What is the maximum current and electric field in a charged wire?

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haroharo
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Homework Statement



Two 10-cm-diameter metal plates are 1.0 cm apart. They are charged to +/-12.5 nC. They are suddenly connected together by a 0.224-mm-diameter copper wire stretched taut from the center of one plate to the center of the other.

d)What is the total amount of energy dissipated in the wire?


Homework Equations


What is the maximum current in the wire?
4.17*105 A

What is the largest electric field in the wire?
1.80*105 V/m

U=qV
U=1/2(Ad)E2ε0

The Attempt at a Solution


I figure that you can't find the energy density and find the volume the wire occupies in the capacitor because the wire has a certain resistance restricting the flow, thus affecting the potential energy.

Attempt1:
U=qV where q = 1.6*10-19 or 12.5*10-9 & V =1.8*103
=2.88*10-16 or 2.25*10-5


Attempt 2:
U=1/2(Ad)E2ε0
=(.5)(0.0001122*pi)(0.01)(1.80*105)2*(8.854*10-12)
=5.64*10-11J

Helps Appreciated!
 
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Well, it seems to me that all the energy in the capacitor will be dissipated.
Have you got a formula for the energy of a capacitor (with charge in it)?
 
actually i do
awesome, thanks for the clarification!
 
haroharo said:

Homework Statement



Two 10-cm-diameter metal plates are 1.0 cm apart. They are charged to +/-12.5 nC. They are suddenly connected together by a 0.224-mm-diameter copper wire stretched taut from the center of one plate to the center of the other.

d)What is the total amount of energy dissipated in the wire?


Homework Equations


What is the maximum current in the wire?
4.17*105 A

What is the largest electric field in the wire?
1.80*105 V/m

U=qV
U=1/2(Ad)E2ε0

The Attempt at a Solution


I figure that you can't find the energy density and find the volume the wire occupies in the capacitor because the wire has a certain resistance restricting the flow, thus affecting the potential energy.

Attempt1:
U=qV where q = 1.6*10-19 or 12.5*10-9 & V =1.8*103
=2.88*10-16 or 2.25*10-5


Attempt 2:
U=1/2(Ad)E2ε0
=(.5)(0.0001122*pi)(0.01)(1.80*105)2*(8.854*10-12)
=5.64*10-11J

Helps Appreciated!


Hey how'd u get the max current and the largest electric field in the wire?
Help would be much appreciated.Thanks in advance :)