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 10-19C  Charge comes in two types, which are called Positive Negative o The total charge in any closed system cannot be changed o 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 direction o The form of the force law is: F = k e q 1 q 2 /r 2 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 of field lines falls off like 1/r 2 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 = q 1 E  where E is the electric field the charge q 1 is in. eg E = k e q 2 /r 2 where r is the distance from q 2 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 of lines passing through a sphere surrounding it cannot 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 = k e q/r 2 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 field o Another example of a vector field is o V where this is the velocity vector for airflow at any point in a volume Consider an open door with air blowing in and out from both 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, d  through a small area dA as d  = E.dA where the direction of A is perpendicular to the surface dA So d  = E.dA And 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  = ∫ s d  ∫E.dA but Evaluate  for A dipole configuration Through 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 Potential Potential Energy associated with Electric Field Consider gravity Potential Energy, U U = mgh

Remember where this comes from: F = Gm e m t /r 2 The work done, WD is force times distance WD = ∫Fdr = ∫dr Gm t m e /r 2 r e +h rere = Gm t (1/(r e +h)-1/r e )m e ≈ m t {Gm e /r e }h = m t gh The formula arises because the strength of the gravitational field is approximately 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 constant These are the values of the gravitational potential When you multiply the potential by the mass you get the potential energy

+ Voltage = V 0 Voltage = 0 Between two metal plates which 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 = V 0 Voltage = 0 + D h So the potential Energy,U U = q {V 0 /D} h compare with gravity U = m {Gm e /r e } h