討厭普物的第N天

Ch16 波 (Waves I)

1. (1) 繩波的張力與質量

In Fig., a string, tied to a sinusoidal (正弦) oscillator (震盪器) at P and running over a support at Q, is stretched by a block of mass m. Separation $L = 1.20$ m, linear density $\mu = 1.6$ g/m and the oscillator frequency $f = 140$ Hz. The amplitude of the motion at P is small enough for that point to be considered a node; a node also exists at Q.

(a) What mass $m$ allows the oscillator to set up the fourth harmonic on the string?

本小題考多種才能形成四諧波

尋找第四諧波 ($n=4$) 的條件：

波長 $\lambda = \frac{2L}{n} = \frac{2(1.20)}{4} = 0.60$ m。

繩波波速 $v=f\lambda=140\times 0.60=84$ m/s。

根據波速公式 $v=\sqrt{\frac{T}{\mu}}$ 可推得所需張力 $T=v^2\mu=84^2\times(1.6\times10^-3)=11.2896$ N。

張力由懸掛重物提供 $T=mg$ 故所需質量 $m\approx1.15$ kg。

(b) What standing wave mode, if any, can be set up if $m = 1.00$ kg?

本小題考駐波的形成要件（$L=n\frac{\Lambda}{2}$）

驗證 $m = 1.00$ kg 時的狀態：

此時張力 $T = 1.00 \times 9.8 = 9.8$ N。

實際波速 $v = \sqrt{\frac{9.8}{1.6 \times 10^{-3}}} \approx 78.26$ m/s。

若要形成頻率為 140 Hz 的駐波，波長應為 $\lambda = \frac{v}{f} = \frac{78.26}{140} \approx 0.559$ m。

對應的諧波數 $n = \frac{2L}{\lambda} = \frac{2(1.20)}{0.559} \approx 4.29$。

因為 $n$ 不是整數，無法滿足兩端為節點的邊界條件，因此無法產生駐波。

2. (11) 複合繩上的駐波

In Fig., an aluminum wire, of length $L_{1} = 60.0$ cm, cross-sectional area $1.00 \times 10^{-2} \text{ cm}^{2}$, and density $2.60 \text{ g/cm}^{3}$ is joined to a steel wire of density $7.80 \text{ g/cm}^{3}$ and the same cross-sectional area. The compound wire, loaded with a block of mass $m = 30.0$ kg, is arranged so that the distance $L_{2}$ from the joint to the supporting pulley is $86.6$ cm. Transverse waves are set up on the wire by an external source of variable frequency; a node is located at the pulley.

(a) Find the lowest frequency that generates a standing wave having the joint as one of the nodes.

找有駐點的最低頻率

先計算兩條金屬線的密度

$\mu_{1} = \rho_{1} A = (2.60 \text{ g/cm}^{3})(1.00 \times 10^{-2} \text{ cm}^{2}) = 0.026 \text{ g/cm} = 0.0026 \text{ kg/m}$

$\mu_{2} = \rho_{2} A = (7.80 \text{ g/cm}^{3})(1.00 \times 10^{-2} \text{ cm}^{2}) = 0.078 \text{ g/cm} = 0.0078 \text{ kg/m}$。

可以看出 $\mu_{2} = 3\mu_{1}$。

兩條線所受張力皆為 $T = mg = 30.0 \times 9.8 = 294$ N。

鋁線波速：$v_{1} = \sqrt{\frac{T}{\mu_{1}}} = \sqrt{\frac{294}{0.0026}} \approx 336.3$ m/s。

鋼線波速：$v_{2} = \sqrt{\frac{T}{\mu_{2}}} = \frac{v_{1}}{\sqrt{3}} \approx 194.2$ m/s。

要讓接點處形成節點，兩邊的駐波頻率必須相同。

$f = \frac{n_{1}v_{1}}{2L_{1}} = \frac{n_{2}v_{2}}{2L_{2}} \Rightarrow \frac{n_{2}}{n_{1}} = \left(\frac{L_{2}}{L_{1}}\right) \left(\frac{v_{1}}{v_{2}}\right) = \left(\frac{86.6}{60.0}\right) \times \sqrt{3} \approx 1.443 \times 1.732 \approx 2.5 = \frac{5}{2}$。

最小整數解為 $n_{1} = 2, n_{2} = 5$。

帶入求最低頻率 $f = \frac{2 \times 336.3}{2 \times 0.60} \approx 560.5$ Hz。

(b) How many nodes are observed at this frequency?

總節點數計算：

鋁線有 $n_{1} = 2$ 個波腹，貢獻 3 個節點。

鋼線有 $n_{2} = 5$ 個波腹，貢獻 6 個節點。

中間接點共用，因此總節點數為 $(n_{1}+1) + (n_{2}+1) - 1 = n_{1} + n_{2} + 1 = 2 + 5 + 1 = 8$ 個節點。

3. (17) 駐波的波方程式推導

Two sinusoidal waves with the same amplitude and wavelength travel through each other along a string that is stretched along an x axis. Their resultant wave is shown twice in Fig. 16.14, as the antinode A travels from an extreme upward displacement to an extreme downward displacement in 6.0 ms. The tick marks along the axis are separated by 5.0 cm; height H is 2.00 cm. Let the equation for one of the two waves be of the form $y(x,t) = y_{m}\sin(kx + \omega t)$.

In the equation for the other wave, what are (a) $y_{m}$, (b) k, (c) $\omega$, and (d) the sign in front of $\omega$?

駐波是由兩個振幅、波長、頻率相同但行進方向相反的波疊加而成。

(a) 總振幅為 $H/2 = 1.00$ cm。

由於駐波波腹的振幅是單一波的兩倍 ($2y_{m}$)，故 $y_{m} = 0.50$ cm。

(b) 相鄰兩個波節 (節點) 的距離是 2 格 = $10.0$ cm，這等於半個波長 $\lambda/2$。

所以波長 $\lambda = 20.0$ cm $= 0.20$ m。

wavenumber $k = \frac{2\pi}{\lambda} = \frac{2\pi}{0.20} = 10\pi \approx 31.4$ rad/m。

(c) 從最高點到最低點經歷了半個週期 $T/2 = 6.0$ ms，所以 $T = 12.0$ ms $= 0.012$ s。

角頻率 $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.012} = \frac{500\pi}{3} \approx 524$ rad/s。

(d) 已知其中一個波向 $-x$ 傳播 (符號為 $+ \omega t$)，另一個波必須向 $+x$ 傳播，所以它前面的符號應為 負號 ($-$)。

4. (25) 駐波粒子位移與速度

For a particular transverse standing wave on a long string, one of the antinodes is at $x = 0$ and an adjacent node is at $x = 0.10$ m. The displacement $y(t)$ of the string particle at $x = 0$ is shown in Fig., where the scale of the y axis is set by $y_{s} = 4.0$ cm.

When $t = 0.50$ s, what is the displacement of the string particle at (a) $x = 0.15$ m and (b) $x = 0.30$ m?

What is the transverse velocity of the string particle at $x = 0.15$ m at (c) $t = 0.50$ s and (d) $t = 1.0$ s?

根據題目條件建立方程式：

波腹在 $x=0$，相鄰節點在 $x=0.10$，這表示 $\lambda/4 = 0.10 \Rightarrow \lambda = 0.40$ m。

$k = \frac{2\pi}{\lambda} = 5\pi$ rad/m。

由圖，當 $x=0$ 時，最大振幅 $y_{s} = 4.0$ cm，且經歷一個完整向下的波週期需要 $2.0$ s，故 $T = 2.0$ s。

$\omega = \frac{2\pi}{T} = \pi$ rad/s。

從圖表得知，$t=0$ 時質點在平衡位置並準備往下掉，故時間項為 $-\sin(\pi t)$。

駐波方程式可寫作：$y(x,t) = -4.0 \cos(5\pi x) \sin(\pi t)$ cm。

(a) $t=0.50$ s, $x=0.15$ m：
$y = -4.0 \cos(0.75\pi) \sin(0.5\pi) = -4.0 (-\frac{\sqrt{2}}{2}) (1) \approx 2.83$ cm。

(b) $t=0.50$ s, $x=0.30$ m：
$y = -4.0 \cos(1.5\pi) \sin(0.5\pi) = -4.0 (0) (1) = 0$ cm。

(c) 速度方程式 $v_{y} = \frac{\partial y}{\partial t} = -4.0\pi \cos(5\pi x) \cos(\pi t)$ cm/s。

代入 $t=0.50, x=0.15$：$\cos(0.5\pi) = 0$，故 $v_{y} = 0$ cm/s。

(d) 代入 $t=1.0, x=0.15$：

$v_{y} = -4.0\pi \cos(0.75\pi) \cos(\pi) = -4.0\pi (-\frac{\sqrt{2}}{2}) (-1) \approx -8.89$ cm/s。

5. (27) 不同線密度的繩波傳遞

In Fig., string 1 has a linear density of $3.00$ g/m and string 2 has a linear density of $5.00$ g/m. They are under tension due to the hanging block of mass $M = 800$ g.

Calculate the wave speed on (a) string 1 and (b) string 2.

(a) & (b) 計算波速：

當繩子迴圈掛在滑輪上時，兩端的張力總和等於重物的重力，因此 $2T = Mg \Rightarrow T = \frac{0.8 \times 9.8}{2} = 3.92$ N。

String 1 波速：$v_{1} = \sqrt{\frac{T}{\mu_{1}}} = \sqrt{\frac{3.92}{0.003}} \approx 36.1$ m/s。

String 2 波速：$v_{2} = \sqrt{\frac{T}{\mu_{2}}} = \sqrt{\frac{3.92}{0.005}} \approx 28.0$ m/s。

Next the block is divided into two blocks (with $M_{1}+M_{2}=M$) and the apparatus is rearranged as shown in Fig. 16.19b

Find (c) $M_{1}$ and (d) $M_{2}$ such that the wave speeds in the two strings are equal.

(c) & (d) 分配質量使兩繩波速相等：

若 $v_{1} = v_{2}$，則 $\sqrt{\frac{M_{1}g}{\mu_{1}}} = \sqrt{\frac{M_{2}g}{\mu_{2}}} \Rightarrow \frac{M_{1}}{M_{2}} = \frac{\mu_{1}}{\mu_{2}} = \frac{3}{5}$。

且已知總質量不變 $M_{1} + M_{2} = 0.8$ kg。

依比例計算：
$M_{1} = 0.8 \times \frac{3}{8} = 0.3$ kg ($300$ g)。

$M_{2} = 0.8 \times \frac{5}{8} = 0.5$ kg ($500$ g)。

Ch17 聲波 (Waves II)

6. (5) 聲音的反射與週期頻率

A handclap on stage in an amphitheater sends out sound waves that scatter from terraces of width $w = 0.60$ m (Fig. 17.9). The sound returns to the stage as a periodic series of pulses, one from each terrace ; the parade of pulses sounds like a played note.

(a) Assuming that all the rays in Fig. 17.9 are horizontal, find the frequency at which the pulses return.

(a) 尋找回音的頻率：

相鄰階梯反射回來的聲音，其路徑差為來回各一次的寬度，即 $\Delta L = 2w = 1.20$ m。

聲波傳遞這個路徑差所需的時間 $\Delta t = \frac{\Delta L}{v} = \frac{1.20}{343}$ s。

所以聽到的脈衝頻率 $f = \frac{1}{\Delta t} = \frac{343}{1.20} \approx 286$ Hz。

(b) If the width w of the terraces were smaller, would the frequency be higher or lower?

(b) 階梯寬度的影響：
若寬度 $w$ 變小，路徑差 $\Delta L$ 變小，時間間隔 $\Delta t$ 會縮短，因此感知到的頻率 $f$ 會變高 (higher)。

7. (8) 都卜勒效應與風速影響

A girl is sitting near the open window of a train that is moving at a velocity of 10.00 m/s to the east. The girl's uncle stands near the tracks and watches the train move away. The locomotive whistle emits sound at frequency 600.0 Hz. The air is still.

(a) What frequency does the uncle hear?

叔叔聽到的頻率 (音源遠離靜止觀察者)：
$f' = f_{0} \left(\frac{v}{v + v_{s}}\right) = 600 \times \left(\frac{343}{343 + 10}\right) \approx 583.0$ Hz。

(b) What frequency does the girl hear?
A wind begins to blow from the east at 10.00 m/s.

(b) 女孩聽到的頻率：
女孩跟著火車一起移動，與聲源相對靜止，因此聽到的就是原頻率 $f' = 600.0$ Hz。

(c) What frequency does the uncle now hear?

(c) 刮風時叔叔聽到的頻率：

火車向東開 (遠離叔叔)，所以叔叔在火車西方。聲音向西傳遞。風從東方吹來，表示風也是向西吹。

有風時，等效聲速變為 $v' = v + v_{wind} = 343 + 10 = 353$ m/s。

$f' = f_{0} \left(\frac{v'}{v' + v_{s}}\right) = 600 \times \left(\frac{353}{353 + 10}\right) \approx 583.5$ Hz。

(d) What frequency does the girl now hear?

(d) 刮風時女孩聽到的頻率：

雖然有風，但女孩與聲源依舊是相對靜止的，沒有相對運動造成的都卜勒效應，因此仍為 $600.0$ Hz。

8. (9) 水中聲源的方向判斷

One clue used by your brain to determine the direction of a source of sound is the time delay $\Delta t$ between the arrival of the sound at the ear closer to the source and the arrival at the farther ear. Assume that the source is distant so that a wavefront from it is approximately planar when it reaches you, and let D represent the separation between your ears.

(a) If the source is located at angle $\theta$ in front of you (Fig. 17.11), what is $\Delta t$ in terms of D and the speed of sound v in air?

(a) 空氣中的時間延遲：

若聲源與前方夾角為 $\theta$，波前到達兩耳的路徑差為 $D\sin\theta$，故 $\Delta t = \frac{D\sin\theta}{v}$。

(b) If you are submerged in water and the sound source is directly to your right, what is $\Delta t$ in terms of D and the speed of sound $v_{w}$ in water?

(b) 水中的時間延遲：

聲源在正右方，表示角度為 $90^{\circ}$，路徑差為兩耳距離 $D$。水中的聲速為 $v_{w}$，因此 $\Delta t = \frac{D}{v_{w}}$。

(c) Based on the time-delay clue, your brain interprets the submerged sound to arrive at an angle from the forward direction. Evaluate $\theta$ for fresh water at $20^{\circ}C$.

(c) 計算大腦錯判的角度：
由於大腦習慣以空氣中的聲速來判斷，會將水中的時間差誤認為是某個空氣中的角度 $\theta$：
$\frac{D\sin\theta}{v} = \frac{D}{v_{w}} \Rightarrow \sin\theta = \frac{v}{v_{w}}$。

代入空氣聲速 $343$ m/s 與淡水聲速 $1482$ m/s：$\sin\theta = \frac{343}{1482} \approx 0.231$。

故大腦感知的角度為 $\theta = \arcsin(0.231) \approx 13.4^{\circ}$。

9. (20) 弦樂器的波速與張力

(a) Find the speed of waves on a violin string of mass 800 mg and length 22.0 cm if the fundamental frequency is 900 Hz.
(b) What is the tension in the string?
For the fundamental, what is the wavelength of (c) the waves on the string and (d) the sound waves emitted by the string?

首先轉換單位：質量 $m = 800$ mg $= 8 \times 10^{-4}$ kg，長度 $L = 0.22$ m。基頻 $f = 900$ Hz。

(a) 弦上的波速：
基礎諧波 (fundamental) 的波長為 $\lambda = 2L = 0.44$ m。
$v = f\lambda = 900 \times 0.44 = 396$ m/s。

(b) 弦的張力：

透過 $v = \sqrt{\frac{T}{\mu}}$，推得 $T = \mu v^{2} = \left(\frac{m}{L}\right)v^{2} = \left(\frac{8 \times 10^{-4}}{0.22}\right) \times 396^{2} \approx 570.24$ N。

(c) 弦上的波長：已在 (a) 計算出為 $0.44$ m。

(d) 發出聲音的波長：

聲音在空氣中傳遞，頻率一樣是 $900$ Hz，但必須使用空氣聲速 ($343$ m/s)。
$\lambda_{sound} = \frac{v_{air}}{f} = \frac{343}{900} \approx 0.381$ m。

10. (25) 雙點波源干涉

Figure 17.19 shows two point sources $S_{1}$ and $S_{2}$ that emit sound of wavelength $\lambda = 2.00$ m. The emissions are isotropic and in phase, and the separation between the sources is $d = 14.0$ m. At any point P on the x axis, the wave from $S_{1}$ and the wave from $S_{2}$ interfere.
When P is very far away ($x \approx \infty$), what are (a) the phase difference between the arriving waves from $S_{1}$ and $S_{2}$ and (b) the type of interference they produce?
Now move point P along the x axis toward $S_{1}$. (c) Does the phase difference between the waves increase or decrease?
At what distance x do the waves have a phase difference of (d) 0.50$\lambda$, (e) 1.00$\lambda$, and (f) 1.50$\lambda$?

假設 $S_{1}$ 在原點 $(0,0)$，P 點在 x 軸上 $(x,0)$，$S_{2}$ 距離 $S_{1}$ 為 $d=14.0$ m，位於 y 軸上 $(0,-d)$。

P 點到 $S_{1}$ 距離 $r_{1} = x$，到 $S_{2}$ 距離 $r_{2} = \sqrt{x^{2} + d^{2}}$。

路徑差 $\Delta L = \sqrt{x^{2} + d^{2}} - x$。

(a) 當 $x \to \infty$ 時，兩條路徑幾乎平行，$\Delta L \to 0$。因此相位差為 0。

(b) 相位差為 0，代表波峰對波峰，產生 完全建設性干涉 (Fully constructive interference)。

(c) 當 P 移向 $S_{1}$ ($x$ 變小) 時，$\Delta L$ 的值會變大，所以相位差會 增加 (Increase)。

(d)(e)(f) 求特定路徑差對應的 $x$：

設路徑差 $\Delta L = k\lambda$。
$\sqrt{x^{2} + d^{2}} - x = k\lambda \Rightarrow \sqrt{x^{2} + d^{2}} = x + k\lambda$。

兩邊平方：$x^{2} + d^{2} = x^{2} + 2xk\lambda + (k\lambda)^{2} \Rightarrow x = \frac{d^{2} - (k\lambda)^{2}}{2k\lambda}$。

代入已知 $d=14.0, \lambda=2.00$。

(d) 當 $k = 0.50$：$x = \frac{196 - (1)^{2}}{2(1)} = 97.5$ m。

(e) 當 $k = 1.00$：$x = \frac{196 - (2)^{2}}{2(2)} = 48.0$ m。

(f) 當 $k = 1.50$：$x = \frac{196 - (3)^{2}}{2(3)} = \frac{187}{6} \approx 31.2$ m。

Ch21 電荷 (Electric Charge)

11. (10) 靜電力的計算與平衡位置

The charges and coordinates of two charged particles held fixed in an xy plane are $q_{1}=+2.0~\mu C$ $x_{1}=3.5$ cm, $y_{1}=0.50$ cm, and $q_{2}=-4.0~\mu C$ $x_{2}=-2.0$ cm, $y_{2}=1.5$ cm. Find the (a) magnitude and (b) direction of the electrostatic force on particle 2 due to particle 1. At what (c) x and (d) y coordinates should a third particle of charge $q_{3}=+4.0~\mu C$ be placed such that the net electrostatic force on particle 2 due to particles 1 and 3 is zero?

(a) 計算靜電力大小：

兩者距離的向量 $\vec{r}_{12} = \vec{r}_{2} - \vec{r}_{1} = (-5.5, 1.0)$ cm。
距離平方 $r^{2} = (-5.5)^{2} + (1.0)^{2} = 31.25 \text{ cm}^{2} = 0.003125 \text{ m}^{2}$。

$F = \frac{k |q_{1}q_{2}|}{r^{2}} = \frac{(8.99 \times 10^{9})(2.0 \times 10^{-6})(4.0 \times 10^{-6})}{0.003125} \approx 23.0$ N。

(b) 力的方向：

因為 $q_{1}$ 是正電、$q_{2}$ 是負電，兩者互相吸引。粒子 2 受到粒子 1 的拉力方向是指向粒子 1。

方向向量為 $\vec{r}_{1} - \vec{r}_{2} = (5.5, -1.0)$。

角度 $\theta = \tan^{-1}(\frac{-1.0}{5.5}) \approx -10.3^{\circ}$ (相對於正 x 軸)。

(c) & (d) 放置第三個粒子使受力為零：

$q_{3}$ 是正電，也會吸引粒子 2。為抵消粒子 1 的拉力，粒子 3 的拉力必須跟粒子 1 在完全相反的方向，所以粒子 3 要放在與粒子 1 反方向的直線上。

根據庫倫定律，平衡條件為 $\frac{k |q_{1}q_{2}|}{r_{12}^{2}} = \frac{k |q_{3}q_{2}|}{r_{32}^{2}} \Rightarrow \frac{q_{1}}{r_{12}^{2}} = \frac{q_{3}}{r_{32}^{2}}$。

解出 $r_{32} = r_{12}\sqrt{\frac{q_{3}}{q_{1}}} = r_{12}\sqrt{\frac{4.0}{2.0}} = \sqrt{2} r_{12}$。

粒子 3 距離粒子 2 是粒子 1 的 $\sqrt{2}$ 倍，且方向相反：

$\vec{r}_{3} - \vec{r}_{2} = \sqrt{2} (\vec{r}_{2} - \vec{r}_{1}) = \sqrt{2} (-5.5, 1.0) \approx (-7.78, 1.41)$ cm。

所以 $\vec{r}_{3} = \vec{r}_{2} + (-7.78, 1.41) = (-2.0 - 7.78, 1.5 + 1.41) = (-9.78, 2.91)$ cm。

x 座標為 -9.78 cm，y 座標為 2.91 cm。

12. (21) 帶電粒子的力平衡

Figure 21.20 shows electrons 1 and 2 on an x axis and charged ions 3 and 4 of identical charge $-q$ and at identical angles $\theta$. Electron 2 is free to move; the other three particles are fixed in place at horizontal distances R from electron 2 and are intended to hold electron 2 in place. For physically possible values of $q \le 5e$ what are the (a) smallest, (b) second smallest, and (c) third smallest values of $\theta$ for which electron 2 is held in place?

為了讓中央的電子 2 保持靜止，它所受的淨力必須為零。

電子 1 (帶電 $-e$) 把它往右推，排斥力為 $F_{1} = \frac{ke^{2}}{R^{2}}$。

離子 3 與 4 (皆帶電 $-q$) 把它往左推，這兩個力的水平分量總和為：$F_{34,x} = 2 \times \frac{kqe}{R^{2}}\cos\theta$。

左右力平衡：$\frac{ke^{2}}{R^{2}} = \frac{2kqe\cos\theta}{R^{2}} \Rightarrow e = 2q\cos\theta \Rightarrow \cos\theta = \frac{e}{2q}$。

題目提到物理上可行的電荷，代表 $q$ 是基本電荷 $e$ 的整數倍，故可能的值有 $e, 2e, 3e, 4e, 5e$。

(a) 最小的角度 $\theta$ 對應最大的 $\cos\theta$。代入 $q = e$ 得到 $\cos\theta = \frac{1}{2}$，求得 $\theta = 60^{\circ}$。

(b) 第二小的角度，代入 $q = 2e$ 得到 $\cos\theta = \frac{1}{4}$，求得 $\theta \approx 75.5^{\circ}$。

(c) 第三小的角度，代入 $q = 3e$ 得到 $\cos\theta = \frac{1}{6}$，求得 $\theta \approx 80.4^{\circ}$。

13. (29) 兩固定電荷對第三電荷的靜電力

In Fig. 21.24a, particle 1 (of charge $q_{1}$) and particle 2 (of charge $q_{2}$) are fixed in place on an x axis, 4.00 cm apart. Particle 3 (of charge $q_{3}=+8.00\times10^{-19}$ C) is to be placed on the line between particles 1 and 2 so that they produce a net electrostatic force $\vec{F}_{3}$ net on it. Figure 21.24b gives the x component of that force versus the coordinate x at which particle 3 is placed. The scale of the x axis is set by $x_{s}=8.0$ cm. What are (a) the sign of charge $q_{1}$ and (b) the ratio $q_{2}/q_{1}$?

(a) 判斷 $q_{1}$ 的正負號：

觀察圖表，當 $x \to 0$ (極度靠近 $q_{1}$) 時，靜電力 $F \to +\infty$ (向右推)。因為測試電荷 $q_{3}$ 是正電荷，代表 $q_{1}$ 正在排斥它，故 $q_{1}$ 必定為正號 (Positive)。

(b) 求兩電荷比值：

圖表中的 X 軸比例尺 $x_{s} = 8.0$ cm，中間劃分為 4 格，故每格代表 $2.0$ cm。

根據圖表，當 $x = 2.0$ cm 時，$F = 0$。

這代表在距離 $q_{1}$ 為 2.0 cm 的地方 (也就是兩顆固定電荷的正中央，總長 4.00 cm) 達到了靜電平衡。

由庫倫定律：$\frac{kq_{1}q_{3}}{x^{2}} = \frac{kq_{2}q_{3}}{(4-x)^{2}}$

代入 $x=2$：$\frac{q_{1}}{2^{2}} = \frac{q_{2}}{(4-2)^{2}} \Rightarrow \frac{q_{1}}{4} = \frac{q_{2}}{4}$。

可推得比值 $q_{2}/q_{1} = 1$。

14. (33) 等邊三角形的靜電力

In Fig. 21.26a, particles 1 and 2 have charge 20.0 $\mu$C each and are held at separation distance $d = 0.50$ m.
(a) What is the magnitude of the electrostatic force on particle 1 due to particle 2?

In Fig. 21.26b, particle 3 of charge 20.0 $\mu$C is positioned so as to complete an equilateral triangle.
(b) What is the magnitude of the net electrostatic force on particle 1 due to particles 2 and 3?

(a) 粒子 1 受粒子 2 的作用力大小：

兩者電荷相同 $q_{1} = q_{2} = 20.0 \times 10^{-6}$ C。

庫倫力 $F = \frac{k q_{1}q_{2}}{d^{2}} = \frac{(8.99 \times 10^{9})(20.0 \times 10^{-6})^{2}}{0.50^{2}} = \frac{(8.99 \times 10^{9})(400 \times 10^{-12})}{0.25} \approx 14.384$ N。

(b) 加上第三顆粒子後的淨力：

加入粒子 3 形成等邊三角形，代表粒子 2 與粒子 3 對粒子 1 施加的排斥力大小皆為 $F = 14.384$ N，且兩力之間的夾角為 $60^{\circ}$。

依照向量相加法則，淨力大小 $F_{net} = 2F \cos(30^{\circ}) = 2(14.384)(\frac{\sqrt{3}}{2}) = 14.384\sqrt{3} \approx 24.9$ N。