Ch24_p34 - 34(a According to the result of problem 28 the electric potential at a point with coordinate x is given by V= At x = d we obtain(in SI

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34. (a) According to the result of problem 28, the electric potential at a point with coordinate x is given by 0 ln . 4 Qx L V Lx ε §· = ¨¸ ©¹ π At x = –d we obtain (in SI units) 91 5 0 3 (8.99 10 )(43.6 10 ) 0.135 ln ln 1 4 0.135 0.135 (2.90 10 V)ln 1 . Qd L V Ld d d × § · == + ¨ ¸ © ¹ + π (b) We differentiate the potential with respect to x to find the x component of the electric field (in SI units): 2 00 0 5 4 1 ln 44 4 ( ) (8.99 10 )(43.6 10 ) (3.92 10 ) , ( 0.135) ( 0.135) x V Q xL Q x Q E x x L x L x x x x L xx εε −− ∂∂ § · =− ¨ ¸ © ¹ ×× × ++ ππ π or 4 (3.92 10 ) || . ( 0.135) x E × = + (c) Since 0 x E < , its direction relative to the positive x axis is 180 . ° (d) At x = –d we obtain (in SI units) 4 (3.92 10 ) | | 0.0321 N/C. (0.0620)(0.0620 0.135) x E
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This note was uploaded on 01/25/2011 for the course PHYSICS 17029 taught by Professor Rebello,nobels during the Spring '10 term at Kansas State University.

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