Irodov 3.15: Equal Angles Give Equal Fields – Amit & Deepali Sinha Foundation

Problem Irodov 3.15

A thread with uniform charge $\lambda$ per unit length is bent in one of two ways, each with a circular bend of radius $R$ about a center $O$ , and with straight parts much longer than $R$ that continue the bend smoothly. (a) An L shape: two perpendicular straight arms joined by a quarter circle. (b) A hairpin: two parallel straight arms, pointing the same way, joined by a half circle. In each case find the magnitude of the electric field at $O$ .

(a) Two perpendicular arms joined by a quarter circle of radius R about O. — (a) Two perpendicular arms joined by a quarter circle of radius $R$ about $O$ .

(b) Two parallel arms joined by a half circle of radius R about O. — (b) Two parallel arms joined by a half circle of radius $R$ about $O$ .

Given

charge $\lambda$ per unit length, uniform
bend radius $R$ about $O$
arms much longer than $R$

Find

the magnitude of

\vect E

O

for (a) and (b)

Solution

Step 1Every element gives the same field per unit angle

In both shapes every straight arm continues the arc along its tangent, so the perpendicular from $O$ to each arm has length $R$ and meets the arm where it joins the arc. Take an element of an arm seen from $O$ at the angle $\alpha$ from that perpendicular; it lies at distance $R\tan\alpha$ from the join and at distance $r = R/\cos\alpha$ from $O$ .

An element dq of one arm of (a), seen from O at the angle α from the perpendicular R to the arm, and its field d E at O, along the line from the element; drawn for λ > 0. — An element $dq$ of one arm of (a), seen from $O$ at the angle $\alpha$ from the perpendicular $R$ to the arm, and its field $d\vect E$ at $O$ , along the line from the element; drawn for $\lambda > 0$ .

Differentiating $R\tan\alpha$ gives the element’s length, and Coulomb’s law its field at $O$ :

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

An element of the arc that subtends $d\alpha$ at $O$ has length $R\,d\alpha$ and lies at distance $R$ , so it gives the same:

dE = \frac{1}{4\pi\varepsilon_0}\,\frac{\lambda R\,d\alpha}{R^2} = \frac{\lambda\,d\alpha}{4\pi\varepsilon_0 R}

So label each element by the direction $\psi$ , the polar angle from $O$ to the element, measured from the $x$ axis to the right. For $\lambda > 0$ its field points from the element through $O$ , along $\psi + \pi$ :

d\vect E = -\frac{\lambda}{4\pi\varepsilon_0 R}\left(\cos\psi\,\vect i + \sin\psi\,\vect j\right)d\psi \tag{1}

and the total field is this integrated over every direction in which $O$ sees the thread.

Step 2(a) The L shape covers three quarters of a turn

With $O$ at the origin, the vertical arm is the line $x = -R$ , $y \ge 0$ , the arc runs through the third quadrant and the horizontal arm is $y = -R$ , $x \ge 0$ . Seen from $O$ , the vertical arm fills the directions from $\psi = \pi/2$ (its far end) to $\pi$ (the join), the arc from $\pi$ to $3\pi/2$ , and the horizontal arm from $3\pi/2$ to $2\pi$ (its far end). Integrating (1) over $\pi/2 \le \psi \le 2\pi$ :

\begin{aligned} \vect E &= -\frac{\lambda}{4\pi\varepsilon_0 R}\left(\Big[\sin\psi\Big]_{\pi/2}^{2\pi}\vect i + \Big[-\cos\psi\Big]_{\pi/2}^{2\pi}\vect j\right)\\ &= -\frac{\lambda}{4\pi\varepsilon_0 R}\left(-\vect i - \vect j\right) = \frac{\lambda}{4\pi\varepsilon_0 R}\left(\vect i + \vect j\right) \end{aligned}

The field points along the bisector of the corner, away from the bend, and its magnitude is

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

(2)

Step 3(b) The hairpin covers a full turn

Now the arms are $x = \pm R$ , $y \ge 0$ , and the half circle fills the lower half plane. Seen from $O$ , the left arm fills $\pi/2$ to $\pi$ , the arc $\pi$ to $2\pi$ , and the right arm $0$ to $\pi/2$ (from its join at $\psi = 0$ to its far end). Together these are every direction exactly once, so (1) integrates over a full turn:

\vect E = -\frac{\lambda}{4\pi\varepsilon_0 R}\left(\Big[\sin\psi\Big]_{0}^{2\pi}\vect i + \Big[-\cos\psi\Big]_{0}^{2\pi}\vect j\right) = 0

E = 0

(3)

Check

Piece by piece for (a), with $k = \lambda/(4\pi\varepsilon_0R)$ . Integrating (1) over the quarter arc ( $\pi$ to $3\pi/2$ ), the vertical arm ( $\pi/2$ to $\pi$ ) and the horizontal arm ( $3\pi/2$ to $2\pi$ ) separately:

\begin{aligned} \vect E_{\mathrm{arc}} &= -k\left[(-1)\,\vect i + (-1)\,\vect j\right] = k(\vect i + \vect j)\\ \vect E_{\mathrm{vert}} &= -k\left[(-1)\,\vect i + (1)\,\vect j\right] = k(\vect i - \vect j)\\ \vect E_{\mathrm{horiz}} &= -k\left[(1)\,\vect i + (-1)\,\vect j\right] = k(-\vect i + \vect j) \end{aligned}

The two arms cancel exactly, and the whole field comes from the arc, of magnitude $\sqrt{2}\,k$ , as found. Each arm alone is a half-infinite thread seen from the perpendicular through its end, with equal components along and across it.

Remark

Seen from $O$ , the shape of the thread does not matter, only the directions it fills, provided every element’s tangent line passes at the distance $R$ . The hairpin fills every direction once, exactly as a uniform ring of radius $R$ does seen from its center, and so it leaves $O$ field-free for the same reason the ring does.