1. Chapter 3. Gauss’ law, Divergence
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2. Displacement flux : Faraday’s Experiment
charged sphere
(+Q) +
insulator
metal
Two concentric conducting spheres are separated by an insulating material.
The inner sphere is charged to +Q. The outer sphere is initially uncharged.
The outer sphere is grounded momentarily.
The charge on the outer sphere is found to be -Q.
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3. Faraday concluded there was a “displacement” from the charge on the inner sphere through the inner sphere through the insulator to the outer sphere.
The electric displacement (or electric flux) is equal in magnitude to the charge that produces it, independent of the insulating material and the size of the spheres. +Q -Q +Q
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4. Gauss’ law
D : electric displacement flux density. The electric field lines come out of positive charges and induce another charge distribution.
Gauss’ law : The integral of electric flux density over a closed surface is equal to the sum of charges inside the surface.
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5. Integration surface deformability
If the integrand follows an inverse square law, the integration surface is deformable as long as the position of the charge remains inside or outside the surface during deformation. The result of the integration does not change after deformation.
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6. Proof : Gaussian surface contains a charge
The contribution of the integrand on da1 is equal to that on da2
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7. S1 S2
Because S1 and S2 are similar, the ratio of the area S1 to that of S2 is
where
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8. EMLAB

9. Proof : Gaussian surface contains no charge
An integration surface is deformable as long as the position of the charge remains outside the surface S. The result of the surface integral remains the same.
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10. Gauss’ law
Gauss’ law holds for the case with multiple charges.
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11. Gauss law usefulness
Gaussian surface
The integration in Gauss’ law becomes trivial when there exists a symmetry in the problem geometry.
‘Dr’ can be taken out of the integral thanks to the symmetry.
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12. Example 1: electric field due to spherically symmetric volume charges
Gaussian surface
Charge enclosed in the Surface integral
(r<a)
(r>a)
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13. Example 2: E-field due to an axially symmetric charge distribution
Usually, the total charge on the outer surface is equal to the negative of that on the inner surface.
D is determined by the charges inside the Gaussian surface regardless of the charges outside the surface.
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14. Gauss’ law in differential form
An integral form of Gauss’ law only states that a surface integral of D over a closed surface is equal to the total charge inside. We cannot obtain the behavior of D at one specific point.
To observe D at a point, an infinitesimally small integration surface is chosen for the Gauss’ law.
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15. Gauss’ law on an infinitesimally small surface
‘Divergence’ D
The coordinate of the center of the cube is (x,y,z)
The differential form of Gauss’ law states that the sum of partial derivatives of Dx, Dy, Dz with respect to x, y, z is equal to the charge density at that point.
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16. Divergence in cylindrical coordinate
Divergence in spherical coordinate
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17. Boundary Conditions on the Electric Field at the Surface of a Metallic Conductor- - - - -
E = 0 + + + + +
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18. Del operator : rectangular coordinate
Divergence theorem
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19. Example 3.5
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20. How to obtain D using a differential form Gauss’ law
Due to boundary condition
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