Kvant Physics Problem 189

A point charge $q = 10^{-8},\text{C}$ is uniformly distributed 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 a subtended 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 point charge $q = 10^{-8},\text{C}$ is uniformly distributed along a circular arc of radius $R = 1,\text{cm} = 10^{-2},\text{m}$. Two cases are considered for the subtended central angle: $\theta = \pi$ and $\theta = \frac{2}{3}\pi$.

The electric field is required at the center of the circle. The charge distribution is assumed to lie in a plane, with uniform linear charge density along the arc. The medium is vacuum, so the Coulomb constant is $k = \frac{1}{4\pi\varepsilon_0} = 9.0 \times 10^{9},\text{N}\cdot\text{m}^2/\text{C}^2$. Edge effects and any external fields are neglected.

Physical Principles

The electric field contribution from a point charge element $dq$ is given by Coulomb’s law in vector form,

$d\mathbf{E} = k \frac{dq}{r^2} \hat{\mathbf{r}}.$

For a continuous charge distribution, the total field is obtained by integration,

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

The linear charge density along the arc is

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

Each charge element $dq = \lambda R,d\varphi$ produces a field at the center of magnitude $dE = k \frac{dq}{R^2}$ directed along the radius toward the center. By symmetry, transverse components cancel and only the component along the bisector of the arc remains.

Derivation

Let the arc be symmetric about the $x$-axis, with angular coordinate $\varphi \in [-\theta/2, \theta/2]$. A charge element is

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

The field magnitude from this element at the center is

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

Only the component along the symmetry axis survives after integration. The perpendicular components cancel by antisymmetry, and the axial component is

$dE_x = dE \cos\varphi = k \frac{q}{\theta R^2} \cos\varphi , d\varphi.$

Integration over the arc gives

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

Evaluating the integral,

$\int_{-\theta/2}^{\theta/2} \cos\varphi , d\varphi = 2\sin\left(\frac{\theta}{2}\right).$

Thus the field magnitude at the center is

$E = \frac{2kq}{\theta R^2}\sin\left(\frac{\theta}{2}\right).$

Result

For $\theta = \pi$,

$E_1 = \frac{2kq}{\pi R^2}\sin\left(\frac{\pi}{2}\right) = \frac{2kq}{\pi R^2}.$

Substitution gives

$E_1 = \frac{2 \cdot 9.0 \times 10^{9},\text{N}\cdot\text{m}^2/\text{C}^2 \cdot 10^{-8},\text{C}}{\pi \cdot (10^{-2},\text{m})^2}.$

Using $R^2 = 10^{-4},\text{m}^2$,

$E_1 = \frac{180}{\pi \cdot 10^{-4}},\text{N/C} \approx 5.7 \times 10^{5},\text{N/C}.$

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

$E_2 = \frac{2kq}{\left(\frac{2\pi}{3}\right)R^2}\sin\left(\frac{\pi}{3}\right).$

Using $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$,

$E_2 = \frac{2kq}{\left(\frac{2\pi}{3}\right)R^2} \cdot \frac{\sqrt{3}}{2} = \frac{3kq\sqrt{3}}{2\pi R^2}.$

Substitution yields

$E_2 = \frac{3 \cdot 9.0 \times 10^{9} \cdot 10^{-8} \cdot \sqrt{3}}{2\pi \cdot 10^{-4}},\text{N/C} \approx 7.4 \times 10^{5},\text{N/C}.$

Final results:

$\boxed{E_1 \approx 5.7 \times 10^{5},\text{N/C}}, \quad \boxed{E_2 \approx 7.4 \times 10^{5},\text{N/C}}.$

Sanity Checks

The expression $E = \frac{2kq}{\theta R^2}\sin\left(\frac{\theta}{2}\right)$ has dimensions of electric field since $kq/R^2$ already carries units of $\text{N/C}$ and the remaining factor is dimensionless.

In the limit of small $\theta$, $\sin(\theta/2) \approx \theta/2$, giving $E \approx kq/R^2$, consistent with the field of a point charge when the arc collapses toward a single location.

For increasing arc length, the factor $\sin(\theta/2)/\theta$ decreases, reflecting partial cancellation of vector contributions from widely separated charge elements. The numerical values satisfy this trend since the shorter arc $\theta = \frac{2}{3}\pi$ produces a slightly larger field at the center than the semicircular case.