Irodov 3.14: Half a Thread Tilts the Field – Amit & Deepali Sinha Foundation

Problem Irodov 3.14

A very long straight thread carries a uniform charge $\lambda$ per unit length. A point lies on the line drawn through one end of the thread perpendicular to it, at a distance $y$ from the thread. Find the magnitude and the direction of the electric field at that point.

The thread of density λ, starting at the origin and running to the right without end, and the point P on the perpendicular through its end at distance y; drawn for λ > 0. — The thread of density $\lambda$ , starting at the origin and running to the right without end, and the point $P$ on the perpendicular through its end at distance $y$ ; drawn for $\lambda > 0$ .

Given

thread density $\lambda$
one end, unbounded otherwise
$P$ on the perpendicular through the end
distance $y$

Find

the magnitude and direction of

\vect E

P

Solution

Step 1The field of one element

Put the end at the origin, the thread along $+x$ and $P$ at height $y$ on the $y$ axis. Label each element by the angle $\alpha$ at $P$ between the perpendicular and the line to the element; $\alpha$ runs from $0$ at the end to $\pi/2$ far along the thread. The element lies at $x = y\tan\alpha$ , at distance $y/\cos\alpha$ from $P$ .

One element dq, seen from P at the angle α from the perpendicular y, and its field d E at P with its components. — One element $dq$ , seen from $P$ at the angle $\alpha$ from the perpendicular $y$ , and its field $d\vect E$ at $P$ with its components.

Differentiating $x = y\tan\alpha$ gives the element’s length, and Coulomb’s law its field at $P$ :

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

For $\lambda > 0$ the element’s field points from the element through $P$ , which has components $-\sin\alpha$ along $x$ and $\cos\alpha$ along $y$ :

dE_x = -\frac{\lambda\sin\alpha\,d\alpha}{4\pi\varepsilon_0 y},\qquad dE_y = \frac{\lambda\cos\alpha\,d\alpha}{4\pi\varepsilon_0 y} \tag{1}

Step 2Sum over the thread

Integrating (1) from $\alpha = 0$ to $\pi/2$ :

\begin{aligned} E_x &= -\frac{\lambda}{4\pi\varepsilon_0 y}\int_0^{\pi/2}\sin\alpha\,d\alpha = -\frac{\lambda}{4\pi\varepsilon_0 y}\Big[-\cos\alpha\Big]_0^{\pi/2} = -\frac{\lambda}{4\pi\varepsilon_0 y}\\ E_y &= \frac{\lambda}{4\pi\varepsilon_0 y}\int_0^{\pi/2}\cos\alpha\,d\alpha = \frac{\lambda}{4\pi\varepsilon_0 y}\Big[\sin\alpha\Big]_0^{\pi/2} = \frac{\lambda}{4\pi\varepsilon_0 y} \end{aligned}

\vect E = \frac{\lambda}{4\pi\varepsilon_0 y}\left(\vect j - \vect i\right)

(2)

The two components are equal in size, so the magnitude is

E = \frac{\sqrt{2}\,\lambda}{4\pi\varepsilon_0 y}

and $\vect E$ makes $45^\circ$ with the thread (and with the perpendicular), leaning away from the side the thread lies on, for $\lambda > 0$ ; for $\lambda < 0$ it is reversed.

The total field E at P, at 45° to the perpendicular through the end, leaning away from the thread; drawn for λ > 0. — The total field $\vect E$ at $P$ , at $45^\circ$ to the perpendicular through the end, leaning away from the thread; drawn for $\lambda > 0$ .

Check

Units: $\lambda/(\varepsilon_0 y)$ is the form of the infinite thread’s field $\lambda/(2\pi\varepsilon_0 y)$ . Superposition: a second half-thread on the other side of the perpendicular gives, by mirror symmetry, $E_x = +\lambda/(4\pi\varepsilon_0 y)$ and the same $E_y$ ; the two together make $E_y = \lambda/(2\pi\varepsilon_0 y)$ with no $x$ component, exactly the field of the whole infinite thread.

Remark

The $45^\circ$ is independent of $y$ : the half-thread looks the same at every scale, so the field’s direction cannot depend on the one length in the problem. The fan picture makes it visible: seen from $P$ the elements fill a quarter-turn of directions, from straight down to straight along the thread, so their pushes, each pointing back from its element through $P$ , fill a quarter-turn too, and equal pushes spread evenly over a quarter-turn add up along its bisector.