Question

A very long, straight wire has charge per unit length 3.20 x 10^-10 C/m. At what distance from the wire is the electric field magnitude equal to 2.50 N/C?

Short answer

Answer

The required distance is 2.30 m.

Step-by-step solution

Important formula

The equation which relates electric field and linear charge density for an infinite line of charge is:

$$\begin{gathered} \mathbf{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{o}}}\frac{\mathbf{\lambda }}{\mathbf{r}} \\ \mathit{r}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{\circ }}}\frac{\mathbf{\lambda }}{\mathbf{E}} \end{gathered}$$

where E=electric field, =linear charge density and$\frac{1}{4\pi {\epsilon }_{\circ }}=9.0\times {10}^9 {\text{N.m}}^2/{\text{C}}^2$

Substitution

Substitute the values to find r:

$$\begin{gathered} r=2(9.0\times {10}^9 {\text{N.m}}^2/{\text{C}}^2)\left(\frac{3.20\times {10}^{-10} {\text{C/m}}^{}}{2.50 \text{N/C}}\right) \\ =2.30 \text{m} \end{gathered}$$

The electric field magnitude will be equal to 2.50 N/C at a distance of 2.30 m.