. . . plates?¹

ANSWER: The electric field is reduced by a factor of $\kappa$, giving E = 100/310 or . For simple effects like this, a long explanation is unnecessary. Derivations need not be reproduced if you understand their results.

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. . . resistances?²

ANSWER: R₁ + R₂ = 16 [1] or R₂ = 16 - R₁ [1a]. R₁^-1 + R₂^-1 = 1/3 [2] or ${\displaystyle {R_1 + R_2 \over R_1 R_2} = {1\over3}}$ [2a] or, substituting [1], $3 \times 16 = R_1 R_2$ [2b] or, substituting [1a], 48 = R₁ (16 - R₁) [2c] or, collecting terms by powers of R₁, R₁² - 16 R₁ + 48 = 0 [2d], from which the Quadratic Theorem gives ${\displaystyle R_1 = {16 \pm \sqrt{16^2 - 4 \times 48} \over 2}}$ so that $R_1 = 8 \pm 4$. Pick one sign for R₁ and the other for R₂: and . Quite a few people jumped over all the algebra to the correct answers. I would be grateful if someone could explain to me how this is possible.

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. . . closed?³

ANSWER: All six capacitors see the same voltage (1 V) and have the same capacitance (1 F) so each has the same charge (1 C) and the total charge is .

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. . . $e \equiv 2.71828\cdots$]⁴

ANSWER: By symmetry, no capacitor can ``get ahead'' of any of the others, so we have basically one big 6 F capacitor being drained through 3 resistors in parallel on each side. Each set of 3 resistors has an effective resistance of R/3 and they are in series, so the overall effective resistance is 2R/3. The time constant is thus $\tau = R_{\rm eff} C_{\rm eff} = (2R/3)(6C) = 4RC$ or .

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. . . Explain.⁵

ANSWER: . That capacitor would have a different time constant from the others, and the others would be affected as well. This was intended as a hint for the preceding question, where symmetry is essential to the argument. Surprisingly, many people gave self-contradictory answers; it seems the phrase, ``Would there be an unique answer . . . '' was frequently misunderstood. Again, I would be grateful if someone could explain this confusion to me.

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. . . 1 C.⁶

ANSWER: . The capacitor is already fully charged to $Q = C {\cal E}$.

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. . . uncharged.⁷

ANSWER: An uncharged capacitor is ``like a short'' so the current initially flows as if it weren't there: $I(t=0) = {\cal E}/R = 1$ A. Then as the capacitor charges up, its charge asymptotically approaches Q<sub>f</sub> = 1 C as $t \to \infty$. The time constant is $\tau = RC = 1$ s. Algebraically, . Many people omitted all quantitative information, despite the provision of numerical values for all relevant quantities and the emphasis on ``in detail''. I have emphasized the importance of doing one's calculations in terms of algebraic symbols until the very end, but then you should recognize whether a quantitative answer is called for.

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. . . Tesla]?⁸

ANSWER: $\vec{\mbox{\boldmath$F$\unboldmath }} = q (\vec{\mbox{\boldmath$E$\unboldmath } . . . . . . box{\boldmath$v$\unboldmath }} \times \vec{\mbox{\boldmath$B$\unboldmath }} = 0$ (no deflection) if $\vec{\mbox{\boldmath$E$\unboldmath }} = - \vec{\mbox{\boldmath$v$\unboldmath }} \times \vec{\mbox{\boldmath$B$\unboldmath }}$. Since all three vectors are mutually perpendicular, this means the magnitudes must obey E = v B or $B = E/v = 3 \times 10^4 / 2 \times 10^5$ or .

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. . . constant.⁹

ANSWER: This is actually a trivial question, embedded in a complicated context to see if you can remember the simple ideas under pressure. Since the Lorentz force is proportional to the cross product of the velocity with the magnetic field, it is perpendicular to both; thus a magnetic field can change the direction of the velocity, but never its magnitude. Without a change in speed, the kinetic energy cannot change either.

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. . . words.)¹⁰

ANSWER: By Ampère's Law, the field outside the whole cable (r>R) is zero, because the same current links any Ampèrian loop in both directions, canceling out. Inside the outer shell (r<R) only the central conductor's current links the loop, so the magnetic field goes around the wire in circles in a direction given by the Right Hand Rule (RHR): if the thumb of your right hand points along the current in the inner wire, its fingers curl around the wire in the direction of the magnetic field - i.e. down in front and up on the back side.

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. . . . ¹¹

ANSWER: Since the outer conductor plays no role for an Ampèrian loop with r<R, in this region the field is given by the familiar result for a simple wire: . Outside, as stated earlier, the field is zero: .

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. . . meters).¹²

ANSWER:

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. . . motion.¹³

This is not the usual orientation in which a pole vaulter would hold the pole during a ``run up,'' but this sprinter is performing a Physics experiment, so unconventional behaviour is expected!

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. . . other?¹⁴

ANSWER: This can be solved several ways.
You can imagine the pole to be part of a rectangular Faraday loop where the opposite side is stationary; so the area of the loop is increasing at a rate dA/dt = Lv where L=6 m and v=7 m/s and the magnetic flux through the loop is therefore increasing at a rate $d\Phi/dt = LvB_v$ where $B_v = (0.5 \times 10^{-4}) (\sin 40^\circ) = 0.3214 \times 10^{-4}$ T. Thus the potential developed around the loop (i.e. across the pole) is $6 \times 7 \times 0.3214 \times 10^{-4}$ = .
Or you can imagine a charge q in the wire in the pole, moving along at horizontal speed v through a vertical magnetic field component $B \sin \theta$, and thus experiencing a Lorentz force $F = q v B \sin \theta$ along the pole. If this force were to move the charge from one end of the pole to the other, it would do work $W = q L v B \sin \theta$; thus the potential created by the Lorentz force is $W/q = L v B \sin \theta$, giving the same answer as before.
Quite a few people got their trigonometry mixed up (or failed to read the question carefully); I gave more marks for understanding the basic principles than for flawless algebra or arithmetic.

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