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P132_ch24

# P132_ch24 - Chapter 24 Gauss Law Gauss Law relates the net...

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R. Kass P132 Sp04 Ch24 1 Chapter 24: Gauss’ Law Gauss’ Law relates the net amount of electric charge enclosed by a surface to the electric field on that surface. For certain situations Gauss’ Law provides an easier way to calculate the electric field than the integration methods discussed in chapter 23. These cases most always involve a situation with a high degree of symmetry, e.g. a sphere or cylinder. To be specific, Gauss’ Law is: ε 0 Φ =q enc e 0 is our friend the permittivity constant=8.85x10 -12 C 2 /(N-m 2 ). q enc is the net electric charge enclosed by our surface. This is the algebraic sum of all enclosed charges and can be +, -, or zero. Φ is the electric flux through the surface that encloses our charge(s). few words about the “surface”: 1) it is not a real surface, i.e. something we can touch. 2) We try to choose the appropriate surface to simplify the problem. 3) The surface must fully surround the charge(s), i.e. it must be CLOSED. Example: a rectangular box with 6 sides gives a closed surface. If we take the top of the box off it is no longer a closed surface. 4) We call it a “ Gaussian surface ”. = Φ A d E . surface closed entire the over taken is n integratio the that means symbol The Flux is a scalar even though E and dA are vectors. Flux has units Nm 2 /C The integral contains the dot product of 2 vectors. Karl Friedrich Gauss 1777-1855. A true math-physics genius. The electric flux through the Gaussian surface is:

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R. Kass P132 Sp04 Ch24 2 Electric Flux Calculations = Φ A d E So, how do we calculate the electric flux? This integral looks worse than chapter 23’s! Let’s calculate the electric flux through the box in the figure. Since the box is closed the Gaussian surface is the 6 box sides. We really have 6 integrals to worry about!! 1 2 3 4 5 6 + + + + + = = Φ 6 5 4 3 2 1 A d E A d E A d E A d E A d E A d E A d E For this example the electric field is uniform along the +z-axis. The direction of dA is always normal to a surface and points outward. Ex: for surface #1 dA points in the +z direction. For #6 dA points in the –z direction. = = = = = = = = = = = = 6 6 5 5 4 4 3 3 2 2 1 1 ) 180 cos( 0 ) 270 cos( 0 ) 270 cos( 0 ) 90 cos( 0 ) 90 cos( ) 0 cos( EA EdA A d E EdA A d E EdA A d E EdA A d E EdA A d E EA EdA A d E HRW fig. CP1 So, the flux through the Gaussian surface is: Φ =EA+0+0+0+0-EA=0, The total flux is zero even though the flux through sides 1 and 6 is non-zero. According to Gauss’ Law there is zero charge enclosed in the box. This should make sense since if we had enclosed a positive charge in the box we would have electric field lines pointing outward on each of the 6 surfaces. Likewise if we had a negative charge in the box we would have field lines pointing inward on each of the 6 surfaces. The only way to have an electric field that points only in the +z direction is to have the charge(s) outside the box.
R. Kass P132 Sp04 Ch24 3 Electric Flux Calculations continued What would the flux be if I rotated the box 45 0

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P132_ch24 - Chapter 24 Gauss Law Gauss Law relates the net...

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