1 / 18

2). Gauss’ Law and Applications

F ij. q i. r i -r j. r i. r j. q j. O. 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

weed
Download Presentation

2). Gauss’ Law and Applications

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. 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

  2. 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

  3. 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?

  4. 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

  5. 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

  6. 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

  7. .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)

  8. 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

  9. + + + + + + + + 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??

  10. 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)

  11. 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

  12. 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

  13. 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

  14. 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

  15. 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

  16. Electrostatic energy of charge distribution • For a continuous distribution

  17. 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:

  18. 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

More Related