Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Home exercise 3.3.1

Finite line source

A line charge of uniform density $ρ_{L}$ lies on the $x$ -axis from $x = - L$ to $x = L$ . Find the electrostatic potential at the point $P (0, y, 0)$ , and from it the electric field at $P$ .

Source: D. K. Cheng, Field and Wave Electromagnetics, 2nd ed., Problem P.3-16.

Try it yourself first.

Strategy

Superposition of charges: one scalar integral of $ρ_{L} d x^{'} / (4 π ε_{0} R)$ along the line, then $E = - \nabla V$ . Symmetry tells us that on the $y$ -axis only $E_{y}$ survives, so the derivative along the axis is enough (see the mistake box on differentiating a potential known only on a line, Section 3.3.1).

Potential

A source point on the line is $Q (x^{'}, 0, 0)$ , so the displacement vector is

R = r - r^{'} = - x^{'} a_{x} + y a_{y} \Rightarrow R = (x^{'})^{2} + y^{2} .

With $d q^{'} = ρ_{L} d x^{'}$ the potential integral is

V = \frac{1}{4 π ε _{0}} \int_{x^{'} = - L}^{L} \frac{ρ _{L} d x ^{'}}{( x ^{'} ) ^{2} + y ^{2}} = \frac{ρ _{L}}{4 π ε _{0}} [ln (x^{'} + (x^{'})^{2} + y^{2})]_{x^{'} = - L}^{L},

using the standard integral $\int d u / u^{2} + c^{2} = ln (u + u^{2} + c^{2})$ . This gives

Answer

V (0, y, 0) = \frac{ρ _{L}}{4 π ε _{0}} ln (\frac{L ^{2} + y ^{2} + L}{L ^{2} + y ^{2} - L}) [V]

Field

On the $y$ -axis the field has only a $y$ -component: the contributions of the elements at $+ x^{'}$ and $- x^{'}$ cancel along $x$ , and all charges lie in the plane $z = 0$ . The $y$ -derivative can therefore be taken along the $y$ -axis itself. Writing $s = L^{2} + y^{2}$ , so that $d s / d y = y / s$ ,

E = - \nabla V = - \frac{d V}{d y} a_{y} = - a_{y} \frac{ρ _{L}}{4 π ε _{0}} \frac{d}{d y} [ln (s + L) - ln (s - L)] = - a_{y} \frac{ρ _{L}}{4 π ε _{0}} \frac{y}{s} [\frac{1}{s + L} - \frac{1}{s - L}] .

The bracket equals $- 2 L / (s^{2} - L^{2}) = - 2 L / y^{2}$ , so

E = - a_{y} \frac{ρ _{L}}{4 π ε _{0}} \frac{y}{s} (- \frac{2 L}{y ^{2}}),

Answer

E = a_{y} \frac{ρ _{L}}{2 π ε _{0}} \frac{L}{y L ^{2} + y ^{2}} [V/m]

Home exercise 3.3.1: the potential on the y-axis for half-lengths L=1, 3 and 10 (same length unit as y, same \rho_L). Making the line longer adds charge and raises the whole curve, by about 2\ln of the length ratio for y\ll L. For an infinite line there is no finite potential with V(\infty)=0. — **Figure 3.3.25.** Home exercise 3.3.1: the potential on the $y$ -axis for half-lengths $L = 1$ , $3$ and $10$ (same length unit as $y$ , same $ρ_{L}$ ). Making the line longer adds charge and raises the whole curve, by about $2 ln$ of the length ratio for $y ≪ L$ . For an infinite line there is no finite potential with $V (\infty) = 0$ .

Check

Far away ( $y ≫ L$ ), $s ≃ y$ and $ln \frac{s + L}{s - L} ≃ \frac{2 L}{y}$ , so

V \to \frac{2 L ρ _{L}}{4 π ε _{0} y}, E \to \frac{2 L ρ _{L}}{4 π ε _{0} y ^{2}} :

a point charge $Q = 2 L ρ_{L}$ , as expected. The unit of $ρ_{L} / ε_{0}$ is $(C/m) / (F/m) = V$ .

Remark

This is the result of Home exercise 3.1.1, obtained there by integrating $d E$ directly. With the potential only one scalar integral is needed. For $L \to \infty$ the field tends to $ρ_{L} / (2 π ε_{0} y)$ , the infinite line charge of Problem 3.1.1. The potential, however, grows without bound (Figure 3.3.25): since $(s + L) / (s - L) = (s + L)^{2} / y^{2} \to 4 L^{2} / y^{2}$ , $V \to \frac{ρ _{L}}{4 π ε _{0}} ln (4 L^{2} / y^{2})$ . For an infinitely long line the reference $V (\infty) = 0$ cannot be used.

Home exercise 3.3.1: Finite line source