## Materials Science and Engineering: An Introduction

$E = 8.0 \times 10^{-3} V/m$
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: $E = \frac{V}{l} = \frac{0.04 V}{5 m} = 8.0 \times 10^{-3} V/m$