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Exercise:
The charge density in a straight wire of length L is given by lambdax mu absx quad textfor -L/ leq x leq +L/ where mu is given by the length and the charge Q on the wire: mu frac QL^ Derive a formal expression for the electric field at a po on the perpicular bisector to the wire.

Solution:
The solution is similar to that of part a of "Charged Wire" see link with the difference being that the charge density is not constant. Therefore the electric field can be written as Er k_C r _-L/^+L/ fraclambdaxleftr^+x^right^/ textdx k_C r _-L/^+L/ fracmu absxleftr^+x^right^/ textdx The egrand is symmetrical with respect to the erval -L/ +L/ so we only have to calculate on half of the egral: Er k_C r _^+L/ fracmu xleftr^+x^right^/ textdx k_C r mu _^+L/ fracxleftr^+x^right^/ textdx The antiderivative of the egrand is Fx -fracleftr^+x^right^/ so the electric field is Er - k_C r mu left fracleftr^+x^right^/ right_^L/ k_C r mu left fracr - fracsqrtr^+L/^ right k_C fracQL^ left -fracrsqrtr^+L/^ right frac k_C QL^ left-fracsqrt+L/r^ right For a long wire L gg r the second term in the parentheses is much smaller than the first one so the expression can be approximated as Er &approx frac k_C QL^
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Exercise:
The charge density in a straight wire of length L is given by lambdax mu absx quad textfor -L/ leq x leq +L/ where mu is given by the length and the charge Q on the wire: mu frac QL^ Derive a formal expression for the electric field at a po on the perpicular bisector to the wire.

Solution:
The solution is similar to that of part a of "Charged Wire" see link with the difference being that the charge density is not constant. Therefore the electric field can be written as Er k_C r _-L/^+L/ fraclambdaxleftr^+x^right^/ textdx k_C r _-L/^+L/ fracmu absxleftr^+x^right^/ textdx The egrand is symmetrical with respect to the erval -L/ +L/ so we only have to calculate on half of the egral: Er k_C r _^+L/ fracmu xleftr^+x^right^/ textdx k_C r mu _^+L/ fracxleftr^+x^right^/ textdx The antiderivative of the egrand is Fx -fracleftr^+x^right^/ so the electric field is Er - k_C r mu left fracleftr^+x^right^/ right_^L/ k_C r mu left fracr - fracsqrtr^+L/^ right k_C fracQL^ left -fracrsqrtr^+L/^ right frac k_C QL^ left-fracsqrt+L/r^ right For a long wire L gg r the second term in the parentheses is much smaller than the first one so the expression can be approximated as Er &approx frac k_C QL^
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Branches
Electrostatics
Tags
charge density, electric field, integral
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Difficulty
(4, default)
Points
0 (default)
Language
ENG (English)
Type
Calculative / Quantity
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Decoration