Homework1 - Homework 1 Answers, 95.657 Fall 2011,...

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Homework 1 Answers, 95.657 Fall 2011, Electromagnetic Theory I Dr. Christopher S. Baird, UMass Lowell Problem 1 Considering only the fields in a two-dimensional plane, draw the electrostatic field lines as accurately as possible for the following systems: (a) A triangle-shaped object with a total charge + Q fixed uniformly over its surface. (b) A point charge -2 Q placed a distance d away from the triangle already described. (c) The point charge and triangle already described placed inside a thin, hollow conducting circular shell with no net charge. SOLUTION: When constructing field lines diagrams, we should keep several points in mind. Very near point charges, the field lines extend out radially. Also, field lines are perpendicular to the surface of conductors at the surface. We should remember that far away from an object, the object looks like a point charge, so the field line pattern far away should become radial. When we draw field lines, they should start on positive charges and end on negative charges or lead off the page. They should never cross or have unnatural kinks. Lastly, conducting shells shield the interior field pattern, and therefore all internal information from the outside world, except for the total net internal charge. This is because the charges on the shell's inner surface move around until they cancel the fields. (a) (b) + Q + Q -2 Q
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(c) Problem 2 A thin wire ring of radius R is centered on the origin, lies in the x - y plane, and contains a total charge + Q fixed permanently and uniformly along its extent. An external field is applied, E x = ( E 0 / L 2 )( x 2 + y 2 )2 z/L , E y = 3 E 0 ( z/L +1) 3 , E z = E 0 sin( z/L ). Here, L is a length. What is the total force that the ring experiences? SOLUTION: First we note that the the uniform charge is permanently fixed, so that the introduction of the external field does not alter the charge distribution. We must first find a mathematical expression for the ring charge in terms of Dirac deltas. Because the ring is round and centered on the origin, the most natural coordinate system is spherical coordinates ( r , θ , ϕ ). The ring is infinitely thin in the radial direction and the polar angle direction, so that we must use Dirac deltas. The ring is constant in shape and in charge density in the azimuthal direction, so our expression for the charge density should not include the azimuthal angle in any way. The general expression is: δ( x x ' )=δ( u u ' )δ( v v ' w w ' ) U V W For spherical coordinates, u = r , v = θ, w = ϕ and U = 1, V = 1/ r , W = 1/( r sin θ ) so that we have: x x ' r r ' ) δ(θ−θ) r δ(ϕ−ϕ ' ) r sin θ Apply this to our case and multiply by some constant A so that the total charge works out: + Q -2 Q
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ρ( x )= A δ( r R ) δ(θ−π/ 2 ) r Now we determine A by forcing the total charge to be equal to + Q . + Q = ρ dV + Q = 0 2 π 0 π 0 r , θ , ϕ) r 2 sin θ dr d θ d ϕ + Q = 0 2 π 0 π 0 [ A r R ) 2 ) r ] r 2 sin θ dr d θ d ϕ + Q = 2 π R A A = Q 2 π R We have our final form for the charge density. Note that once the Dirac deltas are applied, their
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This note was uploaded on 02/13/2012 for the course PHYSICS 95.657 taught by Professor Staff during the Fall '11 term at UMass Lowell.

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Homework1 - Homework 1 Answers, 95.657 Fall 2011,...

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