Irodov 3.24: A Flux That Counts Only Length – Amit & Deepali Sinha Foundation

Problem Irodov 3.24

An electric field depends on the coordinates $x$ and $y$ only, as $\vect{E} = a(x\vect{i} + y\vect{j})/(x^2 + y^2)$ , with $a$ a constant and $\vect{i}$ , $\vect{j}$ the unit vectors along $x$ and $y$ . Find the flux of $\vect{E}$ through a sphere of radius $R$ centred at the origin.

Given

$\vect{E} = a(x\vect{i} + y\vect{j})/(x^2 + y^2)$
sphere of radius $R$ about $O$

Find

the flux

\Phi

\vect E

through the sphere

Solution

Step 1What the field looks like

The vector $x\vect i + y\vect j$ is the part of the position vector perpendicular to the $z$ axis, and its length is the distance from that axis:

\rho_\perp = \sqrt{x^2 + y^2}

So the field points straight away from the $z$ axis (for $a > 0$ ) with magnitude

E = \frac{a\,\rho_\perp}{\rho_\perp^2} = \frac{a}{\rho_\perp}

Step 2The normal component on the sphere

A point P of the sphere at polar angle θ from the z axis: the outward normal n lies along the radius, the field E points away from the axis. — A point $P$ of the sphere at polar angle $\theta$ from the $z$ axis: the outward normal $\vect n$ lies along the radius, the field $\vect E$ points away from the axis.

On the sphere the outward unit normal is the position vector divided by $R$ :

\vect n = \frac{x\vect i + y\vect j + z\vect k}{R}

Taking the scalar product with the field, the $z$ part drops out because $\vect E$ has no $z$ component:

E_n = \vect E\cdot\vect n = \frac{a\,(x^2 + y^2)}{(x^2 + y^2)\,R}

The factor $x^2 + y^2$ cancels:

E_n = \frac{a}{R} \tag{1}

Step 3Add up over the sphere

The normal component (1) is the same at every point of the sphere, so the flux is that constant times the area $4\pi R^2$ :

\Phi = \oint E_n\,dS = \frac{a}{R}\cdot 4\pi R^2

\Phi = 4\pi a R

(2)

Check

The given field is that of a long straight thread on the $z$ axis, $\lambda/(2\pi\varepsilon_0\rho_\perp)$ , with charge per unit length $\lambda = 2\pi\varepsilon_0 a$ . The sphere encloses a length $2R$ of that thread, so Gauss’s law gives the flux independently:

\Phi = \frac{2R\lambda}{\varepsilon_0} = \frac{2R\cdot 2\pi\varepsilon_0 a}{\varepsilon_0} = 4\pi aR

in agreement with the answer.

Remark

The flux grows in proportion to $R$ , not to $R^2$ and not as a constant: it counts the length of the $z$ axis inside the sphere, because the source of this field is spread along that axis. A sphere is not essential: any closed surface cutting the axis over a length $L$ gets the flux $2\pi aL$ .