Kvant Physics Problem 189

A total charge $q = 10^{-8},\text{C}$ is distributed uniformly along a circular arc of radius $R = 1,\text{cm} = 10^{-2},\text{m}$.

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Problem

Charge $q=10^{-8}\text{Кл}$ is uniformly distributed along an arc of a circle of radius $R=1\text{см}$ with an opening angle

  1. $\pi$ radians,
  2. $\dfrac{2}{3}\pi$ radians.

Determine the electric field strength at the center of the circle.

V. G. Svetozarov

Setup and Assumptions

A total charge $q = 10^{-8},\text{C}$ is distributed uniformly along a circular arc of radius $R = 1,\text{cm} = 10^{-2},\text{m}$. The arc lies in a plane and is centered on a circle whose center is the observation point where the electric field is required.

Two cases are considered for the arc opening angle: $\theta = \pi$ and $\theta = \frac{2\pi}{3}$.

The charge distribution is assumed to be static and continuous, with uniform linear charge density along the arc. Edge effects at the ends of the arc are neglected beyond the Coulomb interaction itself. The medium is vacuum, so the Coulomb constant is $k = \frac{1}{4\pi\varepsilon_0} = 9.0 \times 10^9,\text{N}\cdot\text{m}^2/\text{C}^2$.

The goal is the electric field vector at the center of curvature of the arc.

Physical Principles

The electric field produced by a point charge element $dq$ at distance $R$ has magnitude

$dE = \frac{k,dq}{R^2}.$

For a continuous charge distribution, the total field is the vector sum

$\mathbf{E} = \int d\mathbf{E}.$

Uniform distribution along an arc implies a constant linear charge density

$\lambda = \frac{q}{R\theta}.$

A charge element located at angular coordinate $\varphi$ contributes a field at the center directed along the line connecting the element and the center, with magnitude $dE = \frac{k,dq}{R^2}$. Only components along the symmetry axis of the arc survive after integration, while transverse components cancel due to symmetry.

Derivation

The arc is taken symmetric about its bisector. The angular coordinate $\varphi$ is measured from this bisector, ranging from $-\theta/2$ to $+\theta/2$.

An infinitesimal charge element is

$dq = \lambda R,d\varphi = \frac{q}{\theta},d\varphi.$

The field magnitude from this element at the center is

$dE = \frac{k}{R^2} dq = \frac{kq}{R^2\theta},d\varphi.$

The direction of $d\mathbf{E}$ is along the line from the charge element to the center. Its projection onto the symmetry axis equals $dE \cos\varphi$, since the angle between the radial direction of the element and the bisector is $\varphi$.

The total field along the symmetry axis becomes

$E = \int_{-\theta/2}^{\theta/2} \frac{kq}{R^2\theta} \cos\varphi, d\varphi.$

Carrying out the integration,

$E = \frac{kq}{R^2\theta} \left[ \sin\varphi \right]_{-\theta/2}^{\theta/2} = \frac{kq}{R^2\theta} \left( \sin\frac{\theta}{2} - \left(-\sin\frac{\theta}{2}\right) \right).$

This yields

$E = \frac{2kq}{R^2\theta}\sin\frac{\theta}{2}.$

The direction of the field is along the bisector of the arc and points toward the arc for positive charge, since each element produces a field at the center directed from the charge toward the center.

Result

The electric field magnitude at the center is

$E = \frac{2kq}{R^2\theta}\sin\frac{\theta}{2}.$

For $\theta = \pi$,

$E_1 = \frac{2kq}{R^2\pi}\sin\frac{\pi}{2} = \frac{2kq}{\pi R^2}.$

Substituting values $k = 9.0 \times 10^9,\text{N}\cdot\text{m}^2/\text{C}^2$, $q = 10^{-8},\text{C}$, $R^2 = (10^{-2},\text{m})^2 = 10^{-4},\text{m}^2$,

$\frac{kq}{R^2} = \frac{9.0 \times 10^9 \cdot 10^{-8}}{10^{-4}} = 9.0 \times 10^5,\text{N/C}.$

Thus,

$E_1 = \frac{2}{\pi} \cdot 9.0 \times 10^5,\text{N/C} \approx 5.7 \times 10^5,\text{N/C}.$

For $\theta = \frac{2\pi}{3}$,

$E_2 = \frac{2kq}{R^2 (2\pi/3)} \sin\frac{\pi}{3} = \frac{3kq}{\pi R^2} \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3},kq}{2\pi R^2}.$

Substituting numerical values,

$E_2 = \frac{3\sqrt{3}}{2\pi} \cdot 9.0 \times 10^5,\text{N/C} \approx 7.4 \times 10^5,\text{N/C}.$

Final boxed results:

$\boxed{E(\pi) \approx 5.7 \times 10^5,\text{N/C}}, \quad \boxed{E!\left(\frac{2\pi}{3}\right) \approx 7.4 \times 10^5,\text{N/C}}.$

Sanity Checks

The expression $E = \frac{2kq}{R^2\theta}\sin\frac{\theta}{2}$ has dimensions $\frac{\text{N}\cdot\text{m}^2/\text{C}^2 \cdot \text{C}}{\text{m}^2} = \text{N/C}$, confirming dimensional consistency.

For small $\theta$, the approximation $\sin(\theta/2) \approx \theta/2$ gives $E \approx \frac{kq}{R^2}$, matching the field of a concentrated charge at distance $R$, since the arc collapses to a point charge distribution.

Increasing $\theta$ from $\frac{2\pi}{3}$ to $\pi$ reduces the field because symmetry increases and transverse cancellation strengthens, consistent with the computed values $7.4 \times 10^5,\text{N/C}$ and $5.7 \times 10^5,\text{N/C}$.