Flux and Gauss' Law

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PHY110W ELECTRICITY SUPPLEMENTARY GAUSS’ LAW We make things mathematically simpler (!) by turning the normal into the elemental area vector, dA , whose magnitude is equal to the area of the surface element, and whose direction is always out of the surface (at right angles to it, like the normal). So field lines which enter the Gaussian surface (for which θ is larger than 90°) result in negative flux (because cosθ is now negative). And, representing the field by the electric field vector, E , we can now write the equation even more simply (!) as: Φ = E ⋅dA area vector, dA θ electric field vector, E To find the flux through a whole area, we must sum the flux through each of the elements which makes up that area, or (as the size of each element → 0) integrate over the whole area: total Φ = ∫ E ⋅dA If the whole area is a complete Gaussian surface, we write: Φ = ∫ E ⋅dA , where Φ now represents the net flux through the Gaussian surface. The importance of this equation is that Gauss’ Law tells us that the net flux through a Gaussian surface is proportional to the charge enclosed by that surface, or mathematically: ε 0Φ = qenclosed . Which gives us… Gauss’ Law: ε E⋅ dA= q ∫ 0 enclosed …and from this we derive useful equations for the electric field due to various charge distributions. 2
PHY110W ELECTRICITY SUPPLEMENTARY GAUSS’ LAW Types of questions Questions on this section usually fall into one of three categories: 1. Calculating the flux through some given surface (maybe part of a bigger, complete, Gaussian surface). [This type of question is a bit artificial and is really just for practice.] 2. Working with the net flux through an entire Gaussian surface and applying Gauss’ Law to find the enclosed charge (or vice versa). 3. Using the principle of Gauss’ Law to determine the electric field due to some distribution of charge. [Which is the main purpose of this section – determining electric fields – so that we can easily determine the force on a charged particle, and from here the amount of work done in moving it, etc etc.] Examples: * * * * * Type 1. A cubic Gaussian surface is bounded by the planes at x = 0.40 m, y = 0.60 m and z = 0.40 m as shown in the sketch. The electric field in this region is non-uniform and is given by E = (2.00 + 3.00x 2 )î N/C. Calculate the flux through the right hand face (ie the shaded face) of the cube. Solution: Φ= ∫ E ⋅dA Over the whole shaded surface x = 0.40 m, so the electric field is a constant (2.00 + 3.00 × 0.40 2 ) = 2.48 N/C and lies in the positive x direction, ie parallel to the area vector, so cosθ = 1, and we get 0.6 0.2 y z 0.4 0.4 Φ right = E∫ dA = E A = 2.48 × 0.40 2 = 0,397 N⋅m 2 /C [Strictly speaking, you should evaluate the dot product more formally using unit vectors.] x 3
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Flux and Gauss' Law

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