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Electrostatic Formula

Electrostatic Formula. Electric Force. F = 1 q 1 q 2 = k q 1 q 2 4 p e o r 2 r 2 k = 9.0x10 9 N m 2 = 1 = -------1------------ C 2 4 p e o 4 p 8.85x10 -12 C 2 N m 2.

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Electrostatic Formula

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  1. Electrostatic Formula

  2. Electric Force • F = 1 q1q2 = k q1q2 4 p eo r2 r2 k = 9.0x10 9N m2 = 1 = -------1------------ C2 4 p eo 4 p 8.85x10-12C2 N m2

  3. Electric Force • F = 1 q1q2 = k q1q2 4 p eor2 r2 k = 9.0x10 9N m2 = 1 = -------1------------ C2 4 p eo 4 p 8.85x10-12C2 N m2

  4. Electric Force • F = 1 q1q2 = k q1q2 4 p eor2 r2 k = 9.0x10 9N m2 = 1 = -------1------------ C2 4 p eo 4 p 8.85x10-12C2 N m2

  5. Electric Force • F = 1 q1q2 = k q1q2 4 p eor2 r2 k = 9.0x10 9N m2= 1 =-------1------------ C24 p eo4 p 8.85x10-12C2 N m2

  6. Electric Force • F = 1 q1q2 = k q1q2 4 p eor2 r2 k = 9.0x10 9N m2= 1 =-------1------------ C24 p eo4 p 8.85x10-12C2 N m2 Vector sum – x components, y components, sum of x components, sum of y components, pythagorean theorem, and inv tan of (sum of y )/(sum of x )

  7. Electric Force • F = 1 q1q2 = k q1q2 4 p eor2 r2 k = 9.0x10 9N m2= 1 =-------1------------ C24 p eo4 p 8.85x10-12C2 N m2 Vector sum – x components, y components, sum of x components, sum of y components, pythagorean theorem, and inv tan of (sum of y )/(sum of x )

  8. Electric Field associated with point charges • E = F q F = kqq therefore E = Kqq therefore E =kq r2 r2q r2 Vector sum – x components, y components, sum of x components, sum of y components, pythagorean theorem, and inv tan of (sum of y )/(sum of x )

  9. Electric Field associated with point charges • E = F q F = kqq therefore E = Kqq therefore E =kq r2 r2q r2 Vector sum – x components, y components, sum of x components, sum of y components, pythagorean theorem, and inv tan of (sum of y )/(sum of x )

  10. Electric Field associated with point charges • E = F q F = kqq therefore E = Kqq therefore E =kq r2 r2q r2 Vector sum – x components, y components, sum of x components, sum of y components, pythagorean theorem, and inv tan of (sum of y )/(sum of x )

  11. Electric Field associated with point charges • E = F q F = kqqtherefore E = Kqq therefore E =kq r2 r2q r2 Vector sum – x components, y components, sum of x components, sum of y components, pythagorean theorem, and inv tan of (sum of y )/(sum of x )

  12. Electric Field associated with point charges • E = F q F = kqqtherefore E = Kqqtherefore E =kq r2 r2q r2 Vector sum – x components, y components, sum of x components, sum of y components, pythagorean theorem, and inv tan of (sum of y )/(sum of x )

  13. Electric Field associated with point charges • E = F q F = kqqtherefore E = Kqqtherefore E =kq r2 r2q r2 Vector sum – x components, y components, sum of x components, sum of y components, pythagorean theorem, and inv tan of (sum of y )/(sum of x )

  14. Electric Field Associated with Parallel Plates • The electric field is uniform between plates • Work is required to move +q from the – to + W=Fd W=qEd W=qKqd W=KqqW = Kq W =V=Joules= Electric Potential • d2 d q d q Coulomb • W=qEd W = Ed V = Ed V = E • q d

  15. Electric Field Associated with Parallel Plates • The electric field is uniform between plates • Work is required to move +q from the – to + W=Fd W=qEd W=qKqd W=KqqW = Kq W =V=Joules= Electric Potential • d2 d q d q Coulomb • W=qEd W = Ed V = Ed V = E • q d

  16. Electric Field Associated with Parallel Plates • The electric field is uniform between plates • Work is required to move +q from the – to + W=Fd W=qEd W=qKqd W=KqqW = Kq W =V=Joules= Electric Potential • d2 d q d q Coulomb • W=qEd W = Ed V = Ed V = E • q d

  17. Electric Field Associated with Parallel Plates • The electric field is uniform between plates • Work is required to move +q from the – to + W=Fd W=qEd W=qKqd W=KqqW = Kq W =V=Joules= Electric Potential • d2 d q d q Coulomb • W=qEd W = Ed V = Ed V = E • q d

  18. Electric Field Associated with Parallel Plates • The electric field is uniform between plates • Work is required to move +q from the – to + W=Fd W=qEd W=qKqd W=KqqW = Kq W =V=Joules= Electric Potential • d2 d q d q Coulomb • W=qEd W = Ed V = Ed V = E • q d

  19. Electric Field Associated with Parallel Plates • The electric field is uniform between plates • Work is required to move +q from the – to + W=Fd W=qEd W=qKqd W=KqqW = Kq W =V=Joules= Electric Potential • d2 d q d q Coulomb • W=qEd W = Ed V = Ed V = E • q d

  20. Electric Field Associated with Parallel Plates • The electric field is uniform between plates • Work is required to move +q from the – to + W=Fd W=qEd W=qKqd W=KqqW = Kq W =V=Joules= Electric Potential • d2 d q d q Coulomb • W=qEd W = Ed V = Ed V = E • q d

  21. Electric Field Associated with Parallel Plates • The electric field is uniform between plates • Work is required to move +q from the – to + W=Fd W=qEd W=qKqd W=KqqW = Kq W =V=Joules= Electric Potential • d2 d q d q Coulomb • W=qEd W = Ed V = Ed V = E • q d

  22. Electric Field Associated with Parallel Plates • The electric field is uniform between plates • Work is required to move +q from the – to + W=Fd W=qEdW=qKqd W=KqqW = Kq W =V=Joules= Electric Potential • d2d q d q Coulomb • W=qEd W = Ed V = Ed V = E • q d

  23. Electric Field Associated with Parallel Plates • The electric field is uniform between plates • Work is required to move +q from the – to + W=Fd W=qEdW=qKqd W=KqqW = Kq W =V=Joules= Electric Potential • d2d q d q Coulomb • W=qEd W = Ed V = Ed V = E • q d

  24. Electric Field Associated with Parallel Plates • The electric field is uniform between plates • Work is required to move +q from the – to + W=Fd W=qEdW=qKqd W=KqqW= Kq W =V=Joules= Electric Potential • d2d q d q Coulomb • W=qEd W = Ed V = Ed V = E • q d

  25. Electric Field Associated with Parallel Plates • The electric field is uniform between plates • Work is required to move +q from the – to + W=Fd W=qEdW=qKqd W=KqqW = Kq W =V=Joules= Electric Potential • d2d q d q Coulomb • W=qEd W = Ed V = Ed V = E • q d

  26. Electric Field Associated with Parallel Plates • The electric field is uniform between plates • Work is required to move +q from the – to + W=Fd W=qEdW=qKqd W=KqqW = KqW =V=Joules= Electric Potential • d2d q d q Coulomb • W=qEd W = Ed V = Ed V = E • q d

  27. Electric Field Associated with Parallel Plates • The electric field is uniform between plates • Work is required to move +q from the – to + W=Fd W=qEdW=qKqd W=KqqW = KqW =V=Joules= Electric Potential • d2d q d q Coulomb • W=qEd W = Ed V = Ed V = E • q d

  28. Electric Field Associated with Parallel Plates • The electric field is uniform between plates • Work is required to move +q from the – to + W=Fd W=qEdW=qKqd W=KqqW = KqW =V=Joules= Electric Potential • d2d q d q Coulomb • W=qEd W = Ed V = Ed V = E • q d

  29. Electric Field Associated with Parallel Plates • The electric field is uniform between plates • Work is required to move +q from the – to + W=Fd W=qEdW=qKqd W=KqqW = KqW =V=Joules= Electric Potential • d2d q d q Coulomb • W=qEd W = Ed V = Ed V = E • q d

  30. Electric Field Associated with Parallel Plates • The electric field is uniform between plates • Work is required to move +q from the – to + W=Fd W=qEdW=qKqd W=KqqW = KqW =V=Joules= Electric Potential • d2d q d q Coulomb • W=qEd W = Ed V = Ed V = E • q d

  31. Electric Field Associated with Parallel Plates • The electric field is uniform between plates • Work is required to move +q from the – to + W=Fd W=qEdW=qKqd W=KqqW = KqW =V=Joules= Electric Potential • d2d q d q Coulomb • W=qEd W = EdV = EdV = E • q d

  32. Electric Field Associated with Parallel Plates • The electric field is uniform between plates • Work is required to move +q from the – to + W=Fd W=qEdW=qKqd W=KqqW = KqW =V=Joules= Electric Potential • d2d q d q Coulomb • W=qEd W = EdV = EdV = E • q d

  33. Electric Field Associated with Parallel Plates • The electric field is uniform between plates • Work is required to move +q from the – to + W=Fd W=qEdW=qKqd W=KqqW = KqW =V=Joules= Electric Potential • d2d q d q Coulomb • W=qEd W = EdV = EdV = E = Volts • q d meter

  34. Capacitance • C = Q • V • C = KAeo • d • K= dielectric constant • A = area in m2 • e0= 8.85x10-12C2 • N m2 • d = distance between plates in m

  35. Capacitance • C = Q • V • C = KAeo • d • K= dielectric constant • A = area in m2 • e0= 8.85x10-12C2 • N m2 • d = distance between plates in m

  36. Capacitance • C = Q • V • C = KAeo • d • K= dielectric constant • A = area in m2 • e0= 8.85x10-12C2 • N m2 • d = distance between plates in m

  37. Capacitance • C = Q • V • C = KAeo • d • K= dielectric constant • A = area in m2 • e0= 8.85x10-12C2 • N m2 • d = distance between plates in m

  38. Capacitance • C = Q • V • C = KAeo • d • K= dielectric constant • A = area in m2 • e0= 8.85x10-12C2 • N m2 • d = distance between plates in m

  39. Capacitance • C = Q • V • C = KAeo • d • K= dielectric constant • A = area in m2 • e0= 8.85x10-12C2 • N m2 • d = distance between plates in m

  40. Capacitance • C = Q • V • C = KAeo • d • K= dielectric constant • A = area in m2 • e0= 8.85x10-12C2 • N m2 • d = distance between plates in m

  41. Capacitance • W = QV = U ( Potential energy stored ) • For a capacitor • U = ½ QV since it is easier to add charge at first and progressively gets more difficult • C = Q CV = Q therefore U =1/2 QV= ½ CV2 V C = Q V = Q therefore U = ½ QV = ½ Q2 V C C

  42. Capacitance • W = QV= U ( Potential energy stored ) • For a capacitor • U = ½ QV since it is easier to add charge at first and progressively gets more difficult • C = Q CV = Q therefore U =1/2 QV= ½ CV2 V C = Q V = Q therefore U = ½ QV = ½ Q2 V C C

  43. Capacitance • W = QV= U( Potential energy stored ) • For a capacitor • U = ½ QV since it is easier to add charge at first and progressively gets more difficult • C = Q CV = Q therefore U =1/2 QV= ½ CV2 V C = Q V = Q therefore U = ½ QV = ½ Q2 V C C

  44. Capacitance • W = QV= U( Potential energy stored ) • For a capacitor • U = ½ QV since it is easier to add charge at first and progressively gets more difficult • C = Q CV = Q therefore U =1/2 QV= ½ CV2 V C = Q V = Q therefore U = ½ QV = ½ Q2 V C C

  45. Capacitance • W = QV= U( Potential energy stored ) • For a capacitor • U = ½ QV since it is easier to add charge at first and progressively gets more difficult • C = Q CV = Q therefore U =1/2 QV= ½ CV2 V C = Q V = Q therefore U = ½ QV = ½ Q2 V C C

  46. Capacitance • W = QV= U( Potential energy stored ) • For a capacitor • U = ½ QV since it is easier to add charge at first and progressively gets more difficult • C = Q CV = Q therefore U =1/2 QV= ½ CV2 V C = Q V = Q therefore U = ½ QV = ½ Q2 V C C

  47. Capacitance • W = QV= U( Potential energy stored ) • For a capacitor • U = ½ QV since it is easier to add charge at first and progressively gets more difficult • C = QCV = Q therefore U =1/2 QV= ½ CV2 V C = Q V = Q therefore U = ½ QV = ½ Q2 V C C

  48. Capacitance • W = QV= U( Potential energy stored ) • For a capacitor • U = ½ QV since it is easier to add charge at first and progressively gets more difficult • C = QCV = Qtherefore U =1/2 QV= ½ CV2 V C = Q V = Q therefore U = ½ QV = ½ Q2 V C C

  49. Capacitance • W = QV= U( Potential energy stored ) • For a capacitor • U = ½ QV since it is easier to add charge at first and progressively gets more difficult • C = QCV = Qtherefore U =1/2 QV= ½ CV2 V C = Q V = Q therefore U = ½ QV = ½ Q2 V C C

  50. Capacitance • W = QV= U( Potential energy stored ) • For a capacitor • U = ½ QV since it is easier to add charge at first and progressively gets more difficult • C = QCV = Qtherefore U =1/2 QV= ½ CV2 V C = Q V = Q therefore U = ½ QV = ½ Q2 V C C

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