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

Home exercise 3.1.1

Interactive: Finite Line Charge

Home exercise 3.1.1

Finite line source

cf. Problem 3.1.1

Find the electric field at the point $P (0, 0, h)$ due to a finite line charge that lies on the $y$ -axis from $y = - L$ to $y = + L$ and carries a uniform charge density $ρ_{L}$ .

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

Try it yourself first.

Strategy

The recipe with a straight line source, $d q^{'} = ρ_{L} d y^{'}$ , now of finite length. Every element at $+ y^{'}$ has a partner at $- y^{'}$ , so only $E_{z}$ survives. The remaining integral is the second standard integral, taken between finite limits.

Points and displacement vector

The source point is $Q (0, y^{'}, 0)$ with $- L \leq y^{'} \leq L$ . The displacement vector and its magnitude are

R R = (0 - 0) a_{x} + (0 - y^{'}) a_{y} + (h - 0) a_{z} = - y^{'} a_{y} + h a_{z}, = (y^{'})^{2} + h^{2} .

Charge element and integral

With $d l^{'} = d y^{'}$ the electric field at P is

E (P) = \frac{1}{4 π ε _{0}} \int_{y^{'} = - L}^{L} \frac{ρ _{L} ( - y ^{'} a _{y} + h a _{z} )}{[ ( y ^{'} ) ^{2} + h ^{2} ] ^{3/2}} d y^{'} = \frac{ρ _{L}}{4 π ε _{0}} {- a_{y} \int_{- L}^{L} \frac{y ^{'} d y ^{'}}{[ ( y ^{'} ) ^{2} + h ^{2} ] ^{3/2}} + h a_{z} \int_{- L}^{L} \frac{d y ^{'}}{[ ( y ^{'} ) ^{2} + h ^{2} ] ^{3/2}}} .

Integrate

Using $(y^{'})^{2} + h^{2} = t$ for the first integral and $y^{'} = h tan θ$ for the second one, exactly as in Problem 3.1.1 (these are the standard integrals (3.1.30)),

\int_{- L}^{L} \frac{y ^{'} d y ^{'}}{[ ( y ^{'} ) ^{2} + h ^{2} ] ^{3/2}} \int_{- L}^{L} \frac{d y ^{'}}{[ ( y ^{'} ) ^{2} + h ^{2} ] ^{3/2}} = - \frac{1}{( y ^{'} ) ^{2} + h ^{2}}_{y^{'} = - L}^{L} = 0, = \frac{1}{h ^{2}} \frac{y ^{'}}{( y ^{'} ) ^{2} + h ^{2}}_{y^{'} = - L}^{L} = \frac{2 L}{h ^{2} L ^{2} + h ^{2}} .

The $y$ -component vanishes because every element at $+ y^{'}$ has a twin at $- y^{'}$ . Finally,

Answer

E (P) = \frac{ρ _{L}}{4 π ε _{0}} h \frac{2 L}{h ^{2} L ^{2} + h ^{2}} a_{z} = \frac{ρ _{L} L}{2 π ε _{0} h L ^{2} + h ^{2}} a_{z} [V/m]

Check

For $L \to \infty$ , $L / L^{2} + h^{2} \to 1$ and $E \to ρ_{L} / (2 π ε_{0} h) a_{z}$ : the result of Problem 3.1.1 with $ρ = h$ . For $h ≫ L$ , $L^{2} + h^{2} \to h$ and $E \to 2 L ρ_{L} / (4 π ε_{0} h^{2}) a_{z}$ : the field of a point charge $Q = 2 L ρ_{L}$ . In general the finite line gives the infinite-line value multiplied by $L / L^{2} + h^{2} < 1$ , the sine of the angle under which half of the line is seen from P.

Figure 3.1.29 shows the whole curve on logarithmic axes, where both limits appear as straight lines: slope $- 1$ near the line, slope $- 2$ far away.

Home exercise 3.1.1 on logarithmic axes. For h\ll L the finite line is indistinguishable from an infinite line (E\propto1/h), while for h\gg L it acts like a point charge (E\propto1/h^2). Around h\approx L the curve turns from one behaviour to the other, and at h=L the field is 1/\sqrt2\approx71\ \% of the infinite-line value. — **Figure 3.1.29.** Home exercise 3.1.1 on logarithmic axes. For $h ≪ L$ the finite line is indistinguishable from an infinite line ( $E \propto 1/ h$ ), while for $h ≫ L$ it acts like a point charge ( $E \propto 1/ h^{2}$ ). Around $h \approx L$ the curve turns from one behaviour to the other, and at $h = L$ the field is $1/ 2 \approx 71 %$ of the infinite-line value.

Place the field point on the perpendicular bisector of the line and vary its distance: the computed field follows $ρ_{L} / (2 π ε_{0} ρ)$ close to the line and the point-charge field of $Q = 2 L ρ_{L}$ far away.

InteractiveLoading simulation…

Home exercise 3.1.1: Finite line source