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Gauss’s Law

Gauss’s Law. PHYS 2326-6. Review 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

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Gauss’s Law

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  1. Gauss’s Law PHYS 2326-6

  2. Review 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, ΦEthe electric flux is ΦE = EA (Nm^2/C) • Flux is proportional to the number of electric field lines penetrating a surface

  3. Non Perpendicular Surface • Taking a square that is not perpendicular to the surface, the relationship becomes ΦE = EAcosθ (Nm^2/C) where θ is the angle between the surfaces (normals to the surfaces)

  4. General Case

  5. Observations • 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

  6. General Equation

  7. Vector Area dA

  8. Vector Dot Product • E·dA vector dot product between the differential surface normal and Electric field vector. • Magnitude = E dA cos θ • If θ = 0 °, Magnitude = E dA • if θ = 90 °, Magnitude = 0 E θ dA

  9. Gauss’s Law • If there is no charge inside the surface, there is 0 net flux through the surface • ΦE through a closed surface is independent of shape of the surface • Independent of size given a constant charge

  10. Gaussian Surface • Imaginary surface construct to analyze the problem • Useful to exploit symmetry and remove E from the integral • The goal is to establish the situation where every part of the surface satisfies one or more of the following conditions

  11. Useful Surface Conditions • Electric field value argued by symmetry to be a constant over the portion of the surface • The dot product can be simplified to E da because vectors are parallel • The dot product is 0, because the vectors are perpendicular • The electric field is 0 over the portion of the surface

  12. Example 1Cube in E field • In a certain region of space the electric field is given by E = (b+cx)i, where b=2N/C and c=10 N/Cm. What is the electric charge inside a cube 20cm on a side centered at the origin?

  13. Example 1 Given: a=0.2m, b= 2N/C, c = 10 N/Cm εo = 8.85 E-12 C^2/Nm^ 2 Problem Configutation y A6 A3 A1 A2 x A4 E A5 Z

  14. Example 1 Equations Electric field E(x) = (b+cx)i

  15. Example 1 Answers

  16. Example 2 • Prove in electrostatic case E field is 0 inside a conductor • Prove any excess charge on a solid conductor resides on the outside • Prove if a hollow conductor surrounds a charge q, an equal but opposite charge –q resides on the inside surface of the conductor and all excess charge resides on the outside surface

  17. Example 2a • Show E must be 0 in a conductor for the electrostatic case • Answer • Charges move freely in a conductor • If E not 0, charges move – not electrostatic E +

  18. Example 2b • Prove charge must reside on outside of solid conductor Answer • E must be 0 inside conductor • If E = 0, then q/εo = 0, so q must be 0 + + Conductor surface + + Gaussian surface + +

  19. Example 2c • Show if hollow conductor surrounds charge then the inner wall has an equal and opposite charge while outer surface has the rest • Create gaussian surface in the conductor,E must be 0 in the conductor so q must be 0 inside the gaussian surface, the rest must therefore be on the surface + - - + - + - + Conductor surface - - Conductor inner surface + + Gaussian surface +

  20. Example 3 • Prove a point charge electric field is the same at a distance r from q from Gauss’s Law as it is from Coulomb’s law E p r +

  21. Example 3 • Coulomb’s Law • Gauss’s Law E p r + q1 Gaussian Surface

  22. Example 3 Construct a gaussian surface exploiting symmetry E p r + q1 Gaussian Surface

  23. Example 3

  24. Summary • Gauss’s Law can provide the same results as Coulomb’s law but is easier if symmetry can be exploited. • No electric fields in conductors with electrostatic conditions • Excess charge in conductor is on the outer surface • If a charge is inside a hollow conductor there will be a counterbalancing charge on the inside conducting surface to give a net charge of 0 for a Gaussian surface in the conductor with the balance on the outer surface

  25. Useful Conditions for Gauss • Electric field value argued by symmetry to be a constant over the portion of the surface • The dot product can be simplified to E da because vectors are parallel • The dot product is 0, because the vectors are perpendicular • The electric field is 0 over the portion of the surface

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