2). Gauss’ Law and Applications

Fij qi ri-rj ri rj qj O 2). Gauss’ Law and Applications • Coulomb’s Law: force on charge i due to charge j is • Fij is force on i due to presence of j and acts along line of centres rij. If qi qj are same sign then repulsive force is in direction shown • Inverse square law of force

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 qi ar ri is

qj -ve qj +ve 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?

Y dS` da2 dS da1 dS = da1 x da2 |dS| =|da1| |da2|sin(p/2) Flux of a Vector Field • Normal component of vector field transports fluid across element of surface area • Define surface area element as dS = da1 x da2 • Magnitude of normal component of vector field V is V.dS = |V||dS| cos(Y) • For current density j flux through surface S is Cm2s-1

n da2 da1 q f Flux of Electric Field • Electric field is vector field (c.f. fluid velocity x density) • Element of flux of electric field over closed surface E.dS Gauss’ Law Integral Form

n da2 da1 q f Integral form of Gauss’ Law • Factors of r2 (area element) and 1/r2 (inverse square law) cancel in element of flux E.dS • E.dS depends only on solid angle dW Point charges: qienclosed by S q1 q2 Charge distribution r(r) enclosed by S

.V dv V.n dS 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 Differential form of Gauss’ Law (Poisson’s Equation)

dA E + + + + + + + + + + + + + + + + + + + + + + + + E Apply Gauss’ Law to charge sheet • r (C m-3) is the 3D charge density, many applications make use of the 2D density s (C m-2): • Uniform sheet of charge density s = 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 + + + + + + + + + + + + + + + + + + Outside E = ’/2o + ’/2o = ’/o = /2o Inside fields from opposite faces cancel + + + + + + dA Apply Gauss’ Law to charged plate • s’ = Q/2A surface charge density Cm-2 (c.f. Q/A for sheet) • E 2dA = ’ dA/o • E = ’/2o (outside left surface shown) E = 0 (inside metal plate) why??

B E r2 q q dl A r r1 q1 Work of moving charge in E field • FCoulomb=qE • Work done on test charge dW • dW = Fapplied.dl = -FCoulomb.dl = -qE.dl = -qEdl cos q • dl cos q = dr • W is independent of the path (E is conservative field)

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 f(r→∞) as zero of energy

Electrostatic potential • Work done on test charge moving from A to B when charge q1 is at the origin • Change in potential due to charge q1 a distance of rB from B

Electric field from electrostatic potential • Electric field created by q1 at r = rB • Electric potential created by q1 at rB • Gradient of electric potential • Electric field is therefore E= –f

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

q1 q2 r12 r13 r23 r1 r2 r3 O q1 q2 r12 r1 r2 O Electrostatic energy of point charges • Work to bring charge q2 to r2 from ∞ when q1 is at r1 W2 = q2 f2 • NB q2 f2 =q1 f1 (Could equally well bring charge q1 from ∞) • Work to bring charge q3 to r3 from ∞ when q1 is at r1 and q2 is at r2 W3 = q3 f3 • Total potential energy of 3 charges = W2 + W3 • In general

Electrostatic energy of charge distribution • For a continuous distribution

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

Energy in vacuum in terms of E For pair of point charges, contribution of surface term  1/r  -1/r2 dA  r2 overall  -1/r Let r →∞ and only the volume term is non-zero Energy density

2). Gauss’ Law and Applications