A thin ring of radius R = 10 cm carries a charge Q = +2.0 µC, spread evenly round it…

A thin ring of radius $R = 10\ \text{cm}$ carries a charge $Q = +2.0\ \mu\text{C}$ , spread evenly round it. The ring is held fixed in a vertical plane. A thin, smooth, straight insulating rod is fixed along the axis of the ring, so it is horizontal and passes through the ring's centre $O$ , as in the figure. A small bead of mass $1.0\ \text{g}$ and charge $-0.50\ \mu\text{C}$ can slide along the rod. The rod keeps the bead on the axis, and gravity does not act along it. Take $\dfrac{1}{4\pi\varepsilon_0} = 9.0 \times 10^{9}\ \text{N m}^2\,\text{C}^{-2}$ .

A thin ring of radius 10 cm, standing in a vertical plane and seen at a slant, carries a charge of plus 2.0 microcoulombs. A thin horizontal rod runs along the axis of the ring through its centre O. A small bead of mass 1.0 g and charge minus 0.50 microcoulombs sits on the rod, 10 cm from O.

What is the field of the ring on its axis, at a distance $z = R$ from $O$ ?
At what distance from $O$ is the field on the axis greatest?
What is that greatest field?
The bead is released from rest at $z = R$ . What is its speed as it passes $O$ ?
The bead is instead pushed a little way from $O$ and let go. What is the period of its small oscillations?
The bead is replaced by one of the same mass with a charge of $+0.50\ \mu\text{C}$ . It is placed at $O$ and nudged gently along the rod. What speed does it reach far from the ring?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Find the field on the axis, then treat the bead's motion with work and energy. Near the centre the force is proportional to $z$ , so the small oscillations are simple harmonic.

Step 1. The field on the axis (a). Each bit $dQ$ of the ring is $\sqrt{R^2 + z^2}$ away, and only the part of its field along the axis adds up:

\displaystyle E(z) = \frac{kQz}{(R^2 + z^2)^{3/2}}

At $z = R$ : $E = \dfrac{kQ}{2\sqrt2\,R^2} = 6.36 \times 10^{5}\ \text{N/C}$ .

Step 2. The greatest field (b, c). $\dfrac{dE}{dz} = kQ\,\dfrac{R^2 - 2z^2}{(R^2 + z^2)^{5/2}} = 0$ at $z = \dfrac{R}{\sqrt2} = 7.07\ \text{cm}$ , where $E_{\max} = \dfrac{2kQ}{3\sqrt3\,R^2} = 6.93 \times 10^{5}\ \text{N/C}$ .

Step 3. Speed at the centre (d). The bead is pulled towards $O$ . The work done on it from $z = R$ to $O$ is

\displaystyle W = \int_0^R \frac{kQqz}{(R^2 + z^2)^{3/2}}\,dz = kQq\left(\frac1R - \frac{1}{\sqrt2\,R}\right) = 0.0264\ \text{J}

so $v = \sqrt{2W/m} = 7.26\ \text{m/s}$ .

Step 4. Small oscillations (e). For $z \ll R$ the force is $F \approx -\dfrac{kQq}{R^3}z$ . So $\omega = \sqrt{\dfrac{kQq}{mR^3}} = 94.87\ \text{rad/s}$ and $T = \dfrac{2\pi}{\omega} = 66.2\ \text{ms}$ .

Step 5. A positive bead (f). Now the force pushes the bead away from $O$ . All the way out it gains $\int_0^\infty qE\,dz = \dfrac{kQq}{R} = 0.090\ \text{J}$ , so $v = \sqrt{\dfrac{2kQq}{mR}} = 13.42\ \text{m/s}$ .

Answer: (a) $6.36 \times 10^{5}\ \text{N/C}$ . (b) $7.07\ \text{cm}$ . (c) $6.93 \times 10^{5}\ \text{N/C}$ . (d) $7.26\ \text{m/s}$ . (e) $66.2\ \text{ms}$ . (f) $13.42\ \text{m/s}$ .

Check. The field at $z = R$ is a little below the peak, as it must be: $R$ is beyond $R/\sqrt2$ .

Watch out. The release from $z = R$ is not a small oscillation, so its time is not given by (e). And the field at $O$ is zero, yet the bead feels a growing pull as it moves off, because the field grows from zero along the axis.

← All Problems

← A solid insulating sphere of radius R = 10 cm carries charge spread through its volume…A light insulating rod AB, 20 cm long, has a small ball fixed at each end… →

More onElectricity & Magnetism › Electrostatics

Discussion (0)

Loading…