E d e k e dq r 2 ˆ r e de k e dq r 2 the electric

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E = d ! E ! = k e dq r 2 ! ˆ r E = dE ! = k e dq r 2 ! The electric field for this differential charge contribution will be given by: dE = k e dq r 2
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Continuous Distributions For example, let’s determine the magnitude of the electric field at a point P which is at a distance z from a very long (nearly infinite) wire of uniformly distributed charge. Assume z is much smaller than the length of the wire and let λ be the charge per unit length of the wire. Solution : dx dq + + + + + + + + + + z P A small segment of wire dx has a charge of dq= λ dx.
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Continuous Distributions At point P, the differential electric field contribution will be: Let’s handle them separately: dE dx dq + + + + + + + + + + z x r = z 2 + x 2 dE x dE y P θ dE = k e ! dx z 2 + x 2 ( ) dE will have dE x and dE y components. dE = k e dq r 2 cos ! = z z 2 + x 2 We have two variables (x and θ ) dE x = dE sin ! dE y = dE cos ! E y = dE y ! = cos ! dE ! E y = cos ! ( ) !" + " # k e ! dx z 2 + x 2 ( )
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Continuous Distributions Let’s switch from dx to d θ in our formula: Giving us: And we have: cos ! = z z 2 + x 2 tan ! = x z x = z tan ! dx = z d tan ! ( ) = z sec 2 ! ( ) d ! dx = z cos 2 ! d ! E y = ! k e cos " ( ) !" + " # dx z 2 + x 2 ( ) cos 2 ! z 2 = 1 z 2 + x 2 ( ) E y = ! k e cos " ( ) ! ! 2 + ! 2 " cos 2 " z 2 z cos 2 " d " E y = ! k e z cos " ( ) ! ! 2 + ! 2 " d " E y = 2 ! k e z E y = 1 2 !" o # z
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Continuous Distributions The differential contributions for dE x can easily be shown to cancel out via symmetry as you sum over the entire wire (you get sin θ instead of cos θ ).
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