P23_023 - 1 L + a − x ¯ ¯ ¯ ¯ ¯ L = λ 4 πε µ 1 a...

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23. (a) The linear charge density is the charge per unit length of rod. Since the charge is uniformly distributed on the rod, λ = q/L . (b) We position the x axis along the rod with the origin at the left end of the rod, as shown in the diagram. Let dx be an inFnitesimal length of rod at x . The charge in this segment is dq = λdx . The charge dq may be considered to be a point charge. The electric Feld it produces at point P has only an x component and this component is given by dE x = 1 4 πε 0 λdx ( L + a x ) 2 . The total electric Feld produced at P by the whole rod is the integral E x = λ 4 πε 0 Z L 0 dx ( L + a x ) 2 = λ 4 πε
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Unformatted text preview: 1 L + a − x ¯ ¯ ¯ ¯ ¯ L = λ 4 πε µ 1 a − 1 L + a ¶ = λ 4 πε L a ( L + a ) . When − q/L is substituted for λ the result is E x = − 1 4 πε q a ( L + a ) . The negative sign indicates that the Feld is toward the rod. dx • P x L L + a (c) If a is much larger than L , the quantity L + a in the denominator can be approximated by a and the expression for the electric Feld becomes E x = − q 4 πε a 2 . This is the expression for the electric Feld of a point charge at the origin....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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