Solution to Quiz5

# Solution to Quiz5 - R and uniform line charge density-λ ....

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ENEE 380 Fall 2011 September 30, 2011 Discussion Quiz 5 1. Griffiths Problem 3.6: Find the force on the charge + q in the figure below. (The xy plane is the grounded conductor) 2. A circular wire of radius R and uniform line charge density is placed above a grounded infinite conducting plane a distance d from it. a) Find the potential in the region above the plane along the z-axis. b) Find the charge density induced on the conducting plane at z = 0. c) Determine the total induced charge on the conducting plane.

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Problem 1 Easy to see that we need to have two image charges to make xy plane’s potential zero. One is located at z = - d with +2 q , the other at z = - 3 d with - q . It follows from Coulomb’s Law that the force on the charge + q is simply F = q 4 πε 0 ± - 2 q (2 d ) 2 + 2 q (4 d ) 2 + - q (6 d ) 2 ² ˆ z = q 2 4 πε 0 d 2 ³ - 29 72 ´ ˆ z . Problem 2 a) Image charge would be a circular wire with center located at z = - d , radius
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Unformatted text preview: R and uniform line charge density-λ . For a given point on z-axis, the distance between it and the element of charge on each wire r = p R 2 + ( z-d ) 2 r = p R 2 + ( z + d ) 2 . Potential at the given point V ( z ) = 1 4 πε λ 2 πR ± 1 r-1 r ² = λR 2 ε " 1 p R 2 + ( z-d ) 2-1 p R 2 + ( z + d ) 2 # . b) E z =-∂ ∂z V ( z ) =-λR 2 ε µ z + d [ R 2 + ( z + d ) 2 ] 3 / 2-z-d [ R 2 + ( z-d ) 2 ] 3 / 2 ¶ ˆ z . σ | z =0 = ε E z | z =0 =-λRd ( R 2 + d 2 ) 3 / 2 . c) Total induced charge is the same as the image charge(Why?): Q ind =-λ 2 πR. (Hint: take a hemispherical Gaussian surface with the great circle just above the xy-plane and the south pole inside the conducting plane, then let the radius of the hemisphere tend to inﬁnity.) 1...
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## This note was uploaded on 02/02/2012 for the course ELECTRICAL ENEE380 taught by Professor D.romeo during the Fall '11 term at Maryland.

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Solution to Quiz5 - R and uniform line charge density-λ ....

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