Question

The electric field inside a 30-cm-long copper wire is 0.010 V/m.

What is the potential difference between the ends of the wire?

Answer 1

Concepts and reason

The concept needed to solve this problem is definition of electric field in terms of change in potential over a displacement.

Use the definition of electric field in terms of change in potential over a displacement and solve for required potential.

Fundamentals

The magnitude of electric field (E) is equal to amount of change in potential ( $\Delta V$ ) over a displacement ( $\Delta r$ ) along the electric field direction.

$E = \frac{{\Delta V}}{{\Delta r}}$

The expression for the electric field is,

$E = \frac{{\Delta V}}{{\Delta r}}$

Rearrange the equation for $\Delta V$ .

$\Delta V = E\left( {\Delta r} \right)$

The final expression for potential difference between the ends of wire is,

$\Delta V = E\left( {\Delta r} \right)$

Convert the unit of length of the wire from cm to m.

$\begin{array}{c}\\\Delta r = 30{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)\\\\ = 0.30{\rm{ m}}\\\end{array}$

Substitute 0.010 V/m for E and 0.30 m for $\Delta r$ in the equation $\Delta V = E\left( {\Delta r} \right)$ .

$\begin{array}{c}\\\Delta V = \left( {0.010{\rm{ V/m}}} \right)\left( {0.30{\rm{ m}}} \right)\\\\ = {\rm{0}}{\rm{.003 V}}\\\end{array}$

Ans:

The required of potential difference is 0.003 V.