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ch24_p34 - 34(a According to the result of problem 28 the...

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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 Q x L V L x ε § · = ¨ ¸ © ¹ π At x = –d we obtain (in SI units) 9 15 0 3 (8.99 10 )(43.6 10 ) 0.135 ln ln 1 4 0.135 0.135 (2.90 10 V)ln 1 . Q d L V L d d d ε + × × § · § · = = + ¨ ¸ ¨ ¸ © ¹ © ¹ § · = × + ¨ ¸ © ¹ π (b) We differentiate the potential with respect to x to find the x component of the electric field (in SI units): 2 0 0 0 9 15 4 1 ln 4 4 4 ( ) (8.99 10 )(43.6 10 ) (3.92 10 ) , ( 0.135) ( 0.135) x V Q x L Q x x L Q E x L x x L x L x x x x L x x x x ε ε ε § · § · = − = − = − = − ¨ ¸ ¨ ¸ © ¹ © ¹ × × × = − = − + + π π π or 4 (3.92 10 ) | | . ( 0.135) x E x x × = + (c) Since 0 x E < , its direction relative to the positive x axis is 180 . ° (d) At x = –d we obtain (in SI units)
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