Irodov 3.42: Same Strength in Every Direction – Amit & Deepali Sinha Foundation

Problem Irodov 3.42

Two long thin parallel threads carry uniform charges of linear density $\lambda$ and $-\lambda$ , a distance $l$ apart. Let $\vect{l}$ be the vector of length $l$ from the $-\lambda$ thread to the $+\lambda$ thread, perpendicular to both. At a point a distance $r \gg l$ from the midpoint between the threads, in a plane perpendicular to them, along a direction making the angle $\theta$ with $\vect{l}$ , find the potential of the electric field and the magnitude of the field.

The two threads seen end-on, the vector l between them pointing from the -λ thread to the +λ one, and the position r of the field point from their midpoint, at the angle θ to l. — The two threads seen end-on, the vector $\vect{l}$ between them pointing from the $-\lambda$ thread to the $+\lambda$ one, and the position $\vect{r}$ of the field point from their midpoint, at the angle $\theta$ to $\vect{l}$ .

Given

threads $\lambda$ and $-\lambda$
separation $l$
distance $r \gg l$
angle $\theta$ to $\vect{l}$

Find

the potential

\varphi

and the field magnitude

E

Solution

Step 1The potential of one thread

By symmetry the field of a long uniform thread is radial and the same all round it, so Gauss’s law on a coaxial cylinder of radius $\rho$ and length $\ell$ gives it at once:

E\cdot 2\pi\rho\ell = \frac{\lambda\ell}{\varepsilon_0}

Integrating $E = \lambda/(2\pi\varepsilon_0\rho)$ along the radius gives the potential, up to a constant $C$ :

\varphi_1 = -\int\frac{\lambda\,d\rho}{2\pi\varepsilon_0\rho} = -\frac{\lambda}{2\pi\varepsilon_0}\ln\rho + C \tag{1}

Step 2The pair

Call the distances from the field point $P$ to the $+\lambda$ and $-\lambda$ threads $r_+$ and $r_-$ . Adding (1) for $\lambda$ and for $-\lambda$ , the constants cancel, since the total charge is zero:

\varphi = -\frac{\lambda}{2\pi\varepsilon_0}\ln r_+ + \frac{\lambda}{2\pi\varepsilon_0}\ln r_- = \frac{\lambda}{2\pi\varepsilon_0}\ln\frac{r_-}{r_+} \tag{2}

The distances r_+ and r_- from the two threads to the field point P (the separation drawn much larger than it is). — The distances $r_+$ and $r_-$ from the two threads to the field point $P$ (the separation drawn much larger than it is).

Step 3Far from the pair

At $r \gg l$ the lines from the two threads to $P$ are nearly parallel to $\vect{r}$ , and the $+\lambda$ thread is nearer by half the projection of $\vect{l}$ on that direction, the other farther by the same:

r_\pm \approx r \mp \frac{l}{2}\cos\theta

The ratio in (2) is then close to one, and $\ln(1 + x) \approx x$ for small $x$ :

\ln\frac{r_-}{r_+} \approx \ln\left(1 + \frac{l\cos\theta}{r}\right) \approx \frac{l\cos\theta}{r}

\varphi = \frac{\lambda l\cos\theta}{2\pi\varepsilon_0 r}

(3)

Step 4The field and its magnitude

The field is minus the gradient of (3), taken in polar coordinates in the plane perpendicular to the threads: along $\vect{r}$ and along the direction of growing $\theta$ .

\begin{aligned} E_r &= -\frac{\partial\varphi}{\partial r} = \frac{\lambda l\cos\theta}{2\pi\varepsilon_0 r^2}\\ E_\theta &= -\frac{1}{r}\frac{\partial\varphi}{\partial\theta} = \frac{\lambda l\sin\theta}{2\pi\varepsilon_0 r^2} \end{aligned}

The two components are perpendicular, and $\cos^2\theta + \sin^2\theta = 1$ under the root:

E = \sqrt{E_r^2 + E_\theta^2} = \frac{\lambda l}{2\pi\varepsilon_0 r^2}\sqrt{\cos^2\theta + \sin^2\theta}

E = \frac{\lambda l}{2\pi\varepsilon_0 r^2}

(4)

The strength falls as $1/r^2$ and does not depend on the direction at all.

Field lines (crimson) of the +λ, -λ pair and its equipotentials (teal). Away from the pair the lines become circles tangent to l at the midpoint, the equipotentials circles tangent to the midplane. — Field lines (crimson) of the $+\lambda$ , $-\lambda$ pair and its equipotentials (teal). Away from the pair the lines become circles tangent to $\vect{l}$ at the midpoint, the equipotentials circles tangent to the midplane.

Check

Along $\vect{l}$ , $\theta = 0$ , the two thread fields are collinear and subtract, giving

E = \frac{\lambda}{2\pi\varepsilon_0}\left(\frac{1}{r - l/2} - \frac{1}{r + l/2}\right) \approx \frac{\lambda l}{2\pi\varepsilon_0 r^2}

which agrees with (4). Units: $\lambda l$ is a charge, so $\lambda l/(\varepsilon_0 r^2)$ is a field.

Remark

Since $E_r = E\cos\theta$ and $E_\theta = E\sin\theta$ , the field makes the angle $\theta$ with $\vect{r}$ , and so the angle $2\theta$ with $\vect{l}$ : going once round the pair, the field vector turns twice round at constant length. The point dipole in three dimensions is less even, its strength varying by a factor of two between axis and equator. Written with the moment per unit length $\lambda l$ , the potential (3) is the two-dimensional dipole: the far field of any pair of opposite line charges, and, outside it, exactly the field of the charge that a uniform transverse field induces on a long conducting cylinder.