Irodov 3.13: Two Views of a Finite Rod – Amit & Deepali Sinha Foundation

Problem Irodov 3.13

A thin straight rod of length $2a$ in vacuum carries a charge $q$ spread evenly along it. Find the magnitude of the electric field as a function of the distance $r$ from the rod’s center, at points:

on the line through the center perpendicular to the rod;
on the line of the rod itself, outside the rod.

Examine both results for $r \gg a$ .

The rod of length 2a with charge q; A is on the perpendicular through the center and B on the rod's own line, each at distance r from the center; drawn for q > 0. — The rod of length $2a$ with charge $q$ ; $A$ is on the perpendicular through the center and $B$ on the rod’s own line, each at distance $r$ from the center; drawn for $q > 0$ .

Given

rod of length $2a$
charge $q$ , uniform

Find

E(r)

(a) on the perpendicular bisector, (b) on the rod’s line outside the rod; both for

r \gg a

Solution

Step 1The charge per unit length

The charge is spread evenly over the length $2a$ :

\lambda = \frac{q}{2a} \tag{1}

Step 2(a) Elements seen from the perpendicular

Label each element by the angle $\alpha$ at $A$ between the perpendicular to the rod and the line to the element. The element then lies at distance $\xi = r\tan\alpha$ from the center, at distance $r/\cos\alpha$ from $A$ .

One element dq, seen from A at the angle α from the perpendicular r, and its field d E at A with the components dE_x along the rod and dE_y along the perpendicular. — One element $dq$ , seen from $A$ at the angle $\alpha$ from the perpendicular $r$ , and its field $d\vect E$ at $A$ with the components $dE_x$ along the rod and $dE_y$ along the perpendicular.

Differentiating $\xi = r\tan\alpha$ gives the element’s length, and Coulomb’s law its field at $A$ :

\begin{aligned} d\xi &= \frac{r\,d\alpha}{\cos^2\alpha}\\ dE &= \frac{1}{4\pi\varepsilon_0}\,\frac{\lambda\,d\xi}{(r/\cos\alpha)^2} = \frac{\lambda\,d\alpha}{4\pi\varepsilon_0 r} \end{aligned}

The elements at $\alpha$ and $-\alpha$ cancel each other’s components along the rod, so only $dE_y = dE\cos\alpha$ survives. The rod’s ends are seen at $\pm\alpha_0$ , where

\sin\alpha_0 = \frac{a}{\sqrt{a^2 + r^2}} \tag{2}

Summing over the rod:

E = \frac{\lambda}{4\pi\varepsilon_0 r}\int_{-\alpha_0}^{\alpha_0}\cos\alpha\,d\alpha = \frac{\lambda}{4\pi\varepsilon_0 r}\Big[\sin\alpha\Big]_{-\alpha_0}^{\alpha_0} = \frac{2\lambda\sin\alpha_0}{4\pi\varepsilon_0 r}

With (1) and (2):

E = \frac{2}{4\pi\varepsilon_0 r}\cdot\frac{q}{2a}\cdot\frac{a}{\sqrt{a^2 + r^2}}

E = \frac{q}{4\pi\varepsilon_0 r\sqrt{a^2 + r^2}}

(3)

Step 3(b) Elements on the rod’s own line

Now every element’s field at $B$ lies along the rod’s line, pointing away from the rod, so the fields add as plain numbers. The element at coordinate $\xi$ from the center, $-a \le \xi \le a$ , is at distance $r - \xi$ from $B$ .

One element dq at coordinate ξ from the center O, at distance r - ξ from B, and its field d E at B along the rod's line. — One element $dq$ at coordinate $\xi$ from the center $O$ , at distance $r - \xi$ from $B$ , and its field $d\vect E$ at $B$ along the rod’s line.

Summing Coulomb’s law over the rod:

\begin{aligned} E &= \frac{\lambda}{4\pi\varepsilon_0}\int_{-a}^{a}\frac{d\xi}{(r-\xi)^2} = \frac{\lambda}{4\pi\varepsilon_0}\left[\frac{1}{r-\xi}\right]_{-a}^{a}\\ &= \frac{\lambda}{4\pi\varepsilon_0}\left(\frac{1}{r-a} - \frac{1}{r+a}\right) = \frac{\lambda}{4\pi\varepsilon_0}\cdot\frac{2a}{r^2-a^2} \end{aligned}

With (1), $\lambda\cdot 2a = q$ :

E = \frac{q}{4\pi\varepsilon_0 (r^2 - a^2)}

(4)

Step 4Far from the rod

For $r \gg a$ the length $a$ drops out of both denominators:

\begin{aligned} r\sqrt{a^2 + r^2} &= r^2\sqrt{1 + a^2/r^2} \approx r^2\\ r^2 - a^2 &= r^2\left(1 - a^2/r^2\right) \approx r^2 \end{aligned}

so (3) and (4) both reduce to the field of a point charge:

E \approx \frac{q}{4\pi\varepsilon_0 r^2}

(5)

Check

Limits. In (a), as $a \to \infty$ with $\lambda$ fixed, $q/\sqrt{a^2 + r^2} \to 2\lambda$ and (3) becomes $\lambda/(2\pi\varepsilon_0 r)$ , the field of an infinite thread. In (b), as $r \to a$ the point reaches the rod’s end and $E \to \infty$ , as it must next to a line charge. Units: $q/(\varepsilon_0 r^2)$ in both.

Remark

The first correction tells the two directions apart. Expanding to order $a^2/r^2$ , (a) gives $q/(4\pi\varepsilon_0 r^2)\,(1 - a^2/(2r^2))$ and (b) gives $q/(4\pi\varepsilon_0 r^2)\,(1 + a^2/r^2)$ : along the rod’s line the nearer half of the charge gains more than the farther half loses, because $1/d^2$ rises steeply as $d$ shrinks, so the field is stronger than a point charge’s; broadside every element is farther than the center and pushes partly sideways, so the field is weaker.