Съдържание

IPhO 1969, Теория — Задача 3

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от IPhO 1969

3rd International Physics Olympiad 1969, Brno, Czechoslovakia

Условие

A small charged ball of mass $m$ and charge $q$ is suspended from the highest point of a ring of radius $R$ by means of an insulating cord of negligible mass. The ring is made of a rigid wire of negligible cross section and lies in a vertical plane. On the ring there is uniformly distributed charge $Q$ of the same sign as $q$ . Determine the length $l$ of the cord so as the equilibrium position of the ball lies on the symmetry axis perpendicular to the plane of the ring.

Find first the general solution a then for particular values $Q = q = 9.0 \cdot 10^{-8}\ \mathrm{C}$ , $R = 5\ \mathrm{cm}$ , $m = 1.0\ \mathrm{g}$ , $\varepsilon_0 = 8.9 \cdot 10^{-12}\ \mathrm{F/m}$ .

Отговори

Решение

In equilibrium, the cord is stretched in the direction of resultant force of $\vec{G} = m\vec{g}$ and $\vec{F} = q\vec{E}$ , where $\vec{E}$ stands for the electric field strength of the ring on the axis in distance $x$ from the plane of the ring, see Figure 3. Using the triangle similarity, one can write

\frac{x}{R} = \frac{Eq}{mg}\,. \qquad (11)

Side view of the vertical ring (labelled R) with the horizontal symmetry axis; the cord of length l runs from the top of the ring to the ball on the axis at distance x; at the ball the force F points along the axis and the weight G points down, with the dashed parallelogram of their resultant along the cord. — Figure 3:

For the calculation of the electric field strength let us divide the ring to $n$ identical parts, so as every part carries the charge $Q/n$ . The electric field strength magnitude of one part of the ring is given by

\Delta E = \frac{Q}{4\pi\varepsilon_0 l^2 n}\,.

The same side view of the ring (R), cord l, axis distance x and angle α at the ball; the field ΔE of one ring element is split into ΔE_x along the axis and ΔE_⊥ perpendicular to it, with a dashed parallelogram. — Figure 4:

This electric field strength can be decomposed into the component in the direction of the $x$ -axis and the one perpendicular to the $x$ -axis, see Figure 4. Magnitudes of both components obey

\begin{aligned} \Delta E_x &= \Delta E\, \cos\alpha = \frac{\Delta E\, x}{l}\,, \\ \Delta E_\perp &= \Delta E\, \sin\alpha\,. \end{aligned}

It follows from the symmetry, that for every part of the ring there exists another one having the component $\Delta\vec{E_\perp}$ of the same magnitude, but however oppositely oriented. Hence, components perpendicular to the axis cancel each other and resultant electric field strength has the magnitude

E = E_x = n\Delta E_x = \frac{Q\, x}{4\pi\varepsilon_0\, l^3}\,. \qquad (12)

Substituting (12) into (11) we obtain for the cord length

l = \sqrt[3]{\frac{Q\, q\, R}{4\pi\varepsilon_0\, m\, g}}\,.

Numerically

l = \sqrt[3]{\frac{9.0 \cdot 10^{-8} \cdot 9.0 \cdot 10^{-8} \cdot 5.0 \cdot 10^{-2}}{4\pi \cdot 8.9 \cdot 10^{-12} \cdot 10^{-3} \cdot 9.8}}\ \mathrm{m} = 7.2 \cdot 10^{-2}\ \mathrm{m}\,.

Оригинал в Архива: ipho69.pdf