TOPIC : GAUSS;S LAW AND APPLICATION Mahatma Phule Arts, Science And Commerce College, Panvel ( Academic Year 2015 – 2016 ) SEMESTER 5TH T. Y. B. Sc. Physics Paper FOURTH ELECTRODYNAMICS TOPIC : GAUSS;S LAW AND APPLICATION By Dr. Sarode M. T. Department of Physics 05/08/2015

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 O ri rj ri-rj qi qj Fij Dr.M.T.Sarode, dept. of Physics, M.P.A.S.C.College,Panvel

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 Dr.M.T.Sarode, dept. of Physics, M.P.A.S.C.College,Panvel

Dr.M.T.Sarode, dept. of Physics, M.P.A.S.C.College,Panvel Electric Field qj +ve 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? qj -ve Dr.M.T.Sarode, dept. of Physics, M.P.A.S.C.College,Panvel

Dr.M.T.Sarode, dept. of Physics, M.P.A.S.C.College,Panvel 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 da1 da2 dS dS = da1 x da2 |dS| = |da1| |da2|sin(p/2) Y dS` Dr.M.T.Sarode, dept. of Physics, M.P.A.S.C.College,Panvel

Dr.M.T.Sarode, dept. of Physics, M.P.A.S.C.College,Panvel 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 da1 da2 n q f Gauss’ Law Integral Form Dr.M.T.Sarode, dept. of Physics, M.P.A.S.C.College,Panvel

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 da1 da2 n q f Point charges: qi enclosed by S q1 q2 Charge distribution r(r) enclosed by S Dr.M.T.Sarode, dept. of Physics, M.P.A.S.C.College,Panvel

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) Dr.M.T.Sarode, dept. of Physics, M.P.A.S.C.College,Panvel

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 dA Dr.M.T.Sarode, dept. of Physics, M.P.A.S.C.College,Panvel

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 dA E = 0 (inside metal plate) why?? + + + + Outside E = ’/2o + ’/2o = ’/o = /2o Inside fields from opposite faces cancel Dr.M.T.Sarode, dept. of Physics, M.P.A.S.C.College,Panvel

Work of moving charge in E field 04/09/2013 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) A B q1 q r r1 r2 E dl Dr.M.T.Sarode, dept. of Physics, M.P.A.S.C.College,Panvel

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 Dr.M.T.Sarode, dept. of Physics, M.P.A.S.C.College,Panvel

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 Dr.M.T.Sarode, dept. of Physics, M.P.A.S.C.College,Panvel

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 Dr.M.T.Sarode, dept. of Physics, M.P.A.S.C.College,Panvel

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 Dr.M.T.Sarode, dept. of Physics, M.P.A.S.C.College,Panvel

Electrostatic energy of point charges 04/09/2013 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 O q1 q2 r1 r2 r12 O q1 q2 r1 r2 r12 r3 r13 r23 Dr.M.T.Sarode, dept. of Physics, M.P.A.S.C.College,Panvel

Electrostatic energy of charge distribution For a continuous distribution Dr.M.T.Sarode, dept. of Physics, M.P.A.S.C.College,Panvel

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: Dr.M.T.Sarode, dept. of Physics, M.P.A.S.C.College,Panvel

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 Dr.M.T.Sarode, dept. of Physics, M.P.A.S.C.College,Panvel

Dr.M.T.Sarode, dept. of Physics, M.P.A.S.C.College,Panvel THANK YOU Dr.M.T.Sarode, dept. of Physics, M.P.A.S.C.College,Panvel