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Chapter 1, problem 6 in Smythe's Static and Dynamics Electricity (2nd ed) reads

Charges $+4q$, $-q$ are placed at the points $A$, $B$, and $C$ is the point of equilibrium. Prove that the line of force that passes through $C$ meets $AB$ at an angle of $60^{\circ}$ at $A$ and at right angles at $C$. Find the angle at $A$ between $AB$ and the line of force that leaves $B$ at right angles to $AB$.

(Write potential for small region about $C$ in polar coordinates with origin at $C$.)

By an equilibrium point is meant a point in the field where an equipotential surface crosses itself at least twice, so that $\mathbf{E}=-\nabla V=0$ there.

While trying to solve this problem, I found that the point $C$ where $\mathbf E=0$ (excluding points infinitely far away) to be collinear with $A$ and $B$ and the field line is approximately as shown below.

enter image description here

My hang up at the moment is this: how can you prove that the field line passing through $C$ will pass this point perpendicular to the central line?

Intuitively, we might expect that lines of force (in isotropic homogeneous media) cross an axis of symmetry of a system of point charges at oblique angles if and only if there is a point charge on the axis at that location. Then, because the field line passes through an axis of symmetry, and no point charges are present, it must be perpendicular to the central line (neglect the case of a parallel line).

However, throughout the first chapter, Gauss's law in integral form is used to solve problems with field lines, and I'm hoping to find a more rigorous proof along these lines.

Sam Gallagher
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2 Answers2

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I am trying to respond quickly and perhaps not very carefully. (And sorry for my poor english !).

Given the symmetries, we can consider developments of the field around the point of equilibrium of the following form, odd in $r$ for the radial component, even for the $z$ component:

$E_z (r,z)=a_1 (z) r^2+a_2 (z) r^4+⋯$

$E_r (r,z)=b_1 (z)r+b_2 (z) r^3…$

As $dz/dr=E_z/E_r $, we have to find the limit of the slope when one approaches the point of equilibrium,that is the limit of $(a_1 (z) r^2+a_2 (z) r^4…)/(b_1 (z)r+b_2 (z) r^3…)$ when $r$ tends to $0$. As this limit is $dz/dr=0$, the angle is $90°$.

For the other angle, it should be noted that the field lines inside the cone limited by all the lines from A to E return towards the charge $-q$. Only the outer field lines go to infinity.

The flux of the electric field around the charge $4q$ is $(4q/ε_0)$

The solid angle corresponding to the field lines external to the tube which joins A to C is $Ω=2π(1+cos⁡(θ))$ and the corresponding flux is $(4q/ε_0)(Ω/4π)=(2q/ε_ 0)(1 +cos⁡(θ))$ (since, Very close to the $4q$ charge, we can pretend we have a single point charge).

But this flux is also the total flux through a large sphere which surrounds the two charges. So we will have: $(q/ε_0) 2(1+cos⁡(θ) )=3q/ε_0$ or $cos⁡(θ)=1/2$

Hope it can help.

Vincent Fraticelli
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how can you prove that the field line passing through C will pass this point perpendicular to the central line?

enter image description here

The potential at position $X$ is $V_{\rm A}(r_4)+ V_{\rm B}(r_1)$ and $r_1$ and $r_4$ can be written in terms of $a,\,r$ and $\theta$ with $r\ll a$.

You need to show that $\partial V/\partial \theta =0$ for $\theta =\pi/2$ if $r\ll a$.

This method has the advantage of using a scalar function (potential) rather than a vector function (field strength).

Farcher
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