Two copper metal plates separated and charged

[lostinphysics44]

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.

A)What is the maximum current in the wire?
B)What is the largest electric field in the wire?
c)What is the total amount of energy dissipated in the wire?

Homework Equations

really have no clue on this one, i think maybe i have to combine a capacitance equation, with something, or use that somehow to find the current.

i do know that energy=.5C(Vc^2), but i can't see how to translate this to energy that goes into the wire.

The Attempt at a Solution

any advice on how to attempt this would be great

 

[Astronuc]

The largest electric field must coincide with the maximum potential which would be at the instant the wire short circuits the capacitor.

One is basically establishing an RC circuit by placing a copper wire, which has some resistance, across a capacitor.

See -
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/resis.html#c2

http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/rstiv.html#c1

 

[ogb p]

I would approach this problem by first identifying the relevant equations and principles that can be applied to solve it. In this case, we are dealing with a circuit consisting of two charged plates connected by a wire, so we can use principles of electrostatics and basic circuit analysis to find the answers.

To answer part A, we can use Ohm's Law (V=IR) to find the maximum current in the wire. We know the voltage (V) is equal to the potential difference between the two plates, which is given by the charge on each plate (Q) divided by the capacitance (C) of the system. So, V=Q/C. We also know the resistance (R) of the wire, which can be calculated using the wire's diameter and resistivity of copper. Putting all of this together, we can solve for the current (I).

For part B, we can use the fact that the electric field (E) is equal to the voltage divided by the distance between the plates (d). So, E=V/d. We already know the voltage and distance, so we can solve for the electric field.

Finally, for part C, we can use the formula for electrical energy (U=QV/2) to find the total amount of energy dissipated in the wire. We know the voltage and charge, so we can solve for the energy.

In summary, to solve this problem we need to use principles of capacitance, Ohm's Law, and electrical energy. By combining these equations and principles, we can find the maximum current, largest electric field, and total energy dissipated in the wire.