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 laid out in the video transcript.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Capacitance, \( C = 2.0 \, \mu F = 2.0 \times 10^{-6} \, F \)
- Time, \( t = 4.4 \, \mu s = 4.4 \times 10^{-6} \, s \)
- The electric field drops to one third of its initial value.
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