Irodov 3.8: Only the Symmetry Axis Survives – Amit & Deepali Sinha Foundation

Problem Irodov 3.8

A thin wire bent into a semicircle of radius $R = 20$ cm carries a charge $q = 0.70$ nC spread evenly along its length. Find the magnitude of the electric field at the center of the circle the semicircle belongs to.

The half-ring of radius R carrying charge q, and the center O of its circle, where the field is wanted. — The half-ring of radius $R$ carrying charge $q$ , and the center $O$ of its circle, where the field is wanted.

Given

$R = 20$ cm
$q = 0.70$ nC, uniform along the half-ring

Find

the magnitude

E

of the field at the center

O

Solution

Step 1The field of one element

Cut the half-ring into elements labeled by the angle $\varphi$ from one end, with the $x$ axis along the diameter and the $y$ axis along the symmetry line. The charge per unit length is uniform:

\lambda = \frac{q}{\pi R} \tag{1}

One element dq at angle φ and its field d E at O, resolved into dE_x along the diameter and dE_y along the symmetry line. — One element $dq$ at angle $\varphi$ and its field $d\vect E$ at $O$ , resolved into $dE_x$ along the diameter and $dE_y$ along the symmetry line.

The element spans $d\varphi$ , so it has length $R\,d\varphi$ and charge $dq = \lambda R\,d\varphi$ . Every element is the same distance $R$ from $O$ , and Coulomb’s law gives its field there:

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

For positive charge $d\vect E$ points from the element through $O$ , so it makes the angle $\varphi + \pi$ with the $x$ axis.

Step 2Symmetry kills the component along the diameter

Resolving $d\vect E$ along the two axes:

\begin{aligned} dE_x &= -dE\cos\varphi\\ dE_y &= -dE\sin\varphi \end{aligned}

Summing the $x$ components with (2) confirms the cancellation:

E_x = -\frac{\lambda}{4\pi\varepsilon_0 R}\int_0^{\pi}\cos\varphi\,d\varphi = -\frac{\lambda}{4\pi\varepsilon_0 R}\Big[\sin\varphi\Big]_0^{\pi} = 0

The $y$ components all point the same way, away from the half-ring, and their sum is the whole field:

\begin{aligned} E = \lvert E_y\rvert &= \frac{\lambda}{4\pi\varepsilon_0 R}\int_0^{\pi}\sin\varphi\,d\varphi\\ &= \frac{\lambda}{4\pi\varepsilon_0 R}\Big[-\cos\varphi\Big]_0^{\pi} = \frac{2\lambda}{4\pi\varepsilon_0 R} \end{aligned}

E = \frac{\lambda}{2\pi\varepsilon_0 R}

(3)

Step 3In terms of the total charge

Substituting (1) into (3):

E = \frac{1}{2\pi\varepsilon_0 R}\cdot\frac{q}{\pi R}

Numerically, with $\varepsilon_0 = 8.85\times10^{-12}$ F/m:

E = \frac{0.70\times10^{-9}}{2\pi^2\,(8.85\times10^{-12})\,(0.20)^2} = \frac{0.70\times10^{-9}}{6.99\times10^{-12}} \approx 100 \text{ V/m}

E = \dfrac{q}{2\pi^2\varepsilon_0 R^2} \approx 0.10 \text{ kV/m}

(4)

The field points along the symmetry line, away from the half-ring.

Check

Units: $q/(\varepsilon_0R^2)$ has the units of $q/(4\pi\varepsilon_0 r^2)$ , a field. Size: a point charge $q$ at distance $R$ gives $q/(4\pi\varepsilon_0R^2)$ , and the ratio of the two is

\frac{q/(2\pi^2\varepsilon_0R^2)}{q/(4\pi\varepsilon_0R^2)} = \frac{2}{\pi} \approx 0.64,

less than one, as it must be, because the elements near the ends of the half-ring push mostly sideways and those pushes cancel.

Remark

The result is also the field of an infinite straight line of the same density $\lambda$ at distance $R$ , $\lambda/(2\pi\varepsilon_0R)$ , and the match is exact element by element. A thread element that subtends the angle $d\alpha$ at $O$ gives a field $\lambda\,d\alpha/(4\pi\varepsilon_0 h)$ along its line of sight, where $h$ is the distance from $O$ to the element’s tangent line: seen at angle $\alpha$ from the foot of that perpendicular, the element lies at distance $r = h/\cos\alpha$ and has length $dl = h\,d\alpha/\cos^2\alpha$ . On the half-ring and on the line tangent to it at its midpoint, $h = R$ for every element, and the two sweep the same half-turn of directions.