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OPhO 2021, Теория — Задача 4. Missing Energy

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2021 Online Physics Olympiad: Invitational Contest — v1.1 Theoretical Exam · 25 т.

Условие

Part A

1. (3 pts.) Consider a simple circuit with two parallel-plate capacitors of capacitance $C_1$ and $C_2$ connected to each other using purely conducting wires and a switch. One of the capacitors is initially charged to a voltage $V_0$ , while the other one is completely uncharged. The circuit is kept in a square shaped figure of side length $\ell$ throughout the problem, while the diameter of the conducting wires is $D$ . Find the initial total energy of the circuit when the switch is open, given by $E_0$ , and a sufficiently long time after the switch is closed, given by $E_\infty$ . Calculate the remaining energy $E_\Delta=E_0-E_\infty$ . What is $E_\Delta$ for the case $C_2\to\infty$ ?

Complete square capacitor circuit: left C1 and right C2 capacitors, both plates, top open switch and full square conducting wires. — Figure 5: The two parallel plate capacitor-switch circuit.

It seems odd for there to be a difference in energy as the circuit is a closed system. Three young scientists Fermi, Jackson, and Feynman have created different theories to find and verify the correct source of this missing energy.

Thermal Losses: Fermi

To investigate the cause of this missing energy, Fermi assumes that there must be an ohmic resistive load $r$ and a self-inductance $L$ in the circuit responsible for $E_\Delta$ .

2. (3 pts.) Find the current in the circuit $I(t)$ as a function of time and $E_\Delta$ for the circuit.

3. (1 pt.) For small values of $r$ , find the oscillation frequency $\Omega$ of $I(t)$ .

4. (1 pt.) For $L=0$ , can Fermi’s reasoning be correct for any value of $r$ ? If yes, what is this value of $r$ ?

Dipole Radiation Losses: Jackson

Jackson believes that the missing energy is dissipated in the form of dipole radiation losses due to the charges accelerating. He assumes that the electric dipole moment of the system remains constant during the process, but the magnetic moment is allowed to vary. Thus, he seeks to determine the maximal possible radiation losses. For this he uses Larmor’s formula, which states that for small velocities relative to the speed of light $c$ , total power radiated which the radiation power is defined as:

P_r=\frac{\ddot m^2}{4\pi\epsilon_0c^3}

where $m$ is the magnetic moment of the circuit as a function of time. Ignore all relativistic effects and the possible charge accumulation in the wires compared to that on the capacitor plates. Moreover, note that he does not assume any resistance or self-inductance in the circuit in his model.

5. (4 pts.) Find the total energy dissipation $E_r$ due to this radiation.

6. (1 pt.) For what value of time interval $\Delta\tau$ taken by the charges to move from one capacitor to another, can Jackson’s theory be reasoned true?

Kinetic Energy: Feynman

Feynman has the following hypothesis:

The missing energy goes into the kinetic energy of the charge carriers going from $C_1$ to $C_2$ . Assume that the mean free path of collisions of the carriers is $\lambda>2\ell$ .

7. (4 pts.) Find the total kinetic energy $\Delta K$ gained by the carriers during a total charge transfer from $C_1$ to $C_2$ .

8. (1 pt.) Could this be a valid hypothesis to explain the cause of the missing energy? When the charges get completely deposited on the plates of $C_2$ , what happens to this kinetic energy?

Part B

Instead of charging one capacitor using the other, we take an ideal parallel-plate capacitor such that the surface charge density $\pm\sigma$ on its plates is uniform throughout both plates, and that the charges are ‘fixed’ to the surface as the plate expands. The dimensions of the plates are $a,b$ and the plate separation distance is $d$ . The plate is now stretched quasi-statically by a factor of $\kappa$ in one of the dimensions such that the dimensions of the capacitor plates are now $\kappa a,b$ but the plate separation remains $d$ .

1. (6 pts.) Calculate the work $\mathrm dW$ done during stretching the plates of this capacitor. Also write a simplified form of this expression for $d\ll\kappa a,b$ .

2. (1 pt.) In this case, there are no resistive loads, and since the process of plate expansion is quasi-static, there is no gain in kinetic energy of the charge carriers. Where does the energy disappear in this case?

Published organizer clarifications (separate source, pages 2–3)

Quoted from the organizer clarification document:

Part A

There seems to be a numbering issue. This will be fixed when the document is updated.
“Find the energy of the circuit before the switch is open” is a typo. It should properly read: “Find the energy of the circuit when the switch is open”. $C_1$ is initially charged to $V_0$ .
Diameter of the cross-section of the current-carrying wire should be taken as $D$ .
In the first part, “fate of the remaining energy” refers to the calculation of $E_\Delta$ .
Report the value of $r$ in the last part of Fermi’s hypothesis. If the expression for $E_\Delta$ in Fermi’s hypothesis is left as an integral, it may not receive full points.
All quantities must be calculated in terms of $C_1$ and $C_2$ , not $E_\Delta$ unless specified.
For Jackson’s hypothesis, $r=0$ .
Refer to the wikipedia page for the non-relativistic version of Larmor’s frequency, i.e, do not use the formula given in the question paper (this will be updated in the final version). The formula in the version 1.1 of the question paper is off, use $P_r=\frac23\cdot\frac{\ddot m^2}{4\pi\epsilon_0c^5}$ . Perform all calculations in SI units.

Part B

The initial surface charge density of the capacitor plates is $\pm\sigma$ .
By “ideal capacitor”, we mean that the surface charge density of both the plates remains uniform. Also, the charges on the surface remain “fixed” to it as the capacitor plate is stretched.
Calculation of $\mathrm dW$ for $d\ll\kappa a,b$ must be performed after finding an answer in the general scenario. You may submit the general answer in the form of an integral, or otherwise, whichever suitable.

Решение

Внимание: Непълно решение

The complete available printed key is retained. For question5, its two accepted approaches give radiation power/resistance and an instruction to integrate Pr, but no evaluated total radiated energy Er.

Dipole Radiation Losses: Jackson

P_r=\frac23\cdot\frac{\ddot m^2}{4\pi\epsilon_0c^5}

T4: Part A — Problem 1

When the switch is open, there is $q_0=\pm C_1V_0$ charge on one of the capacitors, while the other capacitor is uncharged. At this time, we have

E_0=\frac{q_0^2}{2C_1}=\frac12C_1V_0^2

A long time after the switch is switched on, the potential across both capacitors must be equal. Moreover, by charge conservation we have

q_0=C_1V_0=q_\infty=(C_1+C_2)V_\infty\Rightarrow V_\infty=\frac{C_1V_0}{C_1+C_2}

Hence the energy after a long time is

E_\infty=\frac{q_\infty^2}{2(C_1+C_2)}=\frac12(C_1+C_2)V_\infty^2=\frac12\frac{C_1^2}{C_1+C_2}V_0^2

This means that

E_\Delta=E_0-E_\infty=\frac12\frac{C_1C_2}{C_1+C_2}V_0^2=\frac12C_{\mathrm{eq}}V_0^2

where $C_{\mathrm{eq}}$ is the equivalent capacitance of the circuit. For $C_1=C_2$ , this loss is minimal and equal to $E_0/2$ . For $C_2\to\infty$ , the whole $E_0$ energy is lost. In the following problems, we will investigate why there is a loss of energy and work out different explanations of it.

Problem 2

Let $q_1$ be the charge on the first capacitor at any time $t$ . By conservation of charge, the charge on the second capacitor must be $q_0-q$ . Apply Kirchoff’s Voltage Law on the circuit, we have

L\frac{\mathrm di}{\mathrm dt}+ir+\frac q{C_1}-\frac{q_0-q}{C_2}=0

and

i=-\frac{\mathrm dq}{\mathrm dt}

Differentiating the DE gives us

\frac{\mathrm d^2i}{\mathrm dt^2}+\frac rL\frac{\mathrm di}{\mathrm dt}+\frac1{LC_{\mathrm{eq}}}i=0

To solve this standard second order DE, we write its auxillary quadratic equation in $z$ :

z^2+\frac rLz+\frac1{LC_{\mathrm{eq}}}=0

Upon solving this we get

z=-\frac r{2L}\pm j\sqrt{\frac1{LC_{\mathrm{eq}}}-\frac{r^2}{4L^2}}

Substituting back $i=e^{jz}$ we obtain

i(t)=e^{-\frac r{2L}t}\left(c_1e^{j\sqrt{\frac1{LC_{\mathrm{eq}}}-\frac{r^2}{4L^2}}t}+c_2e^{-j\sqrt{\frac1{LC_{\mathrm{eq}}}-\frac{r^2}{4L^2}}t}\right)

Using Euler’s identity and imposing the boundary conditions $i(t=0)=0$ , $\frac{\mathrm di(t=0)}{\mathrm dt}=\frac{V_0}L$ we get the current of the form

i(t)=\begin{cases}\displaystyle\frac{V_0}{L\sqrt{\frac1{LC_{\mathrm{eq}}}-\frac{r^2}{4L^2}}}e^{-\frac r{2L}t}\sin\left(\sqrt{\frac1{LC_{\mathrm{eq}}}-\frac{r^2}{4L^2}}t\right)&\displaystyle\text{for }\frac1{LC_{\mathrm{eq}}}>\frac r{2L}\\\displaystyle\frac{V_0}Lt e^{-\sqrt{\frac1{LC_{\mathrm{eq}}}-\frac{r^2}{4L^2}}t}&\displaystyle\text{for }\frac1{LC_{\mathrm{eq}}}=\frac r{2L}\\\displaystyle\frac{V_0}{jL\sqrt{\frac1{LC_{\mathrm{eq}}}-\frac{r^2}{4L^2}}}e^{-\sqrt{\frac1{LC_{\mathrm{eq}}}-\frac{r^2}{4L^2}}t}\sinh\left(j\sqrt{\frac1{LC_{\mathrm{eq}}}-\frac{r^2}{4L^2}}t\right)&\displaystyle\text{for }\frac1{LC_{\mathrm{eq}}}<\frac r{2L}\end{cases}

for overdamping, critical damping, and underdamping. For small values of $r$ and $L$ , the system is likely to undergo underdamping. Also note that

E_\Delta=\int_0^\infty(i(t))^2r\mathrm dt

Substituting value of $i(t)$ from any of the cases, we find that

E_\Delta=\frac12\frac{C_1C_2}{C_1+C_2}V_0^2

as found in the first part. This means that Fermi’s explanation may be a valid one, if the experimental values suggest the existence of non-zero $r$ and $L$ .

Problem 3

For small values of $r$ , we must have

\frac1{LC_{\mathrm{eq}}}>\frac r{2L}

On a first degree of approximation for $r\to0^+$ , this gives us

\Omega=\sqrt{\frac1{LC_{\mathrm{eq}}}-\frac{r^2}{4L^2}}\approx\sqrt{\frac1{LC_{\mathrm{eq}}}}

Problem 4

Fermi’s reasoning should hold true for all finite values of $r>0$ for $L\to0^+$ . But note that for zero self-inductance in the circuit, it behaves as a simple RC-circuit with time constant $\tau=rC_{\mathrm{eq}}$ . Since relativistic effects are negligible, we must also simultaneously have $r\gg\frac\ell{cC_{\mathrm{eq}}}$ .

Problem 5

Due to discrepancy in Larmor’s formula during the competition, the following solutions to this problem will be given full credit:

Solution 1:

Magnetic moment of the circuit is $m=iA=i\ell^2$ , and thus $\ddot m=\ell^2\frac{\mathrm d^2i}{\mathrm dt^2}$ . Since the capacitors are arranged symmetrically and the current in the wires flows uniformly, the electric dipole of the system must remain constant. Hence, radiation arises from the magnetic dipole radiation. Assume the voltage drop due to this radiation as $V_r$ . We write

V_r=\frac{P_r(t)}{i(t)}=i(t)R_r=\frac23\frac{\ddot i^2\ell^4}{4\pi\epsilon_0c^5i}

where $R_r=P_r/i^2$ is called the radiation resistance of the circuit and $\ddot i=\mathrm d^2i/\mathrm dt^2$ . From our assumption, since $R_r$ is small, the characteristic time $\tau$ of the circuit must be large compared to the oscillation period. Thus we approximate as follows:

i=i_0e^{-t/\tau}\cos(\Omega t)

\frac{\mathrm d^2i}{\mathrm dt^2}\approx i_0e^{-t/\tau}\cos(\Omega t)

Thus $\ddot i=-\Omega^2i$ . The voltage drop is approximated to

V_r=\frac23\frac{\Omega^4i^2\ell^4}{4\pi\epsilon_0c^5i}=\frac{\Omega^4i\ell^4}{6\pi\epsilon_0c^5}

The dissipated power is given by

P_r(t)=\frac{\Omega^4i^2\ell^4}{6\pi\epsilon_0c^5}

and

R_r=\frac{\Omega^4\ell^4}{6\pi\epsilon_0c^5}

Noting that $\Omega=2\pi c/\lambda$ , we have that $R_r=k(\ell/\lambda)^4$ where $\lambda$ is the wavelength of radiation. In practice, $R_r$ is very small.

Solution 2: The net resistance in the system is zero. We have

a=\frac{\Delta v}\tau

Hence, we write the work energy theorem as follows:

e\Delta V_{12}=e\left[V_0-\frac q{C_{\mathrm{eq}}}\right]=\frac12m\Delta v^2

Substituting $\Delta v$ into Larmor’s formula, the power radiated by each charge is

\mathrm dP_r=\frac23\frac{q^2a^2}{4\pi\epsilon_0c^3}=\frac{q^2\Delta v^2}{4\pi\epsilon_0c^3\tau^2}=\frac{e^2\cdot\frac{2e[V_0-q/C_{\mathrm{eq}}]}m}{4\pi\epsilon_0c^3\tau^2}

We obtain $P_r$ by integrating the same from $q=0$ to $q=q_{\mathrm{total}}$ .

Problem 6

For Jackson’s hypothesis to be true:

\frac12m_e\Delta v^2=P_r\Delta\tau=\frac23\frac{q^2a^2}{4\pi\epsilon_0c^3}\tau

\tau=\frac{e^2}{3\pi\epsilon_0m_ec^3}\approx\frac{r_e}{3\pi\epsilon_0c}

where $r_e$ is the radius of an electron. Thus, all of the missing energy $\frac12C_{\mathrm{eq}}V_0^2$ can appear as dipole radiation iff the charges move from one capacitor to the other in the time in which light travels $1/(3\pi\epsilon_0)$ times the radius of an electron. However, this is theoretically impossible. Thus, Jackson’s hypothesis cannot be true.

Problem 7

Let the charge on the capacitor with capacitance $C_1$ be $q(t)$ . At equilibrium, final charge on $C_1$ is given by $q_0\frac{C_1}{C_1+C_2}$ . The potential difference between the plates after the charge has been transferred is given by:

\Delta V=\frac q{C_1}-\frac{q_0-q}{C_2}

The kinetic energy gained by the charges is then simply

\Delta K=\int_{q_0}^{\frac{C_1}{C_1+C_2}q_0}\Delta V\mathrm dq=\frac{q_0^2C_2}{2C_1(C_1+C_2)}=\frac{q_0^2}{2C_1}\left[\frac{C_2}{C_1+C_2}\right]=\frac12C_{\mathrm{eq}}V_0^2

Problem 8

This could be a possible explanation of the missing energy since the energy gained by the charge carriers comes out to be the same as $E_\Delta$ . When the charge gets deposited on the plate of the second capacitor, it accumulates there and dissipates thermal energy as it interacts with the charges on the plate of this capacitor.

Grading Scheme

A1	Correct answers for $E_\infty,E_\Delta$ and $C\to\infty$ case	2.5
A2	Correct analysis of $C_2\to\infty$ case	0.5
B1	Correct expression for $i(t)$	1.5
B2	Considered underdamping, critical damping and overdamping	0.5
B3	Correct simplified expression of $E_\Delta$	1.0
C1	Correct frequency approximated to the first order	1.0
D1	Noted that all non-zero finite values work	0.5
D2	Second constraint from characteristic time constant	0.5
E1	Correct idea of radiative voltage and resistance	1.0
E2	Correct approximation for large characteristic time constant $\tau$	1.0
E3	Correct calculation of power and energy dissipated	2.0
F1	Correct use of work energy theorem	0.5
F2	Correct final expression for $\tau$	0.5
G1	Exact expression of $\Delta V$	1.0
G2	Correct limits on charge transferred	1.0
G3	Calculation of $\Delta K=\int\Delta V\mathrm dq$	1.0
G4	Correct simplified final expression for kinetic energy	1.0
H1	Correct explanation of why the Feynman scenario is possible	1.0

Problem 9 (question Part B, 1)

Let us assume that the capacitor plates are lying in the $x-y$ plane, and the electric field between the plates is in the $z$ -direction. The initial potential energy of the system is given as $E_0=2\pi\sigma^2abd$ . There are repulsive forces on the charges of the plates parallel to its surface. Usually, these are ignored. But in this case, with the stretching of the plates, work will be performed by these forces. These forces are small near the centre of the plate surface, but as we will find out, appreciable as we go towards the plate edges performing a significant amount of work. The field is thus, fringing.

Full pair of oblique parallel capacitor plates, separation d double arrow, dx1 strip and complete +sigma0/-sigma0 labels.

Since the expansion is done in one dimension, only the $x$ -component of the forces is important. Take a line charge at $x=x_1$ with linear charge density $\sigma\mathrm dx_1$ . At $x=x_2$ on both plates, we also consider line charges. Net force of repulsion due to both the plates is then given by

\begin{aligned}F_0=\frac{\sigma^2}{2\pi\epsilon_0}\Bigg[&\int_0^{2x_1-a}\frac{\sqrt{b^2+(x_1-x_2)^2}-(x_1-x_2)}{x_1-x_2}\mathrm dx_2\\-&\int_0^{2x_1-a}\frac{x_1-x_2}{\sqrt{d^2+(x_1-x_2)^2}}\mathrm dx_2\cdot\frac{b^2+d^2+(x_1-x_2)^2}{\sqrt{d^2+(x_1-x_2)^2}}\sqrt{d^2+(x_1-x_2)^2}\Bigg]\end{aligned}

Now the plates are expanded quasi-statically such that the following transformations occur $\sigma\to\sigma/\kappa$ , $a\to\kappa a$ . As the expansion factor goes from $\kappa$ to $\kappa+\mathrm d\kappa$ , the line of charge at $x_1$ moves a distance $(x-\kappa a/2)\frac{\mathrm d\kappa}\kappa\hat x$ . The work done by the repulsive forces during expansion of both plates is then

\begin{aligned}\mathrm dW=4\times\frac{\sigma^2\mathrm d\kappa}{4\pi\epsilon_0\kappa}\int_0^{\kappa a}(2x_1-\kappa a)\cdot\Bigg(&\sqrt{x_1^2+b^2}-x_1+\sqrt{x_1^2+d^2}-\sqrt{x_1^2+b^2+d^2}\\&-b\ln\left(\frac{x_1}{\sqrt{x_1^2+b^2}-b}\cdot\frac{\sqrt{x_1^2+d^2}-b}{\sqrt{x_1^2+d^2}}\right)\Bigg)\mathrm dx_1\end{aligned}

Now we use the substitution $\Gamma=x_1-x_2$ , solve the indefinite integrals and simplify our result to obtain:

\begin{aligned}\mathrm dW=\frac{\sigma^2\mathrm d\kappa}{\pi\epsilon_0\kappa}\Bigg[&\frac{2d^2}3\left(\sqrt{\kappa^2a^2+d^2}-d+\sqrt{b^2+d^2}-\sqrt{\kappa^2a^2+b^2+d^2}\right)\\&-b^2\left(\sqrt{b^2+d^2}+\sqrt{\kappa^2a^2+b^2}-b+\sqrt{\kappa^2a^2+b^2+d^2}\right)\\&+\frac{\kappa^2a^2}6\left(\sqrt{\kappa^2a^2+d^2}+\sqrt{\kappa^2a^2+b^2}-\kappa a-\sqrt{\kappa^2a^2+b^2+d^2}\right)\\&+\frac{\kappa ab^2}2\ln\left(\frac b{\sqrt{\kappa^2a^2+b^2}-\kappa a}\cdot\frac{\sqrt{\kappa^2a^2+b^2+d^2}-\kappa a}{\sqrt{b^2+d^2}}\right)\\&-\frac{\kappa ad^2}2\ln\left(\frac d{\sqrt{\kappa^2a^2+d^2}-\kappa a}\cdot\frac{\sqrt{\kappa^2a^2+b^2+d^2}-\kappa a}{\sqrt{b^2+d^2}}\right)\\&-bd^2\ln\left(\frac d{\sqrt{d^2+b^2}-b}\cdot\frac{\sqrt{\kappa^2a^2+b^2+d^2}-b}{\sqrt{\kappa^2a^2+d^2}}\right)\\&+\kappa abd\arctan\left(\frac{\kappa ab}{d\sqrt{\sqrt{\kappa^2a^2+b^2+d^2}}}\right)\Bigg]\end{aligned}

This can be written as

\mathrm dW=4\sigma^2abd\frac{\mathrm d\kappa}{\kappa^2}\left(\frac\pi2+\mathcal O\left(\frac d{\kappa a},\frac db,\frac d{\sqrt{\kappa^2a^2+b^2}}\right)\right)

This is our answer for the general case.

Thus, for $d\ll\kappa a,b$ , we have

\mathrm dW=2\pi\sigma^2abd\frac{\mathrm d\kappa}{\kappa^2}=E_0\frac{\mathrm d\kappa}{\kappa^2}

Problem 10 (question Part B, 2)

There is work done against the repulsive forces between the charges on a plate within each capacitor, and the missing energy is used up as work done against them as the plates are expanded quasi-statically.

This analysis is very similar to what we did in Part A. In this case, there was no thermal, radiative losses or gain in kinetic energy due to the quasi-static nature of the process. Instead, the repulsive forces on the capacitor plates were significant and proportional to the plate dimensions. If some other external agency such as “magic tweezers” had been used to pick up charges from one capacitor to the another quasi-statically, the source of missing energy would have been the work done using the magic tweezers.

Grading Scheme

A1	Realize that repulsive forces between charges on plates are significant	1.0
A2	Write forces between line charges considering both plates	1.5
A3	Correct calculation for the work done in expansion	2.5
A4	Correct final simplified expression of $\mathrm dW$ and answer for the $d\ll\kappa a,b$ case	1.0
B1	Correct explanation of work done against repulsive forces	1.0

Organizer clarification document — other notes

Separately quoted from the organizer clarification document:

Other Notes

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Оригинал в Архива: theory.pdf · официални решения: theory_sol.pdf · допълнителни данни: Clarifications.pdf