Irodov 3.69: Pressure over a Projected Disc – Amit & Deepali Sinha Foundation

Problem Irodov 3.69

A metal ball of radius $R = 1.5$ cm carries a charge $q = 10\ \mu$ C. Imagine the ball cut into two halves by a plane through its center. Find the magnitude of the resultant electric force on the charge that lies on one half.

The charged metal ball of radius R, its charge q spread over the surface, and the plane through the center (dashed) that divides it into two halves. — The charged metal ball of radius $R$ , its charge $q$ spread over the surface, and the plane through the center (dashed) that divides it into two halves.

Given

$R = 1.5$ cm
$q = 10\ \mu$ C
metal ball

Find

the magnitude

F

of the force on the charge of one half

Solution

Step 1The outward pressure on the surface

The charge of an isolated ball spreads uniformly, with density $\sigma = q/4\pi R^2$ . The charge on each element of a conductor’s surface feels half the field just outside it, $\sigma/2\varepsilon_0$ , so it is pushed outward with a pressure:

p = \frac{\sigma^2}{2\varepsilon_0} = \frac{1}{2\varepsilon_0}\left(\frac{q}{4\pi R^2}\right)^2 = \frac{q^2}{32\pi^2\varepsilon_0 R^4} \tag{1}

Step 2Adding the pushes over one half

An element of the upper half at polar angle θ from the axis z. The charge on it is pushed outward along the radius by dF; only the part along z survives the sum over the half. — An element of the upper half at polar angle $\theta$ from the axis $z$ . The charge on it is pushed outward along the radius by $d\vect{F}$ ; only the part along $z$ survives the sum over the half.

Every push is radial. By symmetry about the axis $z$ of the upper half, the sideways parts cancel and only the components along $z$ , $dF\cos\theta$ , add up. A ring of the surface between $\theta$ and $\theta + d\theta$ has area $2\pi R^2\sin\theta\,d\theta$ :

\begin{aligned} F &= \int p\cos\theta\,dS = p\int_0^{\pi/2} \cos\theta\;2\pi R^2\sin\theta\,d\theta\\ &= 2\pi R^2 p\left[\frac{\sin^2\theta}{2}\right]_0^{\pi/2} = \pi R^2\,p \end{aligned}

Step 3The force

Substituting (1):

F = \pi R^2\,\frac{q^2}{32\pi^2\varepsilon_0 R^4}

Numerically, with $\varepsilon_0 = 8.85\cdot 10^{-12}$ F/m:

F = \frac{(10\cdot 10^{-6})^2}{32\pi \cdot 8.85\cdot 10^{-12}\cdot(1.5\cdot 10^{-2})^2}\ \mathrm{N}

F = \dfrac{q^2}{32\pi\varepsilon_0 R^2} \approx 0.50\ \mathrm{kN}

(2)

The two halves repel each other with this force.

Check

The same force can be read off the field in the cutting plane alone. Outside the ball the field there, $E = q/4\pi\varepsilon_0 r^2$ , lies in the plane, and a field lying along a plane pushes the two sides of it apart with the pressure $\varepsilon_0 E^2/2$ (the Maxwell stress; inside the ball $E = 0$ ). Summing over the plane outside the ball, in rings of radius $r$ :

\begin{aligned} F &= \int_R^\infty \frac{\varepsilon_0}{2}\left(\frac{q}{4\pi\varepsilon_0 r^2}\right)^2 2\pi r\,dr\\ &= \frac{q^2}{16\pi\varepsilon_0}\int_R^\infty \frac{dr}{r^3}\\ &= \frac{q^2}{16\pi\varepsilon_0}\cdot\frac{1}{2R^2} = \frac{q^2}{32\pi\varepsilon_0 R^2} \end{aligned}

the same as (2).

Remark

Half a kilonewton, the weight of some $50$ kg, between the halves of a ball the size of a marble. The field at its surface is $q/4\pi\varepsilon_0 R^2 = 4\cdot 10^{8}$ V/m, more than a hundred times the breakdown field of air: the charge would leak away in a spark long before the ball felt this force, so the numbers describe a ball in vacuum.