Irodov 3.54: The Image Stretches the Thread – Amit & Deepali Sinha Foundation

Problem Irodov 3.54

A small ball hangs from an insulating elastic thread of stiffness $k$ above an infinite horizontal conducting plane. When the ball is given a charge, it moves down by a distance $x$ and comes to rest at a distance $l$ from the plane. Find the charge of the ball.

The ball on its elastic thread above the conducting plane: the uncharged position (dashed), the descent x on charging and the final distance l from the plane. — The ball on its elastic thread above the conducting plane: the uncharged position (dashed), the descent $x$ on charging and the final distance $l$ from the plane.

Given

thread stiffness $k$
descent $x$
final distance $l$ from the plane

Find

the charge

q

of the ball

Solution

Step 1The plane acts as an image charge

The conducting plane is an equipotential, at the potential of infinity, zero. Above the plane, the field of $q$ and its induced charge is therefore the field of $q$ and an image charge $q' = -q$ at the mirror point, a distance $l$ below the surface: that pair puts every point of the plane at zero potential, because each such point is equally far from both.

\varphi_{\text{plane}} = \frac{1}{4\pi\varepsilon_0}\left(\frac{q}{\rho} - \frac{q}{\rho}\right) = 0

Here $\rho$ is the common distance from a point of the plane to the ball and to the image.

The image q' (dashed) at the mirror point inside the metal and the attraction F it exerts on the ball, pulling it toward the plane. — The image $q'$ (dashed) at the mirror point inside the metal and the attraction $\vect F$ it exerts on the ball, pulling it toward the plane.

The force on the ball is the Coulomb attraction of the image, a distance $2l$ away.

F = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{(2l)^2} = \frac{q^2}{16\pi\varepsilon_0 l^2}

(1)

Step 2The extra stretch of the thread balances the attraction

Before charging, the thread’s tension balanced the weight. The extra stretch $x$ adds a tension $kx$ upward, and in the new equilibrium it balances the attraction (1).

\begin{aligned} kx &= \frac{q^2}{16\pi\varepsilon_0 l^2} \\ q^2 &= 16\pi\varepsilon_0 k x\, l^2 \end{aligned}

Taking the square root:

q = 4l\sqrt{\pi\varepsilon_0 k x}

(2)

The sign of the charge is not fixed: a charge of either sign is attracted by its image.

Check

Units: $\varepsilon_0 k x l^2$ has units $\dfrac{\text{C}^2}{\text{N\,m}^2}\cdot\dfrac{\text{N}}{\text{m}}\cdot\text{m}\cdot\text{m}^2 = \text{C}^2$ , so (2) is a charge. Two limits: no descent, $x \to 0$ , gives $q \to 0$ , as it must, since any charge is pulled toward its image; and at a fixed stretch $kx$ the charge grows in proportion to $l$ , because the image force falls as $1/l^2$ and must be made up by $q^2$ :

\frac{q(2l)}{q(l)} = \frac{4\cdot 2l\sqrt{\pi\varepsilon_0 kx}}{4l\sqrt{\pi\varepsilon_0 kx}} = 2

Remark

The equilibrium is not always stable. Measure the height $s$ of the ball above the plane; the net upward force is the extra tension minus the image pull, and its slope at $s = l$ decides stability.

\begin{aligned} f(s) &= k(l + x - s) - \frac{q^2}{16\pi\varepsilon_0 s^2} \\ f'(l) &= -k + \frac{2q^2}{16\pi\varepsilon_0 l^3} = -k + \frac{2kx}{l} \end{aligned}

A restoring force needs a negative slope, which by the last line is the condition

f'(l) < 0 \iff l > 2x

If the charge pulls the ball down by more than a third of its original height $l + x$ , the image force grows faster than the thread can answer and the ball snaps down onto the plane.