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HW-230B-1_2010a

# HW-230B-1_2010a - ECE 230B HW-1 Winter 2010 Solutions 1 3-D...

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ECE 230B, HW-1, Winter 2010 Solutions 1. 3-D Gauss’ law is obtained after a volume integration of 3-D Poisson’s equation and takes the form of E = ∫∫ dS Q si S ε , 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 (Coulomb’s 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 π r S E , where E is the magnitude of the electric field on S . 3-D Gauss’s law then gives E = Q r si 4 2 πε , which is Coulomb’s law. Since E = - dV / dr , the electric potential at a point on the sphere is V Q r si = 4 πε , if one defines the potential to be zero at infinity. 1

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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 ε , where ε is the permittivity of the medium.
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HW-230B-1_2010a - ECE 230B HW-1 Winter 2010 Solutions 1 3-D...

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