Answer
$E = 8.0 \times 10^{-3} V/m$
Work Step by Step
Given:
For an aluminum wire 5 mm (0.20 in.) in diameter and 5 m (200 in.) in length, the
potential drop across the ends of the wire is 0.04 V
Required:
The magnitude of the electric field across the ends of the wire.
Solution:
$E = \frac{V}{l} = \frac{0.04 V}{5 m} = 8.0 \times 10^{-3} V/m$
