Irodov 3.21: Flux Through a Slice of Ball – Amit & Deepali Sinha Foundation

Problem Irodov 3.21

A ball of radius $R$ carries charge spread uniformly through its volume with density $\rho$ . A plane at distance $r_0 < R$ from the ball’s centre cuts the ball in a flat circular section. Find the flux of the electric field through this section.

The ball in a section through its centre O, perpendicular to the cutting plane; the plane, seen edge-on as the dashed line, is r_0 from O and cuts the ball in a circular disc. — The ball in a section through its centre $O$ , perpendicular to the cutting plane; the plane, seen edge-on as the dashed line, is $r_0$ from $O$ and cuts the ball in a circular disc.

Given

$R$
$\rho$
$r_0 < R$

Find

the flux

\Phi

\vect E

through the section

Solution

Step 1The field inside the ball

By symmetry the field is radial and of the same magnitude all over a concentric sphere of radius $r < R$ , which encloses the charge $4\pi r^3\rho/3$ . Gauss’s law on it:

E\cdot 4\pi r^2 = \frac{4\pi r^3\rho}{3\varepsilon_0}

so at a point with position vector $\vect r$ from the centre

\vect E = \frac{\rho}{3\varepsilon_0}\,\vect r \tag{1}

Step 2The normal component on the section

A point P of the section at distance r from O. The radial field E resolved along the section and along its normal, E_n (drawn for ρ > 0). — A point $P$ of the section at distance $r$ from $O$ . The radial field $\vect E$ resolved along the section and along its normal, $E_n$ (drawn for $\rho > 0$ ).

Take the unit normal $\vect n$ of the section pointing away from the centre. Every point of the section lies at the same distance $r_0$ from the centre along $\vect n$ , so the normal part of its position vector is the same for all of them:

\vect r\cdot\vect n = r_0

Projecting (1) on $\vect n$ :

E_n = \frac{\rho}{3\varepsilon_0}\,\vect r\cdot\vect n = \frac{\rho r_0}{3\varepsilon_0} \tag{2}

Step 3Add up over the section

The section is a disc of radius $\sqrt{R^2 - r_0^2}$ , so its area is

S = \pi\,(R^2 - r_0^2)

With the constant normal component (2), the flux is that component times the area:

\Phi = E_n S = \frac{\rho r_0}{3\varepsilon_0}\cdot\pi\,(R^2 - r_0^2)

\Phi = \frac{\pi\rho\, r_0\,(R^2 - r_0^2)}{3\varepsilon_0}

(3)

For $\rho > 0$ the flux is outward, in the direction away from the centre.

Check

Both ends vanish as they should: at $r_0 = 0$ the section passes through the centre, where the radial field lies in the plane and crosses it nowhere; at $r_0 = R$ the section shrinks to a point. Units: $\rho r_0/\varepsilon_0$ is a field (the uniform-ball field $\rho r/(3\varepsilon_0)$ has the same form), and a field times an area is a flux.

Remark

The flux is largest for an intermediate cut: setting $d\Phi/dr_0$ to zero gives $R^2 - 3r_0^2 = 0$ , so the best cut is at $r_0 = R/\sqrt 3$ . A cut near the centre is large but the field there barely crosses it; a cut near the surface is crossed steeply but has little area.