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F07_hw02a

F07_hw02a - Physics 505 Homework Assignment#2 Solutions...

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Physics 505 Fall 2007 Homework Assignment #2 — Solutions Textbook problems: Ch. 2: 2.2, 2.8, 2.10, 2.11 2.2 Using the method of images, discuss the problem of a point charge q inside a hollow, grounded, conducting sphere of inner radius a . Find a ) the potential inside the sphere; Recall that, if the point charge is outside a grounded conducting sphere, the method of images gives Φ( ~x ) = q 4 π 0 1 | ~x - ~ y | - a/y | ~x - ( a/y ) 2 ~ y | (1) where y = | ~ y | , and ~ y specifies the location of the charge q . Although this expres- sion was derived for y > a , we note that it is equally valid for y < a . After all, in both cases, the potential satisfies the same boundary condition, Φ( | ~x | = a ) = 0. Hence, for q inside the hollow sphere, the potential is also given by (1). In this case, the physical charge is inside the sphere, while the image charge lies outside. The image charge has the opposite sign, and in this case its magnitude is greater than the physical charge. b ) the induced surface-charge density; The induced surface-charge density is given by σ = - 0 Φ ∂n S where in this case the unit normal is pointing into the sphere. Although we can work out this expression with Φ given by (1), it is quicker to note that the result must the be same (up to a sign change) as that for a point charge outside the sphere. The difference in sign is due to the inward pointing normal in this case, as opposed to an outward pointing normal when the point charge is outside. The result is σ = q 4 πa 2 a y 1 - ( a/y ) 2 (1 + ( a/y ) 2 - 2( a/y ) cos γ ) 3 / 2 where γ is the angle between ~x and ~ y . Note that σ has the opposite sign as q . This is because the numerator in the above expression is actually negative for y < a . If desired, this sign can be made explicit by rewriting the above as σ = - q 4 πa 2 1 - ( y/a ) 2 (1 + ( y/a ) 2 - 2( y/a ) cos γ ) 3 / 2

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By Gauss’ law, the total charge induced on the inside surface of the conducting sphere must be - q . This can also be seen by integrating the surface-charge density Q = Z σa 2 d Ω = 2 πa 2 Z 1 - 1 σ d cos γ = - q 2 (1 - ( y/a ) 2 ) Z 1 - 1 d cos γ (1 + ( y/a ) 2 - 2( y/a ) cos γ ) 3 / 2 = - q 2 a y (1 - ( y/a ) 2 ) 1 | 1 - y/a | - 1 | 1 + y/a | = - q where we have used the fact that 0 < y/a < 1 when simplifying the absolute value quantities. c ) the magnitude and direction of the force acting on q . The force acting on q is essentially the force between q and its image. Again, the magnitude of the force is given by the familiar expression F = 1 4 π 0 q 2 a 2 a y 3 1 (1 - ( a/y ) 2 ) 2 Since the charge q is attracted to its image, the direction of the force is given by ˆ y . For y < a , it is convenient to rewrite the force as ~ F = 1 4 π 0 q 2 a 3 ~ y (1 - ( y/a ) 2 ) 2 This demonstrates that, for y a , the force is linear, ~ F ~ y . Because of the positive sign, however, this is the opposite of a restoring force. This demonstrates that the center of the conducting sphere is a point of unstable equilibrium for the charge q .
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F07_hw02a - Physics 505 Homework Assignment#2 Solutions...

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