Kvant Physics Problem 71

Consider a straight line segment $AB$ carrying a uniform linear charge density $\lambda$.

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Problem

F71. A uniformly charged line segment $AB$ is given. In what direction is the electric field intensity created by this segment at the point $C$ directed: along the median of triangle $ABC$? along its angle bisector? along its altitude? or along none of these lines?

MIPT Physics and Mathematics Olympiad, 1970.

Setup and Assumptions

Consider a straight line segment $AB$ carrying a uniform linear charge density $\lambda$. Place coordinates so that $A(-L/2,0)$ and $B(L/2,0)$, hence the segment lies on the $x$-axis and is symmetric with respect to the origin.

Let the observation point be $C(x_0,y_0)$ with $y_0\neq 0$. The goal is to determine the direction of the electric field $\mathbf E$ at $C$ and compare it with the median, angle bisector, and altitude of triangle $ABC$.

The field is computed in vacuum with permittivity $\varepsilon_0$.

Physical Principles

The contribution of an infinitesimal charge element $dq=\lambda,dx$ located at position $x$ is given by Coulomb’s law,

$$d\mathbf E=\frac{1}{4\pi\varepsilon_0}\frac{dq}{r^3}\mathbf r,$$

where $\mathbf r=(x_0-x,y_0)$ is the vector from the charge element to $C$ and $r^2=(x_0-x)^2+y_0^2$.

The total field is

$$\mathbf E=\frac{\lambda}{4\pi\varepsilon_0}\int_{-L/2}^{L/2}\frac{(x_0-x,;y_0)}{\big((x_0-x)^2+y_0^2\big)^{3/2}},dx.$$

Correct Symmetry Structure

The decisive property is not a naive geometric asymmetry argument but the pairing of charge elements under reflection about the midpoint of the segment.

Introduce the substitution $x\mapsto -x$ while keeping the observation point fixed. For every element at $x$ there is a corresponding element at $-x$ with equal charge $dq=\lambda,dx$.

For the $x$-component of the field,

$$dE_x(x)=\frac{\lambda}{4\pi\varepsilon_0}\frac{x_0-x}{\big((x_0-x)^2+y_0^2\big)^{3/2}},dx,$$

$$dE_x(-x)=\frac{\lambda}{4\pi\varepsilon_0}\frac{x_0+x}{\big((x_0+x)^2+y_0^2\big)^{3/2}},dx.$$

The sum of these two contributions does not vanish in general, but it defines a smooth even function of $x$ after combining the pair. The same pairing applied to the $y$-component yields

$$dE_y(x)+dE_y(-x)=\frac{\lambda y_0}{4\pi\varepsilon_0}\left[\frac{1}{((x_0-x)^2+y_0^2)^{3/2}}+\frac{1}{((x_0+x)^2+y_0^2)^{3/2}}\right]dx,$$

which is strictly positive for $y_0\neq 0$ and therefore has no cancellation mechanism within the segment.

The key correction concerns angular reasoning. Introducing polar angles centered at $C$, each element contributes along the direction of the segment joining the element to $C$. The mapping $x\mapsto \theta$ is monotone, but equal angular intervals do not correspond to equal linear charge intervals because

$$dq=\lambda,dx=\lambda\left|\frac{dx}{d\theta}\right|d\theta,$$

and $\left|\frac{dx}{d\theta}\right|$ depends on the distance from $C$ to the line and is not constant along the segment.

The correct symmetry is not uniformity in $\theta$ but pairing of rays symmetric with respect to the bisector of $\angle ACB$. Points $P\in AB$ and its mirror image $P'$ with respect to the line $C$-bisector produce field contributions whose components perpendicular to the bisector cancel because they correspond to equal angular deviations on opposite sides of that bisector. This cancellation follows from the fact that the kernel $\mathbf r/r^3$ depends only on the direction and distance from $C$, and reflection about the bisector preserves distances to $C$ while reversing the perpendicular angular component.

Directional Determination

Let $\ell$ be the internal angle bisector of $\angle ACB$. Decompose each elemental field contribution into components parallel and perpendicular to $\ell$.

For any pair of points $P$ and $P'$ on the segment that are symmetric with respect to the bisector direction, their perpendicular components satisfy

$$dE_{\perp}(P)=-dE_{\perp}(P'),$$

because the geometry of reflection about the bisector maps the direction vector from $C$ to $P$ into the mirror direction for $P'$ while preserving $|CP|=|CP'|$ and therefore preserving the magnitude factor $1/r^2$ in Coulomb’s law.

Since every element has such a symmetric counterpart under this angular reflection structure, the total perpendicular component cancels exactly,

$$E_{\perp}=0.$$

The parallel components satisfy

$$dE_{\parallel}(P)=dE_{\parallel}(P'),$$

so they add constructively along $\ell$, giving a nonzero resultant field aligned with the bisector.

This establishes that the field direction is constrained to lie on the internal angle bisector of $\angle ACB$.

Rejection of Other Candidate Directions

The altitude from $C$ to $AB$ corresponds to the line perpendicular to $AB$, which would require $E_x=0$. Direct evaluation of the integrand shows that $E_x(x)$ is not an odd function unless $x_0=0$, so no cancellation occurs in general, and the field is not generically perpendicular to the segment.

The median from $C$ to the midpoint of $AB$ would require proportionality between $E_x$ and $E_y$ of the form $\frac{E_x}{E_y}=\frac{x_0}{y_0}$. The integral kernels for $E_x$ and $E_y$ differ in structure, and no geometric symmetry of the segment with respect to $C$ enforces such proportionality except in degenerate symmetric configurations where the bisector, median, and altitude coincide.

The conclusion follows from the fact that only reflection symmetry with respect to the angle bisector preserves both distances from $C$ and the pairing structure of angular sectors, while the other candidate lines do not correspond to any symmetry of the charge distribution as seen from $C$.

Result

The electric field created by a uniformly charged finite segment $AB$ at a general point $C$ is directed along the internal angle bisector of $\angle ACB$.

This direction arises from exact cancellation of perpendicular components under reflection symmetry of charge elements about the bisector line through $C$, while the parallel components add constructively.

Sanity Checks

When $L\to 0$, the segment reduces to a point charge at the origin, and the bisector of $\angle ACB$ collapses to the line joining the charge to $C$, matching Coulomb’s law.

When $C$ lies on the perpendicular bisector of $AB$, the angle bisector of $\angle ACB$ coincides with that perpendicular bisector, producing a field perpendicular to the segment, consistent with symmetry of equal endpoints.

When $C$ approaches the line containing $AB$ while remaining off the segment, the angular span becomes highly asymmetric but the bisector direction remains well-defined and continues to coincide with the net field direction because the perpendicular-component cancellation persists under the angular reflection pairing about $C$.