IPhO 2004 · Theory problem 1

“Ping-Pong” Resistor

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A capacitor consists of two circular parallel plates both with radius \(R\) separated by distance \(d\), where \(d\ll R\), as shown in Fig. 1.1(a). The top plate is connected to a constant voltage source at a potential \(V\) while the bottom plate is grounded. Then a thin and small disk of mass \(m\) with radius \(r\) (\(\ll R,d\)) and thickness \(t\) (\(\ll r\)) is placed on the center of the bottom plate, as shown in Fig. 1.1(b).

Let us assume that the space between the plates is in vacuum with the dielectric constant \(\varepsilon_0\); the plates and the disk are made of perfect conductors; and all the electrostatic edge effects may be neglected. The inductance of the whole circuit and the relativistic effects can be safely disregarded. The image charge effect can also be neglected.

Figure 1

Figure 1.1 Schematic drawings of (a) a parallel plate capacitor connected to a constant voltage source and (b) a side view of the parallel plates with a small disk inserted inside the capacitor. (See text for details.)

(a) [1.2 points] Calculate the electrostatic force \(F_{\mathrm p}\) between the plates separated by \(d\) before inserting the disk in-between as shown in Fig. 1.1(a).

(b) [0.8 points] When the disk is placed on the bottom plate, a charge \(q\) on the disk of Fig. 1.1(b) is related to the voltage \(V\) by \(q=\chi V\). Find \(\chi\) in terms of \(r\), \(d\), and \(\varepsilon_0\).

(c) [0.5 points] The parallel plates lie perpendicular to a uniform gravitational field \(g\). To lift up the disk at rest initially, we need to increase the applied voltage beyond a threshold voltage \(V_{\mathrm{th}}\). Obtain \(V_{\mathrm{th}}\) in terms of \(m\), \(g\), \(d\), and \(\chi\).

(d) [2.3 points] When \(V>V_{\mathrm{th}}\), the disk makes an up-and-down motion between the plates. (Assume that the disk moves only vertically without any wobbling.) The collisions between the disk and the plates are inelastic with the restitution coefficient \(\eta\equiv(v_{\mathrm{after}}/v_{\mathrm{before}})\), where \(v_{\mathrm{before}}\) and \(v_{\mathrm{after}}\) are the speeds of the disk just before and after the collision respectively. The plates are stationarily fixed in position. The speed of the disk just after the collision at the bottom plate approaches a “steady-state speed” \(v_{\mathrm s}\), which depends on \(V\) as follows:

\[ v_{\mathrm s}=\sqrt{\alpha V^2+\beta}. \tag{1.1} \]

Obtain the coefficients \(\alpha\) and \(\beta\) in terms of \(m\), \(g\), \(\chi\), \(d\), and \(\eta\). Assume that the whole surface of the disk touches the plate evenly and simultaneously so that the complete charge exchange happens instantaneously at every collision.

(e) [2.2 points] After reaching its steady state, the time-averaged current \(I\) through the capacitor plates can be approximated by \(I=\gamma V^2\) when \(qV\gg mgd\). Express the coefficient \(\gamma\) in terms of \(m\), \(\chi\), \(d\), and \(\eta\).

(f) [3 points] When the applied voltage \(V\) is decreased (extremely slowly), there exists a critical voltage \(V_{\mathrm c}\) below which the charge will cease to flow. Find \(V_{\mathrm c}\) and the corresponding current \(I_{\mathrm c}\) in terms of \(m\), \(g\), \(\chi\), \(d\), and \(\eta\). By comparing \(V_{\mathrm c}\) with the lift-up threshold \(V_{\mathrm{th}}\) discussed in (c), make a rough sketch of the \(I-V\) characteristics when \(V\) is increased and decreased in the range from \(V=0\) to \(3V_{\mathrm{th}}\).

Official English text of the problem.

Answers

Here \(q=\chi V\) denotes the magnitude of the disk charge, consistently with \(qV\gg mgd\) in (e). Its signed charge is \(-q\) after contact with the bottom plate and \(+q\) after contact with the top plate. The shuttling results below assume \(0<\eta<1\).

  • (a) The force is attractive, with magnitude \[ \boxed{F_{\mathrm p}=\frac{\varepsilon_0\pi R^2V^2}{2d^2}}. \]

  • (b) \[ \boxed{\chi=\frac{\varepsilon_0\pi r^2}{d}}. \]

  • (c) \[ \boxed{V_{\mathrm{th}}=\sqrt{\frac{2mgd}{\chi}}}. \]

  • (d) \[ \boxed{\alpha=\frac{2\chi\eta^2}{m(1-\eta^2)},\qquad \beta=\frac{2gd\,\eta^2}{1+\eta^2}}. \]

  • (e) \[ \boxed{\gamma=\frac{1}{d} \sqrt{\frac{\chi^3}{2m}\frac{1+\eta}{1-\eta}}}. \]

  • (f) \[ \boxed{V_{\mathrm c} =\sqrt{\frac{mgd}{\chi}\frac{1-\eta^2}{1+\eta^2}} =V_{\mathrm{th}}\sqrt{\frac{1-\eta^2}{2(1+\eta^2)}}}, \] \[ \boxed{I_{\mathrm c} =\frac{2\eta g\sqrt{m\chi}}{1+\eta^2} \sqrt{\frac{1-\eta}{1+\eta}}}. \] On increasing \(V\), the current remains zero until \(V_{\mathrm{th}}\), then jumps to the conducting branch. On decreasing \(V\), it follows that branch down to \((V_{\mathrm c},I_{\mathrm c})\), then drops to zero. At high voltage, the branch approaches \(I=\gamma V^2\).

The exactly sticking endpoint \(\eta=0\) is singular: applying the contact condition after every bottom impact gives \[ \boxed{V_{\mathrm c}=V_{\mathrm{th}},\qquad I_{\mathrm c}=(3-\sqrt3)\,g\sqrt{m\chi}} \] for the limiting current on the conducting side, with no hysteresis. At \(\eta=1\), there is no finite steady state.

Solution

Throughout, neglect the disk thickness compared with \(d\), and use the stated approximations concerning edges and image charges. The applied field has magnitude

\[ E=\frac{V}{d} \]

and points downward for \(V>0\).

(a)

The surface charge density on either capacitor plate has magnitude

\[ \sigma=\varepsilon_0E. \]

The force on a plate must exclude its own field. The field produced by the opposite plate is \(E/2\), so the electrostatic pressure is

\[ p=\sigma\frac{E}{2}=\frac{\varepsilon_0E^2}{2}. \]

Multiplying by the plate area gives

\[ \boxed{F_{\mathrm p}=p\pi R^2 =\frac{\varepsilon_0\pi R^2V^2}{2d^2}}. \]

The oppositely charged plates attract.

(b)

While the disk rests on the bottom plate, it forms part of that conductor. Its exposed upper face carries surface charge density \(-\varepsilon_0E\). Thus the magnitude of its charge is

\[ q=\varepsilon_0E\pi r^2 =\frac{\varepsilon_0\pi r^2}{d}V. \]

Consequently,

\[ \boxed{\chi=\frac{\varepsilon_0\pi r^2}{d}}. \]

If \(q\) were instead interpreted as the signed bottom-contact charge, its coefficient would be negative; the subsequent use of \(qV\) as positive electrical work establishes the magnitude convention used here.

(c)

At rest on the lower plate, the disk is part of the plate surface. Its upward electric force is therefore the pressure found in (a), multiplied by the disk area:

\[ F_{\mathrm{contact}} =\frac{\varepsilon_0E^2}{2}\pi r^2 =\frac{qE}{2} =\frac{\chi V^2}{2d}. \]

If \(N\) is the upward normal reaction, equilibrium requires

\[ N+\frac{\chi V^2}{2d}=mg. \]

Lift-off becomes possible when the required normal reaction reaches zero. Hence

\[ \frac{\chi V_{\mathrm{th}}^2}{2d}=mg, \]

giving

\[ \boxed{V_{\mathrm{th}}=\sqrt{\frac{2mgd}{\chi}}}. \]

The factor \(1/2\) is essential here. Once the disk is in flight, neglecting image forces as instructed, its net force in the uniform applied field is \(qE\). The contact force and the flight force must therefore be treated separately.

(d)

Define the magnitude of the electric acceleration during flight by

\[ a_{\mathrm e}=\frac{qE}{m}=\frac{\chi V^2}{md}. \]

After leaving the bottom plate, the disk has charge \(-q\), so its upward acceleration is \(a_{\mathrm e}-g\). After touching the top plate, its charge becomes \(+q\), so its downward acceleration is \(a_{\mathrm e}+g\).

Let \(v_n\) be the upward speed just after the \(n\)-th bottom collision. Let \(u_n\) be the speed immediately before the following top collision, and \(w_n\) the speed immediately before the next bottom collision.

The work–energy theorem on the upward flight gives

\[ u_n^2=v_n^2+2d(a_{\mathrm e}-g). \]

The top collision reduces the speed to \(\eta u_n\). The downward flight therefore gives

\[ w_n^2=\eta^2u_n^2+2d(a_{\mathrm e}+g). \]

Finally, the bottom collision gives \(v_{n+1}=\eta w_n\). Combining these relations,

\[ v_{n+1}^2 =\eta^4v_n^2 +2d\eta^2\left[(1+\eta^2)a_{\mathrm e} +(1-\eta^2)g\right]. \]

For \(0<\eta<1\), deviations of \(v_n^2\) from its fixed point decrease by a factor \(\eta^4\) per cycle. At steady state,

\[ (1-\eta^4)v_{\mathrm s}^2 =2d\eta^2\left[(1+\eta^2)a_{\mathrm e} +(1-\eta^2)g\right]. \]

Thus

\[ v_{\mathrm s}^2 =\frac{2d\eta^2a_{\mathrm e}}{1-\eta^2} +\frac{2d\eta^2g}{1+\eta^2} =\frac{2\chi\eta^2}{m(1-\eta^2)}V^2 +\frac{2gd\eta^2}{1+\eta^2}. \]

Comparing with Eq. (1.1),

\[ \boxed{\alpha=\frac{2\chi\eta^2}{m(1-\eta^2)},\qquad \beta=\frac{2gd\eta^2}{1+\eta^2}}. \]

(e)

In one complete cycle, the disk changes from \(-q\) to \(+q\) at the top plate and from \(+q\) to \(-q\) at the bottom plate. Thus a charge \(2q\) passes through the source per complete cycle. If the period is \(T\),

\[ I=\frac{2q}{T}. \]

When \(qV\gg mgd\), we have \(a_{\mathrm e}\gg g\), so gravity may be neglected. The two flights then have the same duration \(\tau\), with \(T=2\tau\).

Let \(w\) be the speed immediately before either impact. The speed immediately after impact is \(\eta w\). During each flight,

\[ w^2-\eta^2w^2=2a_{\mathrm e}d, \qquad w=\sqrt{\frac{2a_{\mathrm e}d}{1-\eta^2}}. \]

Because the acceleration is constant, the mean speed during a flight is \((1+\eta)w/2\). Therefore

\[ \tau=\frac{2d}{(1+\eta)w}, \]

and

\[ I=\frac{q}{\tau} =\frac{\chi V(1+\eta)}{2d} \sqrt{\frac{2\chi V^2}{m(1-\eta^2)}}. \]

For \(V\ge0\),

\[ I=\frac{1}{d} \sqrt{\frac{\chi^3}{2m}\frac{1+\eta}{1-\eta}}\,V^2. \]

Hence

\[ \boxed{\gamma=\frac{1}{d} \sqrt{\frac{\chi^3}{2m}\frac{1+\eta}{1-\eta}}}. \]

(f)

The high-voltage approximation in (e) cannot be used to determine cessation. Gravity must be retained.

Let \(u\) and \(w\) denote the steady-state speeds immediately before the top and bottom impacts, respectively. From (d), \(v_{\mathrm s}=\eta w\), and

\[ w^2=2d\left( \frac{a_{\mathrm e}}{1-\eta^2} +\frac{g}{1+\eta^2} \right). \]

Using the upward-flight energy equation,

\[ u^2=v_{\mathrm s}^2+2d(a_{\mathrm e}-g), \]

we obtain

\[ u^2=2d\left( \frac{a_{\mathrm e}}{1-\eta^2} -\frac{g}{1+\eta^2} \right). \]

As the voltage is reduced, the limiting periodic trajectory is the one that just reaches the upper plate with zero speed. Setting \(u=0\) gives

\[ a_{\mathrm e,c}=g\frac{1-\eta^2}{1+\eta^2}. \]

Since \(a_{\mathrm e}=\chi V^2/(md)\),

\[ \boxed{V_{\mathrm c} =\sqrt{\frac{mgd}{\chi}\frac{1-\eta^2}{1+\eta^2}}}. \]

Below this voltage, the disk cannot maintain flights reaching the upper plate. Once it misses that plate, its charge does not reverse, and successive bottom collisions dissipate its remaining mechanical energy.

At the critical voltage,

\[ u_{\mathrm c}=0,\qquad w_{\mathrm c}=2\sqrt{\frac{gd}{1+\eta^2}},\qquad v_{\mathrm s,c}=\eta w_{\mathrm c}. \]

The upward flight starts at speed \(\eta w_{\mathrm c}\) and ends at zero; the downward flight starts at zero and ends at \(w_{\mathrm c}\). Using the mean speeds,

\[ T_{\mathrm c} =\frac{2d}{\eta w_{\mathrm c}}+\frac{2d}{w_{\mathrm c}} =\frac{1+\eta}{\eta} \sqrt{\frac{d(1+\eta^2)}{g}}. \]

Therefore the current immediately before cessation is

\[ I_{\mathrm c} =\frac{2\chi V_{\mathrm c}}{T_{\mathrm c}} =\frac{2\eta\chi V_{\mathrm c}}{1+\eta} \sqrt{\frac{g}{d(1+\eta^2)}}. \]

Substitution of \(V_{\mathrm c}\) yields

\[ \boxed{I_{\mathrm c} =\frac{2\eta g\sqrt{m\chi}}{1+\eta^2} \sqrt{\frac{1-\eta}{1+\eta}}}. \]

For every fixed \(0<\eta<1\), this current is positive, so cessation produces a finite downward jump.

To specify the sketch precisely, the full conducting branch can be written without neglecting gravity. Its two flight times are

\[ \tau_\uparrow=\frac{2d}{\eta w+u}, \qquad \tau_\downarrow=\frac{2d}{w+\eta u}. \]

Thus

\[ \boxed{ I_{\mathrm{on}}(V)= \frac{2\chi V}{ \displaystyle\frac{2d}{\eta w+u} +\displaystyle\frac{2d}{w+\eta u} }}, \]

where \(u(V)\) and \(w(V)\) are given above. These expressions remain valid when the upward flight is decelerated.

Comparison with (c) gives

\[ \boxed{\frac{V_{\mathrm c}}{V_{\mathrm{th}}} =\sqrt{\frac{1-\eta^2}{2(1+\eta^2)}}<1}. \]

The requested sketch therefore has the following features:

  • Increasing voltage: follow \(I=0\) from the origin to \(V_{\mathrm{th}}\). Just above \(V_{\mathrm{th}}\), the disk lifts off and the steady current jumps to \(I_{\mathrm{on}}(V_{\mathrm{th}})\). Continue upward along the conducting branch to \(3V_{\mathrm{th}}\).
  • Decreasing voltage: retrace the conducting branch through \(V_{\mathrm{th}}\) and continue to \((V_{\mathrm c},I_{\mathrm c})\). Just below \(V_{\mathrm c}\), drop to \(I=0\), then follow the voltage axis back to the origin.
  • High-voltage shape: the conducting branch approaches the parabola \(I=\gamma V^2\). At \(3V_{\mathrm{th}}\), \(qV=18mgd\), so a suitable approximate endpoint for the rough sketch is \(I\approx9\gamma V_{\mathrm{th}}^2\).

Equivalently, away from the jump points:

Voltage range Increasing \(V\) Decreasing \(V\)
\(0\le V<V_{\mathrm c}\) \(0\) \(0\)
\(V_{\mathrm c}<V<V_{\mathrm{th}}\) \(0\) \(I_{\mathrm{on}}(V)\)
\(V_{\mathrm{th}}<V\le3V_{\mathrm{th}}\) \(I_{\mathrm{on}}(V)\) \(I_{\mathrm{on}}(V)\)

The interval \(V_{\mathrm c}<V<V_{\mathrm{th}}\) is the hysteresis interval: a resting disk stays at rest, while an already moving disk continues to shuttle.

For completeness, exactly \(\eta=0\) requires separate treatment. Every bottom impact leaves the disk at rest, so the contact lift-off condition from (c) must be met anew. Consequently, the current stops at \(V_{\mathrm{th}}\), rather than at the limit of the positive-\(\eta\) critical-voltage formula. On the conducting side, at \(V\to V_{\mathrm{th}}^+\), \(a_{\mathrm e}\to2g\), and

\[ T\to\sqrt{\frac{2d}{g}}+\sqrt{\frac{2d}{3g}}. \]

Hence

\[ I_{\mathrm c} =\frac{2\chi V_{\mathrm{th}}}{ \sqrt{2d/g}+\sqrt{2d/(3g)} } =\boxed{(3-\sqrt3)\,g\sqrt{m\chi}}. \]

Both switching voltages then coincide, so there is no hysteresis. At the other endpoint, \(\eta=1\), collisions dissipate no energy while the electric field adds \(2qV\) per cycle; a finite steady-state speed cannot exist.

The solution was checked against the official solution.