Electric flux and gauss Law
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Electric flux and gauss Law

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Gauss Law and its applications

Gauss Law and its applications

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  • 1. EMFT/U-2/Part-2 (Electric Flux & Gauss Law) 1 BY: NAVEEN KR. DUBEY EC DEPTT. RKGITW By: Naveen kr Dubey/RKGITW/EC Deptt. 2/10/2014
  • 2. Electric Flux 2 Electric flux is the measure of the ―number of field lines passing through a surface S ‖ For uniform :  E Define: Electric Flux ΦE E A Units: N•m /C 2 S  E  A By: Naveen kr Dubey/RKGITW/EC Deptt. A is the surface area perpendicular to , so Φ=EAcos(θ) 2/10/2014  E
  • 3. Notes: 1) 3 is a scalar called electric flux E 2) Units: N•m2/C 3) E represents the ―number of field lines through surface S.‖ 4) For a closed surface, the area vector points in the outward direction. 5) Flux is zero for a surface parallel to the field (normal is at 90o to E) By: Naveen kr Dubey/RKGITW/EC Deptt. 2/10/2014
  • 4. Example:  4 E 1000 N C S2 S1 30° (rectangle, 1m x 2m) Find: flux (rectangle, 1m x 2m) E S3 (hemisphere, radius 1m) through S1, S2, S3. By: Naveen kr Dubey/RKGITW/EC Deptt. 2/10/2014
  • 5. Faraday‘s Experiment: 5  From this experiment ψ= Q------(1) Now we can find surface charge or flux density D|r=a = ---for inner sphere. Q D|r=b = 4 b a ---for outer sphere 2 r By: Naveen kr Dubey/RKGITW/EC Deptt. 2/10/2014
  • 6. 6  D= Q ar 2 4 r --for a<r<b; now, the electric field at any point in free space E= Q 4 0 r 2 ar hence for free space є D= 0 E By: Naveen kr Dubey/RKGITW/EC Deptt. 2/10/2014
  • 7. Gauss‘s Law 7  The Faraday‘s experiment leads to generalized statement known as Gauss Law “ The Electric flux passing through any closed surface (known as Gaussian surface) is equal to total charge enclosed by the surface.” Mathematically: ∆ψ=flux crossing ∆S = Ds ∆S cosθ= Ds. ∆S By: Naveen kr Dubey/RKGITW/EC Deptt. 2/10/2014
  • 8. 8  the total flux passing through surface: Ψ= d Ds.ds the closed surface is known as Gaussian surface Ψ=charge enclosed =Q By: Naveen kr Dubey/RKGITW/EC Deptt. 2/10/2014
  • 9. 9   now we know that D 0 E   “The total electric flux through any closed surface equals the net charge inside that surface divided by E.ds s o” q 0  This relates an electric field to the charge distribution that creates it. By: Naveen kr Dubey/RKGITW/EC Deptt. 2/10/2014
  • 10. 10 By: Naveen kr Dubey/RKGITW/EC Deptt. 2/10/2014
  • 11. Application of Gauss’s Law: 11   By using gauss law we can determine Electric field or Charge density.   But the solution is easy if we are able to choose a close surface, which satisfies two conditions : 1: Ds is everywhere either normal or tangential to the closed surface, so that Ds.ds become either Dsds or Zer0. 2: One portion of the closed surface for which Ds.ds is not zero, Ds=constant By: Naveen kr Dubey/RKGITW/EC Deptt. 2/10/2014
  • 12. 12   Note: “ Gauss’s Law depends upon symmetry, so that if we can not show symmetry exist then we can not use Gauss law”  For Example : Electric field due to line charge distribution: By: Naveen kr Dubey/RKGITW/EC Deptt. 2/10/2014
  • 13.  Let a cylindrical closed surface around the charged line having radius ‗r‘ side  2/10/2014 By: Naveen kr Dubey/RKGITW/EC Deptt. 13
  • 14. 14  Now apply gauss law Q Ds.ds cyl Ds ds 0. ds 0. side top ds bottom Ds.2 rL Q Ds 2 rL Ds Q E 2 r 0L 0 By: Naveen kr Dubey/RKGITW/EC Deptt. 2 0 r 2/10/2014
  • 15. Exercise 4 Gauss   Q E dA L 0 Area of a cylinder 2 rL L E 2 rL 0 E 2 r 0 2/10/2014 By: Naveen kr Dubey/RKGITW/EC Deptt. 15 0
  • 16. Electric field due to Co-axial cable 16  Let us consider a coaxial cable having inner radius a and outer radius b. and the outer surface of inner cylinder is Ps .  Now we are interested to find field inside and out side..  Green surface: r>b Red Surface: a<r<b By: Naveen kr Dubey/RKGITW/EC Deptt. 2/10/2014
  • 17. 17 Energy and Potential By: Naveen kr Dubey/RKGITW/EC Deptt. 2/10/2014
  • 18. Energy Expended in Moving A Point Charge In An Electric Field 18  The Electric field intensity defined as--- “ The force on a unit test charge at that point at which we wish to find the value of this vector field” Now, if we attempt the test charge against the Electric field, we have to exert a force equal and opposite to that exerted by field. And this requires us to expend energy or do work. F applied By: Naveen kr Dubey/RKGITW/EC Deptt. Q FElectric 2/10/2014
  • 19. 19   Now , suppose we wish to move a charge Q a distance dL in Electric field E. E Q  F FE ˆ QEal al = unit vector in dL direction By: Naveen kr Dubey/RKGITW/EC Deptt. 2/10/2014
  • 20. 20 The applied force: Fapplied ˆ QE.al The work done dW=-QE.dL Net work done: W By: Naveen kr Dubey/RKGITW/EC Deptt. Q final initial E.dL 2/10/2014
  • 21. How to perform line integral: 21 --To perform line integral choose a path , then break it up into large number of very small segments, -- multiply the component of the field along each segment by the length of the segments --and add the results for all the segments. For example : “Let an uniform Electric field for the simplicity and break the path into segments as in figure” By: Naveen kr Dubey/RKGITW/EC Deptt. 2/10/2014
  • 22. 22 By: Naveen kr Dubey/RKGITW/EC Deptt. 2/10/2014
  • 23. 23  Now for final w Q E.dL initial Apply the method: w Q( E1 L1 E2 L2 E3 L3 E4 L4 ) For Uniform field E1=E2=E3=E4=E  W QE( L1 W By: Naveen kr Dubey/RKGITW/EC Deptt. L2 L3 L4 ) QE.LBA 2/10/2014
  • 24. 24  Now if there are infinite number of elements the sum can be converted as integral: w QE. dL  ---here dL called differential length  *Note:    dL= dx i +dy j + dz k-- for cartesian coordinates ˆ ˆ ˆ dL d a d a d z a z --- for cylindrical ˆ ˆ ˆ dL d a rd a r sin d a -- for spherical r r By: Naveen kr Dubey/RKGITW/EC Deptt. 2/10/2014
  • 25.  Potential & Potential Difference: 25  Potential at any point P can be defined as.. “Work done to bring the unit charge from infinity to point P” By: Naveen kr Dubey/RKGITW/EC Deptt. 2/10/2014
  • 26. Potential Difference: 26  “Work done by external source in moving a positive unit test charge from one point to another point (in an electric field)”  Work done: B  E WAB q E.dl B q A A Potential difference: B VAB E.dl A By: Naveen kr Dubey/RKGITW/EC Deptt. 2/10/2014
  • 27. 27   Or We Can say VAB = VA-VB : Where VA—Potential at Point A VB –Potential at point B : UNIT - ―Joule/Coulomb‖ or Volt. By: Naveen kr Dubey/RKGITW/EC Deptt. 2/10/2014
  • 28. 28  Potential difference for 1: In electric field due to line charge distribution. 2: In electric field due to point charge at radial points. By: Naveen kr Dubey/RKGITW/EC Deptt. 2/10/2014
  • 29.  Equi potential Surface 29  Va+dl= Va + (dV/dL)dl Let C1 constant rate of change in potential per unit distance., Now, if Va+dl = Va  dV/dL = 0  V= Constant Hence Equipotntial surface can be defined as: “Surface composed of all those points having the same value of potential. No work is involve in moving a unit test charge” By: Naveen kr Dubey/RKGITW/EC Deptt. 2/10/2014
  • 30. Potential Gradient 30   As we Know the relation V E.dl  But we can make this relation much easy using in reverse order:i.e V E. L for short length  Now, we suppose a region in which a vector is making angle θ with field direction:∆L=∆L aL So we will get the relation: V By: Naveen kr Dubey/RKGITW/EC Deptt. E LCos 2/10/2014
  • 31. 31 By: Naveen kr Dubey/RKGITW/EC Deptt. 2/10/2014
  • 32. 32   Now we can convert this relationship in derivative by applying limits: dV dL E cos  if we want to maximize this relationship the we must move in opposite to electric field(Cosθ=-1) dV dL By: Naveen kr Dubey/RKGITW/EC Deptt. E m ax 2/10/2014
  • 33. Here we observed two characteristics of relationship between E&V: 33 1: The magnitude of the electric field intensity is given by the maximum value of the rate of change of potential with distance. 2: the maximum value is obtained when the direction of distance increment is opposite to E. Or we can say “The direction of E is opposite to the direction in which potential is increasing the most rapidly.” By: Naveen kr Dubey/RKGITW/EC Deptt. 2/10/2014
  • 34. 34  Now , if we consider “Equipotential surface” then we will get the relation: ∆V=-E.∆L=0 Here we know that neither E nor ∆L can be zero. So E must be perpendicular direction of incremental vector ∆L. So electric field intensity can be expressed se… E By: Naveen kr Dubey/RKGITW/EC Deptt. dV dL ˆ aN m ax 2/10/2014
  • 35. 35   so E can be expressed as maximum rate of change in voltage with distance and direction is normal to Equipotential surface.  So, here we conclude “(dV/dl)max occure when ∆L is in the direction of   a” N dV dL m ax E By: Naveen kr Dubey/RKGITW/EC Deptt. dV dN dV ˆ aN dN 2/10/2014
  • 36. 36   The operation on V by which E is obtained is known as gradient. E=-grad V Proof: Let V is an unique function of x,y,z and we take its complete diffrential. dV By: Naveen kr Dubey/RKGITW/EC Deptt. V dx x V dy y V dz z 2/10/2014
  • 37. 37  But we have also dV E.dL E x dx E y dy E z dz   Comparing the both equations: Ex Ey Ez By: Naveen kr Dubey/RKGITW/EC Deptt. V x V y V z 2/10/2014
  • 38. 38  then result may be combined as: V ˆ ax x E    E V ˆ ay y V ˆ az z gradV gradV By: Naveen kr Dubey/RKGITW/EC Deptt. V ˆ ax x V ˆ ay y V ˆ az z 2/10/2014
  • 39. 39 Gradient of V in other coordinate system: V V V ˆ a 1 V V 1 V ˆ ˆ ar a r r By: Naveen kr Dubey/RKGITW/EC Deptt. ˆ a 1 r sin V ˆ a z Cylindrical z V ˆ a Spherical 2/10/2014
  • 40. Electric Dipole  Electric dipole consists of a pair of point charges with equal size but opposite sign separated by a distance d. d + 2/10/2014 - By: Naveen kr Dubey/RKGITW/EC Deptt. 40
  • 41. Electric Dipole  Electric dipole consists of a pair of point charges with equal size but opposite sign separated by a distance d. d + p 2/10/2014 - q d By: Naveen kr Dubey/RKGITW/EC Deptt. 41
  • 42. Electric Dipole  Electric dipole consists of a pair of point charges with equal size but opposite sign separated by a distance d. d + -  p 2/10/2014 By: Naveen kr Dubey/RKGITW/EC Deptt. 42
  • 43. Electric Dipole  Water molecules are electric dipoles  pwater 2/10/2014 + + - By: Naveen kr Dubey/RKGITW/EC Deptt. 43
  • 44. potential due to electric dipole 44 By: Naveen kr Dubey/RKGITW/EC Deptt. 2/10/2014
  • 45. 45  Potential at point P V Q 4 0 1 1 R1 R2 Q R2 R1 4 0 R1R2  R2-R1 may be approximated very easily if R1 and R2 are assumed to be parallel,  R2-R1 = d cos θ   The final result is then: V By: Naveen kr Dubey/RKGITW/EC Deptt. Qd cos 4 0r 2 2/10/2014
  • 46. N0w Electric field: 46  By using gradient relationship for spherical coordinates: E V 1 V ˆ ˆ ar a r r V 1 r sin V ˆ a  So , we will get…. Qd cos ˆ ar 3 2 0r E E By: Naveen kr Dubey/RKGITW/EC Deptt. Qd sin ˆ a 3 4 0r Qd ˆ 2 cos ar 3 4 0r ˆ sin a 2/10/2014