2). Gauss’ Law and Applications Coulomb’s Law: force on charge i due to charge j is F ij is force on i due to presence of j and acts along line of centres r ij. If q i q j are same sign then repulsive force is in direction shown Inverse square law of force O riri rjrj r i -r j qiqi qjqj F ij

Principle of Superposition Total force on one charge i is i.e. linear superposition of forces due to all other charges Test charge: one which does not influence other ‘real charges’ – samples the electric field, potential Electric field experienced by a test charge q i ar r i is

Electric Field Field lines give local direction of field Field around positive charge directed away from charge Field around negative charge directed towards charge Principle of superposition used for field due to a dipole (+ve –ve charge combination). Which is which? q j +ve q j -ve

Flux of a Vector Field Normal component of vector field transports fluid across element of surface area Define surface area element as dS = da 1 x da 2 Magnitude of normal component of vector field V is V.dS = |V||dS| cos(  ) For current density j flux through surface S is Cm 2 s -1 da1da1 da2da2 dSdS dS = da 1 x da 2 |dS| = |da 1 | |da 2 |sin(  /2)  dS`

Electric field is vector field (c.f. fluid velocity x density) Element of flux of electric field over closed surface E.dS da1da1 da2da2 n   Flux of Electric Field Gauss’ Law Integral Form

Factors of r 2 (area element) and 1/r 2 (inverse square law) cancel in element of flux E.dS E.dS depends only on solid angle d  da1da1 da2da2 n   Integral form of Gauss’ Law Point charges: q i enclosed by S q1q1 q2q2 Charge distribution  (r) enclosed by S

Differential form of Gauss’ Law Integral form Divergence theorem applied to field V, volume v bounded by surface S Divergence theorem applied to electric field E V.n dS .V dv Differential form of Gauss’ Law (Poisson’s Equation)

Apply Gauss’ Law to charge sheet  (C m -3 ) is the 3D charge density, many applications make use of the 2D density  (C m -2 ): Uniform sheet of charge density  Q/A By symmetry, E is perp. to sheet Same everywhere, outwards on both sides Surface: cylinder sides + faces perp. to sheet, end faces of area dA Only end faces contribute to integral E E dA

 ’ = Q/2A surface charge density Cm -2 (c.f. Q/A for sheet) E 2dA =  ’ dA/  o E =  ’/2  o (outside left surface shown) Apply Gauss’ Law to charged plate E dA E = 0 (inside metal plate) why?? + + Outside E =  ’/2  o +  ’/2  o =  ’/  o =  /2  o Inside fields from opposite faces cancel

Work of moving charge in E field F Coulomb =qE Work done on test charge dW dW = F applied.dl = -F Coulomb.dl = -qE.dl = -qEdl cos  dl cos  = dr W is independent of the path (E is conservative field) A B q1q1 q r r1r1 r2r2 E dldl 

Potential energy function Path independence of W leads to potential and potential energy functions Introduce electrostatic potential Work done on going from A to B = electrostatic potential energy difference Zero of potential energy is arbitrary –choose  (r→∞) as zero of energy

Electrostatic potential Work done on test charge moving from A to B when charge q 1 is at the origin Change in potential due to charge q 1 a distance of r B from B

Electric field from electrostatic potential Electric field created by q 1 at r = r B Electric potential created by q 1 at r B Gradient of electric potential Electric field is therefore E= –  

Electrostatic energy of charges In vacuum Potential energy of a pair of point charges Potential energy of a group of point charges Potential energy of a charge distribution In a dielectric (later) Potential energy of free charges

Electrostatic energy of point charges Work to bring charge q 2 to r 2 from ∞ when q 1 is at r 1 W 2 = q 2  2 NB q 2  2 = q 1  1 (Could equally well bring charge q 1 from ∞) Work to bring charge q 3 to r 3 from ∞ when q 1 is at r 1 and q 2 is at r 2 W 3 = q 3  3 Total potential energy of 3 charges = W 2 + W 3 In general O q1q1 q2q2 r1r1 r2r2 r 12 O q1q1 q2q2 r1r1 r2r2 r3r3 r 13 r 23

Electrostatic energy of charge distribution For a continuous distribution

Energy in vacuum in terms of E Gauss’ law relates  to electric field and potential Replace  in energy expression using Gauss’ law Expand integrand using identity: .  F = .F + F.  Exercise: write  =  and F =  to show:

Energy in vacuum in terms of E For pair of point charges, contribution of surface term   1/r   -1/r 2 dA  r 2 overall  -1/r Let r → ∞ and only the volume term is non-zero Energy density