charges interact with each other, causing

Repulsive force if charges are the same (++ or --)Attractive forces if charges are opposite (+-)

o The forces are equal and in the opposite directiono The form of the force law is: F = keq1q2/r2

o The nature of the interaction between charge is described using An electric field On a diagram the field is represented by lines

emerge from a source (positive charge) end in a sink (negative charge)

The density of field lines represents the field strength but caution needed – this is fully accurate in 3-D, but in a drawing in 2-D the line density cannot always be exact this is only a problem drawing NOT a problem with the theory

Reminder:

Exact, density offield lines falls offlike 1/r2

Only a representation,Density is falling like 1/r -wrong

o The interaction between charges can be described as an interaction of one of the charges with the field of the other The force law is: F = q1E where E is the electric field the charge q1 is in.

eg E = keq2/r2 where r is the distance from q2

o The electric field is conservative so there is an energy associated with position in the field just like gravity

o Therefore there is a potential associated with position in the field The potential is called the electric potential, measured in volts Equipotentials are drawn on diagrams,

these are accurate representations of 3-D system

Reminder:

Reminder: Gauss’ Law

o Idea behind Gauss’ Law already introduced The field lines come from a charge, so no extra lines appear away from it and none disappear the number of lines is a constant

total number of lines through any closed surface surrounding a charge must be constant

For a single point charge, the number oflines passing through a sphere surrounding itcannot depend on the radius of the sphere.Or in fact on the shape of the surrounding surface

That’s Gauss’ Law

The correct expressions are vectors:

F = qE where F is the force on charge q in field E E = ke q/r2 r is the field created by a point charge q Where r is a vector pointing away from + charges the potential is a scalar field, there is no direction.

Reminder:

Units: F is in Newtons E is in Volts/metre (or Newtons/Coulomb) is in Volts (or Joules/Coulomb)

Note is also often denoted e or V

To define Gauss’ Law more precisely (and accurately!)

need concept of a Flux

o E is a vector fieldo Another example of a vector field iso V where this is the velocity vector for airflow at any point in a volume

Consider an open door withair blowing in and out fromboth sides

o At every point in the door there is a net velocity

magnitude and a direction, ie V – a vector field

oThe net volume of air flowing through the door is the sum of all these vectors

it is called the flux in this case an air volume flux

it is a vector sum it depends on the direction as well as the magnitude

oThe net volume of air flowing through the door is the sum of all these vectors

it is called the flux in this case an air volume flux

it is a vector sum it depends on the direction as well as the magnitude

Air volume/second = ∫∫dxdy V.a

where a is a unit vector perpendicular to the door The quantity ∫∫dxdyV.a is the volume flux of air, we usually write:

= ∫s V.da where ∫s means integration over the surface

= V.da

∫ This symbol means an integral over a closed surface

da means that the integral is being performed as a vector dot product with the local surface direction

∫

is a volume flux it is a measure of how much air is flowing through the door.

For Gauss’ Law we will use

Gauss’ Law

To state it properly, need to define Electric Flux, dthrough a small area dA as d= E.dA where the direction of A is perpendicular to the surface dA

So d = E.dAAnd Gauss’ Law is:

q = 0 E.dA∫Where q is the charge inside the surface

= 0 = q/0 = - q/0

q = 0 E.dA∫Where q is the charge inside the surface, so

d = E.dA

So the total flux through a surface is = ∫sd∫E.dA

but

Evaluate for A dipole configurationThrough a sphere

q = 0 E.dA∫

Gauss’ Law allows cute proofs

eg All excess charge on a conductor is on its surface

o If there was an electric field inside A then free charges would move.oThis means flux through surface is everywhere zero

o because A can be moved arbitrarily close to the conductor surface.

o No flux means no charge inside

Concentric spheres

Around a wire

Planar non-conducting sheet

Electric PotentialPotential Energy associated with Electric Field

Consider gravity

Potential Energy, U

U = mgh

Remember where this comes from:

F = Gmemt/r2

The work done, WD is force times distance

WD = ∫Fdr

= ∫dr Gmtme/r2

re+h

re

= Gmt(1/(re+h)-1/re)me

≈ mt{Gme/re}h

= mtgh

The formula arises because the strength of the gravitational field isapproximately constant for ‘small’ changes in height above earth

Potential Energy, U U = mgh

The equipotential surfaces for the gravitational field above the earth are spheres, which are just lines in this 2-D picture

Potential Energy, U U = mgh

The equipotential surfaces for the gravitational field above the earth are spheres, which are just lines in this 2-D picture

Each line can be given a value = gh

0

2J/kg

4J/kg

6J/kg

8J/kg

10J/kg

12J/kg

14J/kg

16J/kg

18J/kg

These are just the height multiplied by a constantThese are the values of the gravitational potential

When you multiply the potential by the mass you getthe potential energy

+ Voltage = V0

Voltage = 0

Between two metal plateswhich have a voltage V across them,There are equipotential lines,each of which has a voltage

+ A charge has a potential energy U = qV

just like gravity, except that the mass is replaced by q and gh is replaced by the voltage

+ Voltage = V0

Voltage = 0

+D

h

So the potential Energy,U

U = q {V0/D} h

compare with gravity U = m {Gme/re} h