4 Engineering Example 7

4.1 Field strength around a charged line

Problem in words

Find the electric field strength at a given distance from a uniformly charged line.

Mathematical statement of problem

Determine the electric field at a distance $r$ from a uniformly charged line (charge per unit length ${\rho }_L$ ). You may assume that the field points directly away from the line, using symmetry arguments.

Figure 21:

Mathematical analysis

Imagine a cylinder a distance $r$ from the line and of length $l$ . From Gauss’ law

$\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\underset{̲}{E}\cdot \underset{̲}{dS}=\frac{Q}{{\epsilon }_0}$

As the charge per unit length is ${\rho }_L$ , then the right-hand side equals ${\rho }_{L}l∕{\epsilon }_0$ . On the left-hand side, the integral can be expressed as the sum

$\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\underset{̲}{E}\cdot \underset{̲}{dS}=\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_{S\left(ends\right)}\underset{̲}{E}\cdot \underset{̲}{dS}+\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_{S\left(curved\right)}\underset{̲}{E}\cdot \underset{̲}{dS}$

Looking first at the circular ends of the cylinder, the fact that the field lines point (radially) away from the charged line implies that the electric field is in the plane of these circles and has no normal component. Therefore $\underset{̲}{E}\cdot \underset{̲}{dS}$ will be zero.

Next, over the curved surface of the cylinder, the electric field is normal to it, and the symmetry of the problem implies that the strength of the electric field will be constant (here denoted $E$ ). Therefore the integral $=$ Total curved surface area $\times$ Field strength $=2\pi rlE$ .

So, going back to Gauss’ law

$\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_{S\left(ends\right)}\underset{̲}{E}\cdot \underset{̲}{dS}+\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_{S\left(curved\right)}\underset{̲}{E}\cdot \underset{̲}{dS}=\frac{Q}{{\epsilon }_0}$

or

$0+2\pi rlE=\frac{{\rho }_{L}l}{{\epsilon }_0}$

Interpretation

Hence, the field strength $E$ is given by $E=\frac{{\rho }_L}{2\pi {\epsilon }_{0}r}$