Electric field at a point in a wire
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Electric field at a point in a wire

[From: ] [author: ] [Date: 12-05-13] [Hit: ]
but I feel like I dont have enough information.Any ideas?Thanks!And since the voltage is obvously linear w.r.dV/dL = (Vf - Vi)/(xf - xi)= (6 - -4)/(.......
I'm looking at this question: http://i.imgur.com/7TsHH.png?1
I'm thinking about Ohm's Law, but I feel like I don't have enough information.
Any ideas?
Thanks!

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The simpilist way is to note that E is the negative gradient of the voltage;
E = - dV/dL

And since the voltage is obvously linear w.r.t to the length you can find the gradient as;
dV/dL = (Vf - Vi)/(xf - xi) = (6 - -4)/(.1 - 0) = 100 v/m
E = - 100 v/m (the negative means the direction of E is toward lower voltage)

You can also start with Ohm's Law in terms of E and current density J as;
E = (1/D)J = (1/D)(i/A) = (1/D)(V/RA) = (1/D)(V/pLA/A) = V/L

where D = 1/p , conductivity = 1/resistivity

The negative doesn't show up because its a magnitude calculation.
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