Irodov 3.70: The Field Stretches the Ball – Amit & Deepali Sinha Foundation

Problem Irodov 3.70

An uncharged conducting ball of radius $R$ is placed in a uniform external electric field. The charge induced on its surface has density $\sigma = \sigma_0\cos\theta$ , where $\sigma_0$ is a constant and $\theta$ is the polar angle. Find the magnitude of the resultant electric force on all the induced charge of one sign.

The uncharged conducting ball of radius R in the uniform external field E_0, with the induced charge σ_0θ just inside its surface, θ measured from the field's direction (drawn for σ_0 > 0). — The uncharged conducting ball of radius $R$ in the uniform external field $\vect{E}_0$ , with the induced charge $\sigma_0\cos\theta$ just inside its surface, $\theta$ measured from the field’s direction (drawn for $\sigma_0 > 0$ ).

Given

ball radius $R$
induced charge $\sigma = \sigma_0\cos\theta$

Find

the magnitude

F

of the force on the induced charge of one sign

Solution

Step 1The outward pressure at each point

The charge on any element of a conductor’s surface feels half the field just outside it, $\sigma/2\varepsilon_0$ along the outward normal, the external field included, so every element is pushed outward with a pressure:

p(\theta) = \frac{\sigma^2}{2\varepsilon_0} = \frac{\sigma_0^2\cos^2\theta}{2\varepsilon_0} \tag{1}

For $\sigma_0 > 0$ the positive charge covers the half $0 \le \theta < \pi/2$ , facing along the field; the negative half follows the same way.

Step 2Adding the pushes over the positive half

An element of the positive half at polar angle θ, pushed outward along the radius by dF. The dashed plane separates the two signs of charge. — An element of the positive half at polar angle $\theta$ , pushed outward along the radius by $d\vect{F}$ . The dashed plane separates the two signs of charge.

By symmetry about the axis the sideways parts of the radial pushes cancel, and only their components along the axis, $dF\cos\theta$ , add up. A ring between $\theta$ and $\theta + d\theta$ has area $2\pi R^2\sin\theta\,d\theta$ . Using (1):

\begin{aligned} F &= \int_0^{\pi/2} \frac{\sigma_0^2\cos^2\theta}{2\varepsilon_0}\,\cos\theta\;2\pi R^2\sin\theta\,d\theta\\ &= \frac{\pi R^2\sigma_0^2}{\varepsilon_0}\int_0^{\pi/2}\cos^3\theta\,\sin\theta\,d\theta\\ &= \frac{\pi R^2\sigma_0^2}{\varepsilon_0}\left[-\frac{\cos^4\theta}{4}\right]_0^{\pi/2} = \frac{\pi R^2\sigma_0^2}{\varepsilon_0}\cdot\frac{1}{4} \end{aligned}

F = \dfrac{\pi R^2\sigma_0^2}{4\varepsilon_0}

(2)

The positive half is pulled along the field, the negative half against it, each with this force.

Check

The two halves together must feel no net force, since a uniform external field exerts none on an uncharged body, and the field of the induced charge cannot push the ball as a whole. The negative half, $\pi/2 < \theta \le \pi$ , gives along the axis

\begin{aligned} F_- &= \int_{\pi/2}^{\pi} \frac{\sigma_0^2\cos^2\theta}{2\varepsilon_0}\,\cos\theta\;2\pi R^2\sin\theta\,d\theta\\ &= \frac{\pi R^2\sigma_0^2}{\varepsilon_0}\left[-\frac{\cos^4\theta}{4}\right]_{\pi/2}^{\pi} = -\frac{\pi R^2\sigma_0^2}{4\varepsilon_0}, \end{aligned}

which cancels (2) exactly. The units are right too: $\sigma_0^2/\varepsilon_0$ is a pressure, times the area $\pi R^2$ .

Remark

The two halves are pulled apart along the field, so a conductor in a field is stretched along it. For a soft conductor such as a water drop this tension elongates it into a spindle, and in strong enough fields the ends sharpen into cones that spray charge. With the boundary condition of this ball, $\sigma_0 = 3\varepsilon_0 E_0$ , the force reads $F = (9/4)\,\pi\varepsilon_0 R^2 E_0^2$ .