HW-230B-1_2010a

HW-230B-1_2010a - ECE 230B, HW-1, Winter 2010 Solutions 1....

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Unformatted text preview: ECE 230B, HW-1, Winter 2010 Solutions 1. 3-D Gauss law is obtained after a volume integration of 3-D Poissons equation and takes the form of E = dS Q si S , where the LHS is an integral of the normal electric field over a closed surface S , and Q is the net charge enclosed within S . Use it to derive the electric field at a distance r from a point charge Q (Coulombs law). What is the electric potential in this case? Solution: With the point charge Q at the center, construct a closed spherical surface S with radius r . By symmetry, the electric field at every point on S has the same magnitude and points outward perpendicular to the surface. Therefore, E = dS 4 2 r S E , where E is the magnitude of the electric field on S . 3-D Gausss law then gives E = Q r si 4 2 , which is Coulombs law. Since E =- dV / dr , the electric potential at a point on the sphere is V Q r si = 4 , if one defines the potential to be zero at infinity. 1 2. (a) Use Gauss law to show that the electric field at a point above a uniformly-charged sheet of charge density Q s per unit area is Q s /2 , where is the permittivity of the medium....
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This note was uploaded on 10/09/2010 for the course ECE 230 at UCSD.

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HW-230B-1_2010a - ECE 230B, HW-1, Winter 2010 Solutions 1....

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