Съдържание

APhO 2006, Теория — Задача 3C. Part C

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Всички материали от APhO 2006

7th Asian Physics Olympiad — Theoretical Question 3 · 2,5 т.

Условие

C. [2.5 points] Two identical conducting plates $\alpha$ and $\beta$ with charges $-Q$ and $+q$ respectively ( $Q > q > 0$ ) are located parallel to each other at a small distance. Another identical plate $\gamma$ with mass $m$ and charge $+Q$ is situated parallel to the original plates at distance $d$ from the plate $\beta$ (see fig 1). Surface area of the plates is $S$ . The plate $\gamma$ is released and can move freely, while the plates $\alpha$ and $\beta$ are kept fixed. Assume that the collision between the plates $\beta$ and $\gamma$ is elastic, and neglect the gravitational force and the boundary effects. Assume that the charge has enough time to redistribute between plates $\beta$ and $\gamma$ during the collision.

C1) What is the electric field $E_1$ acting on the plate $\gamma$ before the collision with the plate $\beta$ ?

C2) What are the charges of the plates $Q_\beta$ and $Q_\gamma$ after the collision?

C3) What is the velocity $\upsilon$ of the plate $\gamma$ after the collision at the distance $d$ from the plate $\beta$ ?

Three thick vertical parallel plates labelled α, β and γ at the top; at the bottom their charges are marked -Q, +q and +Q. A double-headed arrow labelled d shows the distance between plates β and γ. — Fig. 1

Отговори

Решение

The electric field acting on the plate $\gamma$ before the collision is

(C1)

E_1 = \frac{Q-q}{2\varepsilon_0 S}. \qquad (21)

The force acting on the plate is

F_1 = E_1 Q = \frac{(Q-q)Q}{2\varepsilon_0 S}. \qquad (22)

The work done by the electric field before the collision is

A_1 = F_1 d = \frac{(Q-q)Qd}{2\varepsilon_0 S}. \qquad (23)

The charge will get redistributed between two touching conducting plates during the collision. The values of the charges can be obtained from the condition that the electric field between the touching plates vanishes. If one assumes that the plate $\gamma$ is on the right side, the left surface of the combined plate will have the charge

(C2a)

Q_\beta = Q + q/2, \qquad (24)

and the right surface will have the charge

(C2b)

Q_\gamma = q/2. \qquad (25)

These charges remain on the plates after the collision is over. Now the force acting on the plate $\gamma$ equals $F_2 = E_2 q/2$ , where $E_2 = (q/2)/(2\varepsilon_0 S)$ . The work done by field $E_2$ is

A_2 = F_2 d = \frac{q^2 d}{8\varepsilon_0 S}. \qquad (26)

The total work done by the electric fields is

A = A_1 + A_2 = \frac{d}{2\varepsilon_0 S}\left(Q - \frac{q}{2}\right)^2. \qquad (27)

Velocity at the distance $d$ can be calculated using the following relation:

\frac{m\upsilon^2}{2} = A. \qquad (28)

Substituting (5) into (6), we finally get

(C3)

\upsilon = \left(Q - \frac{q}{2}\right)\sqrt{\frac{d}{m\varepsilon_0 S}}. \qquad (29)

Mark Distribution

No	Total Pt	Partial Pt	Contents
C	2.5	0.5	$E_1$
		0.5	$Q_\beta$
		0.5	$Q_\gamma$
		0.5	Total work by electric fields
		0.5	Answer for final velocity

Оригинал в Архива: 3.pdf · официални решения: 3_sol.pdf