P2213_HW06_solns_Fall2008

# P2213_HW06_solns_Fall2008 - Physics 2213 HW#6 Solutions...

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- 1 - Physics 2213 HW #6 — Solutions Fall 2008 21.96 [Half-Ring of Charge] The x-components of the electric field cancel from the left and right halves of the semicircle, so we only need to calculate the y-component of the field, E y . This y component points in the negative y direction. The charge per unit length of the semicircle is λ = Q π a Consider a small element of the semicircle as shown in the diagram to the right. This element is at angle θ to the +x axis and subtends an arc of angle d θ at the origin. The charge on this element is dQ = λ a d θ . The electric field d E at the origin (point P) due to this element has magnitude |d E | = k dQ a 2 , and is directed away from the charge element. However, we only need to consider the y component of d E : dE y = k dQ sin θ a 2 = k λ a d θ sin θ a 2 = k λ d θ sin θ a The magnitude of the total electric field at P is found by integrating from θ = 0 to π : E = 0 π k λ d θ sin θ a = k λ a | - cos θ | π 0 = k a Q π a (2) = 2kQ π a 2 ; downward . 22.8 [Gauss’s Law] encl Q in Gauss’s law is the algebraic sum of the charges enclosed by each surface. Flux out of the volume is positive and flux into the enclosed volume is negative. (a) 1 9 2 1 0 0 / (4.00 10 C)/ 452 N m /C. S q ε - Φ = = × = (b) 2 9 2 2 0 0 / ( 7.80 10 C)/ 881 N m /C. S q - Φ = = - × = - (c) 3 9 2 1 2 0 0 ( )/ ((4.00 7.80) 10 C)/ 429 N m /C. S q q - Φ = + = - × = - (d) 4 9 2 1 3 0 0 ( )/ ((4.00 2.40) 10 C)/ 723 N m /C. S q q - Φ = + = + × = (e) 5 9 2 1 2 3 0 0 ( )/ ((4.00 7.80 2.40) 10 C)/ 158 N m /C. S q q q - Φ = + + = - + × = -

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- 2 - (f ) All that matters for Gauss’s law is the total amount of charge enclosed by the surface, not its distribution within the surface. 22.37 [Coaxial Cable] (a) a < r < b, E = 1 4 πε 0 2 λ r , radially outward. (b) r c, E = 1 4 0 2 r , radially outward, since again the charge enclosed is the same as in part (a). (c) See diagram at right. (d) The inner and outer surfaces of the outer cylinder must have the same amount of charge on them: l = - inner l inner = - , and outer = . 22.42
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P2213_HW06_solns_Fall2008 - Physics 2213 HW#6 Solutions...

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