M. Course: PHYS X tn PS3-001.pdf x Tn Lecture5 x Tn Lecture4 x Tn Lecture 6.001
ID: 3161738 • Letter: M
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M. Course: PHYS X tn PS3-001.pdf x Tn Lecture5 x Tn Lecture4 x Tn Lecture 6.001 x M [PHYs2015 x Problem set x VA Katz, Physics x New Tab Secure https:// ay16 moodle umn.edu/plugin file.php/1913226/mod resource/content/2/PS3 001.pd PS3 001.pdf 1. In class, we considered the tux through different surfaces of a triangle, as shown in the picture The electric field is constant and uniform, and has magnitude E. Consider the following arguments they like the ones we had in class: (a) Soren: "Well, the area of surface B is greater, so the flux passing though it should be greater!" (b) Nettie: "But, the normal vector of surface B is parallel to the electric field, whereas for surface B its normal vector is at some angle with respect to the electric field. So I say surface A has more flux." Your task is to sort this out! I've already given you the answer that the flux through A is equal to the flux through B. Resolve the competing arguments of Soren and Nettie by proving that the fluxes are equal. 2. Here's a problem asking about the electric field produced by a iforma charge distribution (I haven't shown you explicitly how to do this, but we have done a lot of practice setting problems up! Try to find a method that makes sense. We have a solid, non-conducting sphere with a volume charge density described by p(r) 3(a)"/2. Find the electric field everywhere that means, find E(r) for r inside and outside the sphere. Here's some steps to (a) What's the total charge on the sphere? (b) Use Gauss's law to find the electric field outside the sphere, r a. Does your answer make sense? (c) Use Gauss's law to find the electric field inside the sphere, r a Search all the 5 12:16 PM 2/5/2017Explanation / Answer
1) for the electric flux , as the electric flux is given by formula
flux = E * A * cos(theta)
E is the electric field
A is Area
theta is the angle between electric field and area
Now , in this case , all the electric field lines entering the area A will pass through area B while going out of the triangle.
hence , the electric field lines passing through A and C will be same
and therefore , the electric flux through A and B is equal
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