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IPhO 2004, Теория — Задача 1. “Ping-Pong” Resistor

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Theoretical Question 1: “Ping-Pong” Resistor · 10 т.

Внимание: Бележка към темата

И двете налични архивни издания съдържат една и съща цяла задача с шест свързани подусловия и пълно решение с отговори, алтернативни методи и таблица за оценяване. Печатните несъответствия са запазени: χ е отрицателно в отговор (b), но по-нататък се използва под квадратен корен; в (e13) е отпечатано αV, а етикетите (f21) и (f22) се повтарят. Това са особености на оригиналното решение, а не поправени физични резултати.

Условие

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.

(a) Two circular parallel plates of radius R separated by distance d; the top plate is wired to the + terminal of a battery V, and the bottom plate and the battery are grounded. (b) Side view: top plate at +V, bottom plate grounded, separation d; a thin disk of radius r, thickness t and charge q lies on the bottom plate, with its weight mg drawn downward. — 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 $\eta \equiv (\mathrm{v}_{\mathrm{after}}/\mathrm{v}_{\mathrm{before}})$ , where $\mathrm{v}_{\mathrm{before}}$ and $\mathrm{v}_{\mathrm{after}}$ 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” $\mathrm{v}_{\mathrm{s}}$ , which depends on $V$ as follows:

\mathrm{v}_{\mathrm{s}} = \sqrt{\alpha V^2 + \beta}\,. \qquad (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 $3V_{\mathrm{th}}$ .

Отговори

Решение

1. Answers

(a) $\displaystyle F_{\mathrm{R}} = -\frac{1}{2}\pi R^2\varepsilon_0\frac{V^2}{d^2}$

(b) $\displaystyle \chi = -\varepsilon_0\frac{\pi r^2}{d}$

(d) $\displaystyle \mathrm{v}_{\mathrm{s}} = \sqrt{\alpha V^2 + \beta}$

\alpha = \left(\frac{\eta^2}{1-\eta^2}\right)\left(\frac{2\chi}{m}\right), \quad \beta = \left(\frac{\eta^2}{1+\eta^2}\right)(2gd)

(e) $\displaystyle \gamma = \sqrt{\frac{1+\eta}{1-\eta}}\sqrt{\frac{\chi^3}{2md^2}}$

(f) $\displaystyle V_c = \sqrt{\frac{1-\eta^2}{1+\eta^2}}\sqrt{\frac{mgd}{\chi}}, \quad I_c = \frac{2\eta\sqrt{1-\eta^2}}{(1+\eta)(1+\eta^2)}g\sqrt{m\chi}$

Sketch of current I versus V/V_th: I is zero below z_c; at V/V_th = 1 an upward arrow shows the current jumping up when V is increased, at z_c a downward arrow shows it dropping to zero from I_c (dashed line) when V is decreased; the curve rises as I ~ γV².

2. Solutions

(a) [1.2 points]

The charge $Q$ induced by the external bias voltage $V$ can be obtained by applying the Gauss law:

\varepsilon_0\oint\vec{E}\cdot d\vec{s} = Q \qquad (a1)

Q = \varepsilon_0 E\cdot(\pi R^2) = \varepsilon_0\left(\frac{V}{d}\right)\cdot(\pi R^2), \qquad (a2)

where $V = Ed$ .

The energy stored in the capacitor:

U = \int_0^V Q(V')\,dV' = \int_0^V \varepsilon_0\pi R^2\left(\frac{V'}{d}\right)dV' = \frac{1}{2}\varepsilon_0\pi R^2\frac{V^2}{d}. \qquad (a3)

The force acting on the plate, when the bias voltage $V$ is kept constant:

\therefore F_{\mathrm{R}} = +\frac{\partial U}{\partial d} = -\frac{1}{2}\varepsilon_0\pi R^2\frac{V^2}{d^2}. \qquad (a4)

[An alternative solution:]

Since the electric field $E'$ acting on one plate should be generated by the other plate and its magnitude is

E' = \frac{1}{2}E = \frac{V}{2d}, \qquad (a5)

the force acting on the plate can be obtained by

F_{\mathrm{R}} = QE'. \qquad (a6)

(b) [0.8 points]

The charge $q$ on the small disk can also be calculated by applying the Gauss law:

\varepsilon_0\oint\vec{E}\cdot d\vec{s} = q. \qquad (b1)

Since one side of the small disk is in contact with the plate,

q = -\varepsilon_0 E\cdot(\pi r^2) = -\varepsilon_0\frac{\pi r^2}{d}V = \chi V. \qquad (b2)

Alternatively, one may use the area ratio for $q = -\left(\dfrac{\pi r^2}{\pi R^2}\right)Q$ .

\therefore \chi = -\varepsilon_0\frac{\pi r^2}{d}. \qquad (b3)

The net force, $F_{\mathrm{net}}$ , acting on the small disk should be a sum of the gravitational and electrostatic forces:

F_{\mathrm{net}} = F_{\mathrm{g}} + F_{\mathrm{e}}. \qquad (c1)

The gravitational force: $F_{\mathrm{g}} = -mg$ .

The electrostatic force can be derived from the result of (a) above:

F_{\mathrm{e}} = \frac{1}{2}\varepsilon_0\frac{\pi r^2}{d^2}V^2 = \frac{\chi}{2d}V^2. \qquad (c2)

In order for the disk to be lifted, one requires $F_{\mathrm{net}} > 0$ :

\frac{\chi}{2d}V^2 - mg > 0. \qquad (c3)

\therefore V_{\mathrm{th}} = \sqrt{\frac{2mgd}{\chi}}. \qquad (c4)

(d) [2.3 points]

Let $\mathrm{v}_{\mathrm{s}}$ be the steady velocity of the small disk just after its collision with the bottom plate. Then the steady-state kinetic energy $K_{\mathrm{s}}$ of the disk just above the bottom plate is given by

K_{\mathrm{s}} = \frac{1}{2}m\mathrm{v}_{\mathrm{s}}^2. \qquad (d1)

For each round trip, the disk gains electrostatic energy by

\Delta U = 2qV. \qquad (d2)

For each inelastic collision, the disk lose its kinetic energy by

\Delta K_{\mathrm{loss}} = K_{\mathrm{before}} - K_{\mathrm{after}} = (1-\eta^2)K_{\mathrm{before}} = \left(\frac{1}{\eta^2} - 1\right)K_{\mathrm{after}}. \qquad (d3)

Since $K_{\mathrm{s}}$ is the energy after the collision at the bottom plate and $(K_{\mathrm{s}} + qV - mgd)$ is the energy before the collision at the top plate, the total energy loss during the round trip can be written in terms of $K_{\mathrm{s}}$ :

\Delta K_{\mathrm{tot}} = \left(\frac{1}{\eta^2} - 1\right)K_{\mathrm{s}} + (1-\eta^2)(K_{\mathrm{s}} + qV - mgd). \qquad (d4)

In its steady state, $\Delta U$ should be compensated by $\Delta K_{\mathrm{tot}}$ .

2qV = \left(\frac{1}{\eta^2} - 1\right)K_{\mathrm{s}} + (1-\eta^2)(K_{\mathrm{s}} + qV - mgd). \qquad (d5)

Rearranging Eq. (d5), we have

\begin{aligned} K_{\mathrm{s}} &= \frac{\eta^2}{1-\eta^4}\left[(1+\eta^2)qV + (1-\eta^2)mgd\right] \\ &= \left(\frac{\eta^2}{1-\eta^2}\right)qV + \left(\frac{\eta^2}{1+\eta^2}\right)mgd \\ &= \frac{1}{2}m\mathrm{v}_{\mathrm{s}}^2. \end{aligned} \qquad (d6)

Therefore,

\mathrm{v}_{\mathrm{s}} = \sqrt{\left(\frac{\eta^2}{1-\eta^2}\right)\left(\frac{2\chi V^2}{m}\right) + \left(\frac{\eta^2}{1+\eta^2}\right)(2gd)}. \qquad (d7)

Comparing with the form:

\mathrm{v}_{\mathrm{s}} = \sqrt{\alpha V^2 + \beta}, \qquad (d8)

\alpha = \left(\frac{\eta^2}{1-\eta^2}\right)\left(\frac{2\chi}{m}\right), \quad \beta = \left(\frac{\eta^2}{1+\eta^2}\right)(2gd). \qquad (d9)

[An alternative solution:]

Let $\mathrm{v}_n$ be the velocity of the small disk just after n-th collision with the bottom plate. Then the kinetic energy of the disk just above the bottom plate is given by

K_n = \frac{1}{2}m\mathrm{v}_n^2. \qquad (d10)

When it reaches the top plate, the disk gains energy by the increase of potential energy:

\Delta U_{\mathrm{up}} = qV - mgd. \qquad (d11)

Thus, the kinetic energy just before its collision with the top plate becomes

K_{n-\mathrm{up}} = \frac{1}{2}m\mathrm{v}_{\mathrm{up}}^2 = K_n + \Delta U_{\mathrm{up}}. \qquad (d12)

Since $\eta = \mathrm{v}_{\mathrm{after}}/\mathrm{v}_{\mathrm{before}}$ , the kinetic energy after the collision with the top plate becomes scaled down by a factor of $\eta^2$ :

K'_{n-\mathrm{up}} = \eta^2\cdot K_{n-\mathrm{up}}. \qquad (d13)

Now the potential energy gain by the downward motion is:

\Delta U_{\mathrm{down}} = qV + mgd \qquad (d14)

so that the kinetic energy just before it collides with the bottom plate becomes:

K_{n-\mathrm{down}} = K'_{n-\mathrm{up}} + \Delta U_{\mathrm{down}}. \qquad (d15)

Again, due to the loss of energy by the collision with the bottom plate, the kinetic energy after its $(n+1)$ -th collision can be obtained by

\begin{aligned} K_{n+1} &= \eta^2\cdot K_{n-\mathrm{down}} \\ &= \eta^2(K'_{n-\mathrm{up}} + \Delta U_{\mathrm{down}}) \\ &= \eta^2(\eta^2(K_n + \Delta U_{\mathrm{up}}) + \Delta U_{\mathrm{down}}) \\ &= \eta^2(\eta^2(K_n + qV - mgd) + qV + mgd) \\ &= \eta^4K_n + \eta^2(1+\eta^2)qV + \eta^2(1-\eta^2)mgd. \end{aligned} \qquad (d16)

As $n \to \infty$ , we expect the velocity $\mathrm{v}_n \to \mathrm{v}_{\mathrm{s}}$ , that is, $K_n \to K_{\mathrm{s}} = \frac{1}{2}m\mathrm{v}_{\mathrm{s}}^2$ :

\begin{aligned} K_{\mathrm{s}} &= \frac{1}{1-\eta^4}\left[\eta^2(1+\eta^2)qV + \eta^2(1-\eta^2)mgd\right] \\ &= \left(\frac{\eta^2}{1-\eta^2}\right)qV + \left(\frac{\eta^2}{1+\eta^2}\right)mgd \\ &= \frac{1}{2}m\mathrm{v}_{\mathrm{s}}^2 \end{aligned} \qquad (d17)

(e) [2.2 points]

The amount of charge carried by the disk during its round trip between the plates is $\Delta Q = 2q$ , and the time interval $\Delta t = t_+ + t_-$ , where $t_+$ ( $t_-$ ) is the time spent during the up- (down-) ward motion respectively.

Here $t_+$ ( $t_-$ ) can be determined by

\begin{aligned} \mathrm{v}_{0+}t_+ + \frac{1}{2}a_+t_+^2 &= d \\ \mathrm{v}_{0-}t_- + \frac{1}{2}a_-t_-^2 &= d \end{aligned} \qquad (e1)

where $\mathrm{v}_{0+}$ ( $\mathrm{v}_{0-}$ ) is the initial velocity at the bottom (top) plate and $a_+$ ( $a_-$ ) is the up- (down-) ward acceleration respectively.

Since the force acting on the disk is given by

F = ma_\pm = qE \mp mg = \frac{qV}{d} \mp mg, \qquad (e2)

in the limit of $mgd \ll qV$ , $a_\pm$ can be approximated by

a_0 = a_+ = a_- \approx \frac{qV}{md}, \qquad (e3)

which implies that the upward and down-ward motion should be symmetric. Thus, Eq.(e1) can be described by a single equation with $t_0 = t_+ = t_-$ , $\mathrm{v}_{\mathrm{s}} = \mathrm{v}_{0+} = \mathrm{v}_{0-}$ , and $a_0 = a_+ = a_-$ . Moreover, since the speed of the disk just after the collision should be the same for the top- and bottom-plates, one can deduce the relation:

\mathrm{v}_{\mathrm{s}} = \eta(\mathrm{v}_{\mathrm{s}} + a_0t_0), \qquad (e4)

from which we obtain the time interval $\Delta t = 2t_0$ ,

\Delta t = 2t_0 = 2\left(\frac{1-\eta}{\eta}\right)\frac{\mathrm{v}_{\mathrm{s}}}{a_0}. \qquad (e5)

From Eq. (d6), in the limit of $mgd \ll qV$ , we have

K_{\mathrm{s}} = \frac{1}{2}m\mathrm{v}_{\mathrm{s}}^2 \approx \left(\frac{\eta^2}{1-\eta^2}\right)qV. \qquad (e6)

By substituting the results of Eqs. (e3) and (e6), we get

\Delta t = 2\left(\frac{1-\eta}{\eta}\right)\sqrt{\frac{2\eta^2}{1-\eta^2}}\sqrt{\frac{md^2}{qV}} = 2\sqrt{\frac{1-\eta}{1+\eta}}\sqrt{\frac{2md^2}{\chi V^2}}. \qquad (e7)

Therefore, from $I = \dfrac{\Delta Q}{\Delta t} = \dfrac{2q}{\Delta t}$ ,

I = \frac{2q}{\Delta t} = \chi V\sqrt{\frac{1+\eta}{1-\eta}}\sqrt{\frac{\chi V^2}{2md^2}} = \sqrt{\frac{1+\eta}{1-\eta}}\sqrt{\frac{\chi^3}{2md^2}}V^2. \qquad (e8)

\therefore \gamma = \sqrt{\frac{1+\eta}{1-\eta}}\sqrt{\frac{\chi^3}{2md^2}} \qquad (e9)

[Alternative solution #1:]

Starting from Eq. (e3), we can solve the quadratic equation of Eq. (e1) so that

t_\pm = \frac{\mathrm{v}_{0\pm}}{a_0}\left(\sqrt{1 + \frac{2da_0}{\mathrm{v}_{0\pm}^2}} - 1\right). \qquad (e10)

When it reaches the steady state, the initial velocities $\mathrm{v}_{0\pm}$ are given by

\mathrm{v}_{0+} = \mathrm{v}_{\mathrm{s}} \qquad (e11)

\mathrm{v}_{0-} = \eta\cdot(\mathrm{v}_{\mathrm{s}} + a_0t_+) = \eta\mathrm{v}_{\mathrm{s}}\sqrt{1 + \frac{2da_0}{\mathrm{v}_{\mathrm{s}}^2}}, \qquad (e12)

where $\mathrm{v}_{\mathrm{s}}$ can be rewritten by using the result of Eq. (e6),

\mathrm{v}_{\mathrm{s}}^2 \approx \alpha V = \left(\frac{\eta^2}{1-\eta^2}\right)\frac{2qV}{m} = \left(\frac{\eta^2}{1-\eta^2}\right)2a_0d. \qquad (e13)

As a result, we get $\mathrm{v}_{0-} \cong \eta\mathrm{v}_{\mathrm{s}}\cdot\dfrac{1}{\eta} = \mathrm{v}_{\mathrm{s}}$ and consequently $t_\pm = \dfrac{\mathrm{v}_{\mathrm{s}}}{a_0}\left(\dfrac{1}{\eta} - 1\right)$ , which is equivalent to Eq. (e4).

[Alternative solution #2:]

The current $I$ can be obtained from

I = \frac{2q}{\Delta t} = \frac{2q\bar{\mathrm{v}}}{d}, \qquad (e14)

where $\bar{\mathrm{v}}$ is an average velocity. Since the up and down motions are symmetric with the same constant acceleration in the limit of $mgd \ll qV$ ,

\bar{\mathrm{v}} = \frac{1}{2}\left(\mathrm{v}_{\mathrm{s}} + \frac{\mathrm{v}_{\mathrm{s}}}{\eta}\right). \qquad (e15)

Thus, we have

I = \frac{q}{2d}\left(1 + \frac{1}{\eta}\right)\mathrm{v}_{\mathrm{s}}. \qquad (e16)

Inserting the expression (Eq. (e15)) of $\mathrm{v}_{\mathrm{s}}$ into Eq. (e16), one obtains an expression identical to Eq. (e8).

(f) [3 points]

The disk will lose its kinetic energy and eventually cease to move when the disk can not reach the top plate. In other words, the threshold voltage $V_{\mathrm{c}}$ can be determined from the condition that the velocity $\mathrm{v}_{0-}$ of the disk at the top plate is zero, i.e., $\mathrm{v}_{0-} = 0$ .

In order for the disk to have $\mathrm{v}_{0-} = 0$ at the top plate, the kinetic energy $\overline{K}_{\mathrm{s}}$ at the top plate should satisfy the relation:

\overline{K}_{\mathrm{s}} = K_{\mathrm{s}} + qV_c - mgd = 0, \qquad (f1)

where $K_{\mathrm{s}}$ is the steady-state kinetic energy at the bottom plate after the collision. Therefore, we have

\left(\frac{\eta^2}{1-\eta^2}\right)qV_c + \left(\frac{\eta^2}{1+\eta^2}\right)mgd + qV_c - mgd = 0, \qquad (f2)

or equivalently,

(1+\eta^2)qV_c - (1-\eta^2)mgd = 0. \qquad (f3)

\therefore \quad qV_{\mathrm{c}} = \frac{1-\eta^2}{1+\eta^2}mgd \qquad (f4)

From the relation $q = \chi V_{\mathrm{c}}$ ,

\therefore V_{\mathrm{c}} = \sqrt{\frac{1-\eta^2}{1+\eta^2}}\sqrt{\frac{mgd}{\chi}}. \qquad (f5)

In comparison with the threshold voltage $V_{\mathrm{th}}$ of Eq. (c4), we can rewrite Eq. (f5) by

V_c = z_cV_{\mathrm{th}} \qquad (f6)

where $z_c$ should be used in the plot of $I$ vs. $(V/V_{\mathrm{th}})$ and

z_c = \sqrt{\frac{1-\eta^2}{2(1+\eta^2)}}. \qquad (f7)

[Note that an alternative derivation of Eq. (f1) is possible if one applies the energy compensation condition of Eq. (d5) or the recursion relation of Eq. (d17) at the top plate instead of the bottom plate.]

Now we can setup equations to determine the time interval $\Delta t = t_- + t_+$ :

\mathrm{v}_{0-}t_- + \frac{1}{2}a_-t_-^2 = d \qquad (f8)

\mathrm{v}_{0+}t_+ + \frac{1}{2}a_+t_+^2 = d \qquad (f9)

where the accelerations are given by

a_+ = \frac{qV_{\mathrm{c}}}{md} - g = \left[\frac{1-\eta^2}{1+\eta^2} - 1\right]g = \left(\frac{-2\eta^2}{1+\eta^2}\right)g \qquad (f10)

a_- = \frac{qV_{\mathrm{c}}}{md} + g = \left[\frac{1-\eta^2}{1+\eta^2} + 1\right]g = \left(\frac{2}{1+\eta^2}\right)g \qquad (f11)

\frac{a_+}{a_-} = -\eta^2 \qquad (f12)

Since $\mathrm{v}_{0-} = 0$ , we have $\mathrm{v}_{0+} = \eta(a_-t_-)$ and $t_-^2 = 2d/a_-$ .

t_- = \sqrt{\frac{2d}{a_-}} = \sqrt{(1+\eta^2)\left(\frac{d}{g}\right)}, \qquad (f13)

By using $\mathrm{v}_{0+}^2 = \eta^2(2da_-) = -2da_+$ , we can solve the quadratic equation of Eq. (f9):

t_+ = \frac{\mathrm{v}_{0+}}{a_+}\left(\sqrt{1 + \frac{2da_+}{\mathrm{v}_{0+}^2}} - 1\right) = -\frac{\mathrm{v}_{0+}}{a_+} = \sqrt{\frac{2d}{\lvert a_+\rvert}} = \sqrt{\left(\frac{1+\eta^2}{\eta^2}\right)\left(\frac{d}{g}\right)} = \frac{t_-}{\eta}. \qquad (f14)

\therefore \Delta t = t_- + t_+ = \left(1 + \frac{1}{\eta}\right)\sqrt{(1+\eta^2)\left(\frac{d}{g}\right)} \qquad (f15)

I_{\mathrm{c}} = \frac{\Delta Q_{\mathrm{c}}}{\Delta t} = \frac{2q}{\Delta t} = \frac{2\chi V_{\mathrm{c}}}{\Delta t} = \frac{2\eta\sqrt{1-\eta^2}}{(1+\eta)(1+\eta^2)}g\sqrt{m\chi}. \qquad (f16)

Same I versus V/V_th sketch repeated after Eq. (f16): axis marks 0, z_c, 1; level I_c dashed; downward arrow at z_c, upward arrow at 1; curve labelled I ~ γV².

[A more elaborate Solution:]

One may find a general solution for an arbitrary value of $V$ . By solving the quadratic equations of Eqs. (f8) and (f9), we have

t_\pm = \frac{\mathrm{v}_{0\pm}}{a_\pm}\left[-1 + \sqrt{1 + \frac{2da_\pm}{\mathrm{v}_{0\pm}^2}}\right]. \qquad (f17)

(It is noted that one has to keep the smaller positive root.)

To simplify the notation, we introduce a few variables:

(i) $y = \dfrac{V}{V_{\mathrm{th}}}$ where $V_{\mathrm{th}} = \sqrt{\dfrac{2mgd}{\chi}}$ ,

(ii) $z_c = \sqrt{\dfrac{1-\eta^2}{2(1+\eta^2)}}$ , which is defined in Eq. (f7),

(iii) $w_0 = 2\eta\sqrt{\dfrac{gd}{1-\eta^2}}$ and $w_1 = 2\sqrt{\dfrac{d}{(1-\eta^2)g}}$ ,

In terms of $y$ , $w$ , and $z_c$ ,

a_+ = \frac{qV}{md} - g = g(2y^2 - 1) \qquad (f18)

a_- = \frac{qV}{md} + g = g(2y^2 + 1) \qquad (f19)

\mathrm{v}_{0+} = \mathrm{v}_{\mathrm{s}} = w_0\sqrt{y^2 + z_c^2} \qquad (f20)

\mathrm{v}_{0-} = \eta(\mathrm{v}_{\mathrm{s}} + a_+t_+) = w_0\sqrt{y^2 - z_c^2} \qquad (f21)

t_+ = w_1\frac{\sqrt{y^2 - z_c^2} - \eta\sqrt{y^2 + z_c^2}}{2y^2 - 1} \qquad (f22)

t_- = w_1\frac{\sqrt{y^2 + z_c^2} - \eta\sqrt{y^2 - z_c^2}}{2y^2 + 1} \qquad (f21)

I = \frac{\Delta Q}{\Delta t} = \frac{2q}{t_+ + t_-} = (2\chi V_{\mathrm{th}})\frac{y}{\Delta t} = \frac{\sqrt{8mgd\chi}}{w_1}F(y) \qquad (f22)

where

F(y) = y\left\{\frac{\sqrt{y^2 - z_c^2} - \eta\sqrt{y^2 + z_c^2}}{2y^2 - 1} + \frac{\sqrt{y^2 + z_c^2} - \eta\sqrt{y^2 - z_c^2}}{2y^2 + 1}\right\}^{-1} \qquad (f23)

Computed plot of I divided by the square root of (8mgdχ/w₁²), from 0 to 20, versus V/V_th from 0 to 3: a blue curve starts near V/V_th ≈ 0.55 and rises to about 18 at V/V_th = 3; a red vertical segment at V/V_th = 1 marks the jump from 0 to about 2.

3. Mark Distribution

No.	Total Pt.	Partial Pt.	Contents
(a)	1.2	0.3	Gauss law, or a formula for the capacitance of a parallel plate
		0.5	Total energy of a capacitor at $V$	$E'$ = electrical field by the other plate
		0.4	Force from the energy expression	$F = QE'$
(b)	0.8	0.3	Gauss law	Use of area ratio and result of (a)
		0.5	Correct answer
(c)	0.5	0.1	Correct lift-up condition with force balance
		0.2	Use of area ratio and result of (a)
		0.2	Correct answer
(d)	2.3	0.5	Energy conservation and the work done by the field
		0.5	Loss of energy due to collisions
		0.8	Condition for the steady state: energy balance equation (loss = gain)	Condition for the steady state: recursion relation
		0.5	Correct answer
(e)	2.2	0.2	$\Delta Q = 2q$ per trip
		0.5	Acceleration $a_\pm$ in the limit of $qV \gg mgd$ ; $a_+ = a_-$ by symmetry
		0.3	Kinetic equations for $d$ , $\mathrm{v}$ , $a$ , and $t$ , solutions for $t_+$	By using the symmetry, derive the relation (e4)
		0.4	Expression of $\mathrm{v}_{0\pm}$ and $t_\pm$ in its steady state
		0.4	Solutions of $t_\pm$ in approximation
		0.4	Correct answer
(f)	3.0	0.5	Condition for $V_c$ ; $K_{\mathrm{up}} = 0$ or $\mathrm{v}_{\mathrm{s,up}} = 0$	Using (d8), Recursion relations
		0.3	energy balance equation
		0.3	Correct answer of $V_c$
		0.7	Kinetic equations for $\Delta t$
		0.3	Correct answer of $I_c$
		0.9	Distinction between $V_{\mathrm{th}}$ and $V_c$ , the asymptotic behavior $I = \gamma V^2$ in plots
Total	10

Оригинал в Архива: 1.pdf · официални решения: 1_sol.pdf · допълнителни данни: IPhO_2004_Theoretical Question 1.pdf · допълнителни данни: IPhO_2004_Theo_solution 1.pdf