Irodov 3.22: Strongest Field Between Twin Threads – Amit & Deepali Sinha Foundation

Problem Irodov 3.22

Two long parallel threads, a distance $l$ apart, each carry the same uniform charge $\lambda$ per unit length. Consider the plane of symmetry that lies midway between the threads. Find the largest magnitude that the electric field reaches anywhere in this plane.

The two threads seen end-on, each carrying λ per unit length, a distance l apart; the dash-dot line is the symmetry plane between them, seen edge-on. — The two threads seen end-on, each carrying $\lambda$ per unit length, a distance $l$ apart; the dash-dot line is the symmetry plane between them, seen edge-on.

Given

$\lambda$ on each thread
separation $l$

Find

the maximum of

E

in the symmetry plane

Solution

Step 1The field of one thread

By symmetry the field of a long straight thread is radial and has the same magnitude all round a coaxial cylinder of radius $r$ and length $\ell$ , so Gauss’s law on that cylinder gives

E\cdot 2\pi r\ell = \frac{\lambda\ell}{\varepsilon_0} \quad\Longrightarrow\quad E = \frac{\lambda}{2\pi\varepsilon_0 r} \tag{1}

pointing straight away from the thread for $\lambda > 0$ .

Step 2Add the two fields at a point of the plane

A point P of the symmetry plane at distance x from the plane of the threads. Both threads are at the same distance r from it; their fields E_1 and E_2 add up to E along the symmetry plane. — A point $P$ of the symmetry plane at distance $x$ from the plane of the threads. Both threads are at the same distance $r$ from it; their fields $\vect E_1$ and $\vect E_2$ add up to $\vect E$ along the symmetry plane.

Take a point $P$ of the symmetry plane at distance $x$ from the plane containing the threads. Each thread is at the same distance from it:

r^2 = x^2 + \frac{l^2}{4}

The two fields (1) have equal magnitude. Their components across the symmetry plane are mirror images and cancel; their components along it, each the fraction $x/r$ of the field, add:

E = 2\cdot\frac{\lambda}{2\pi\varepsilon_0 r}\cdot\frac{x}{r} = \frac{\lambda x}{\pi\varepsilon_0 r^2}

Substituting $r^2$ :

E(x) = \frac{\lambda}{\pi\varepsilon_0}\,\frac{x}{x^2 + l^2/4} \tag{2}

Step 3Find the maximum

Differentiate (2) with respect to $x$ and set the derivative to zero:

\frac{dE}{dx} = \frac{\lambda}{\pi\varepsilon_0}\,\frac{(x^2 + l^2/4) - 2x^2}{(x^2 + l^2/4)^2} = \frac{\lambda}{\pi\varepsilon_0}\,\frac{l^2/4 - x^2}{(x^2 + l^2/4)^2} = 0

The numerator vanishes at

x_m = \frac{l}{2}

where the derivative changes sign from positive to negative, so this is the maximum. Putting $x_m$ into (2):

E_{\max} = \frac{\lambda}{\pi\varepsilon_0}\,\frac{l/2}{l^2/4 + l^2/4} = \frac{\lambda}{\pi\varepsilon_0}\,\frac{l/2}{l^2/2}

E_{\max} = \frac{\lambda}{\pi\varepsilon_0 l}

(3)

It is reached at the distance $x_m = l/2$ from the plane of the threads, on either side of it, where the lines to the two threads meet at a right angle.

The field magnitude along the symmetry plane against the distance x from the plane of the threads: zero midway between them, a single maximum at x = l/2, then a fall as 1/x. — The field magnitude along the symmetry plane against the distance $x$ from the plane of the threads: zero midway between them, a single maximum at $x = l/2$ , then a fall as $1/x$ .

Check

Far away, $x \gg l$ , the term $l^2/4$ in (2) is negligible:

E \approx \frac{\lambda}{\pi\varepsilon_0 x} = \frac{2\lambda}{2\pi\varepsilon_0 x}

the field of a single thread carrying $2\lambda$ , as it must look from a distance. Units: $\lambda/(\varepsilon_0 l)$ has the units of the thread field (1).

Remark

At the maximum each thread is seen at $45^\circ$ from the plane of the threads, so each thread is $l/\sqrt 2$ away and each field contributes the fraction $1/\sqrt 2$ of itself along the plane; the two factors of $\sqrt 2$ cancel each other, and adding the two threads removes the 2 in $2\pi\varepsilon_0$ , which is why the answer is so simple. The trade-off is general: moving out reduces each field as $1/r$ while the useful fraction $x/r$ grows, and the right angle is where the two balance.