121A pt2 - Laboratory 202: Numerical Verification of Gauss...

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Laboratory 202: Numerical Verification of Gauss Law In this problem, you will evaluate Gauss Law numerically for different surfaces and ver i fy that the integral over the surface is the enclosed charge . To obtain credit for the Lab, print out the pages of MATHCAD calculations, graphs, etc. for each problem. By definition , flux <D is a dot product of E and area vector A. Area vector A of a surface is defined as a vector with magnitude = area of the surface and direction perpendicular to the surface . For example if you cu t a rectangle a . b on the upper side of a cube area vector A is (a.b) . k where k is a unit vector along z-axis. 1. Consider a point charge q1 :=1.0 at the origin. To calculate electric field E at some arbitrary point (x,y,z) the range of x, y , and z must be defined: x~O y~O z~o Distance from the point (x , y , z) to charge q1 is : r(x,y,z) := /x 2 + y2 + z2 ' 2. To calculate the electric field vector in three dimensions you must find all 3 . components. q 1 . [ x~. ] E(x , y , z ) : = 4 1 · x ( · 3 r x,y,z )
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Look at Figure 2 . The angle between E and the upper surface area vector N is not a constant , i n general. The angle depends on where the E pierces the surface. We can aSsume that if we cut v.v.small rectangle dx.dy from the surface the angle between the E and area vector dx.dy . k is constant ( i n this small rectangle). The flux through the dx . dy surface is then < . p = E(x,y , a) . k . dx . dy = dot product E.k multiplied by area. You get the total flux through the whole surface summing . (= . integrating) along x and y axis . Figure 2 / ~~ { " b envee : t l I ~~~ " I N ( 3. Now evaluate the electric flux through a cube of side 2a (-a to +a) which · is centered on the origin . Since the electric field has radial symmetry, we can calculate the electric flux through anyone side of the cube and multiply by six. So , calculate the electric flux through the face parallel to x-y plane with z=a (yellow face in Figure 2). The size of a must be defined in MCAD . Assume a:=1. Unit vector N (vectors in MCAD are matr i ces) normal to the surface z=a is: k:=l?l (3-row 1 - column matrix).
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I l : = I a I a E (x ,y , Il . ) · k d x d y _ Il . _ Il . 1. Explain why in the above calculation we only need to calculate the flux through one surface and then multiply by 6 (to have total flux). 3. Double the magnitude ofthe charge. By what factor does the flux change? Explain why. 4. Increase the size of the box to a:=2, then a:=10, then a:=100. Does the flux change? Does Gauss's Law appear to hold? Explain. 0 0 i : = 0 j ' - 1 k ' - 0 .- . - 0 0 1 2 . Change the location of the charge to a place outside of t he box. (In this case , you can not assume that the flux through each face of the cube i s equa l . Therefore , you need to evaluate the flux through each surface independently.) Assume that the charge is located now on z-axis at the position z = 2·a. You are not to type z = 2a.
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This note was uploaded on 04/03/2009 for the course PHYS PHYS 121A taught by Professor Any during the Spring '09 term at NJIT.

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121A pt2 - Laboratory 202: Numerical Verification of Gauss...

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