Electrostatics Review

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Electrostatics Review

• PHYS 2326-8

2
Test Coverage

• Primarily on what topics covered in class since
  last test
• Can include questions or parts based upon prior
  classes or courses
• Responsible for the material in the chapters
  covered
• I will put up Constants if asked

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Mechanics

• Calculators allowed
• 1 - 3x5 index card for formulas
• No cell phone or internet devices permitted
  individual efforts only!
• Print clearly and concisely no credit is given
  for right answers that cannot be verified as
  being correct
• Include and keep track of units
• Box in final answers

4
Grading

• Show problem setup, including drawings where
  useful
• Show equations to be used in solution
• Show intermediate steps to solution
• Show Solutions with intermediate values
• This is the least important portion of the
  grading

5
Testing

• Best 4 of the 5 test scores account for 60 of
  the course grade
• Worst one is dropped or test 5 can be skipped
• Test scores exceed 100 points if the top 4 test
  scores average over 100 points, the final can be
  exempted
• Final exam is worth 30 of the course grade

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Test Tips

• Show the problem set up graphically if possible
• Show equations to use
• Provide intermediate steps to solution
• Show solution and intermediate values
• Carefully create your 3x5 index card
• Study book appendix B for math assistance/review

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Useful Constants

• Avogadro 6.022x1023 particles/mol
• me 9.11x10-31kg
• mp 1.67x10-19kg
• e- 1.60x10-27 C
• eo 8.85x10-12 C2/Nm2
• G 6.67x10-11 Nm2/kg2
• ke 9.00x109 Nm2/C2

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Constants

• You need to know what those constants are and how
  and where they are used
• If you cannot read the previous slide, you should
  have those constants on your note card.
• If you dont know where and how those constants
  are used, you should learn before the test, not
  after the test.

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Review

• First Section
• Electrostatics

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Coulombs Law Example

• A hydrogen atom consists of a proton and an
  electron. What is the gravitational force
  between them if they are separated by the Bohr
  radius of 0.5292 x 10-10 m?
• F G m1 m2 / r2
• G 6.67 x 10-11 N m2/kg2
• m1 1.67 x 10-27 kg mass of proton
• m2 9.11 x 10-31 kg mass of electron
• F (6.67 x 10-11 ) (1.67 x 10-27 ) (9.11 x 10-31
  ) / (0.5292 x 10-10 )2
• F 3.623 x 10-47 N

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Coulombs Law Example

• What is the electric field force for the proton
  and electron separated by 0.5292 x 10-10 m,
  electron charge 1.60 x 10 -19 C
• F k q1 q2 / r2 N
• F(9.00 x 109)(1.60 x 10 -19 )(-1.60 x 10 -19
  )/(0.5292 x 10-10 )
• F -8.227 x 10-10 N
• Attractive or repulsive force?
• Opposites attract

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Coulombs Law Example

• How strong is the electric force compared to the
  gravitational force?
• Take a ratio R Fe/Fg
• R 8.227 x 10-10 N/ 3.623 x 10-47 N
• 2.27 x 1039

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Flux

• Electric field lines are tangent to the electric
  field vector at each point along the line
• Lines/unit area are proportional to the electric
  field
• For a uniform electric field perpendicular to a
  rectangular surface, FE the electric flux is FE
  EA (Nm2/C)
• Flux is proportional to the number of electric
  field lines penetrating a surface

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Flux General Case
15
Gausss Law
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Concepts

• Closed surface one that totally surrounds a
  region
• The net number of lines means the total leaving
  the surface minus the total entering the surface
• If more leave than enter, net flux is positive
• If less, net flux is negative
• Electric fields begin on positive charges and end
  on negative charges

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Gaussian SurfacesApplying Gausss Law

• Useful for symmetric charge distributions
• Where the value of the E-field can be argued by
  symmetry to be constant over a portion of the
  surface
• The Dot Product becomes E dA because the vectors
  are parallel
• The Dot Product is 0 because theyre
  perpendicular
• The E-field is zero over a portion of the surface

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Conductors in Equilibrium

• Has the following properties
• E field is 0 inside the conductor
• If an isolated conductor carries a charge, it
  resides on the surface
• Just outside the conductor, the E-field is
  perpendicular to the surface and is s/e
• Irregular shaped conductors have greater charge
  densities at points with smaller radius of
  curvatures