The electric field between the plates of a parallel-plate capacitor of capacitance`2.0(mu)F` drops to one third of its initial value in `(4.4 mu)s` when the plates are connected by a thin wire. Find the resistance of the wire.

Explanation

To solve the problem step by step, we will follow the reasoning presented in the video transcript and apply the relevant formulas.

### Step 1: Understand the problem
We have a parallel-plate capacitor with a capacitance of \( C = 2.0 \, \mu F \) that is discharging through a wire. The electric field between the plates drops to one-third of its initial value in a time of \( t = 4.4 \, \mu s \).

### Step 2: Relate the electric field to charge
The electric field \( E \) between the plates of a capacitor is directly proportional to the charge \( Q \) on the plates:
\[
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