物理学引论 作业三

物理学引论

陆一洲 5140309557

Chapter 24 Question 5

Describe a simple method of measuring $\epsilon_0$ using a capacitor.

使用一个已知电容为 $C$ 的平行板电容器，测量出极板面积 $S$ 与 极板间距 $d$，之后我们即可得 $\epsilon_0=\frac{Cd}{S}$。

Chapter 24 Question 7

A large copper sheet of thickness $l$ is placed between the parallel plates of a capacitor, but does not touch the plates. How will this affect the capacitance?

【思路一】
插入一个铜板，可以看做是，将原来的电容器划分成两个串联的电容器。每个电容器的极板间距为$\frac{1}{2}(d-l)$，电容为

$$C'=\frac{2\epsilon_0S}{d-l}$$ 所以总的电容为 $$C^*=\frac{1}{1/C'_1+1/C'_2}=\frac{\epsilon_0S}{d-l}$$ 相比原电容 $C=\epsilon_0S/d$，电容增大、

【思路二】
由公式 $C=KC_0$，显然 $K_{copper}>K_{air}$，所以该电容器的电容 $C$ 将会增大。

Chapter 24 Question 13

How does the energy stored in a capacitor change when a dielectric is inserted if
(a) the capacitor is isolated so $Q$ does not change;
(b) the capacitor remains connected to a battery so $V$ does not change?

(a) 因为 $W_e=\frac{Q^2}{2C}$，$Q$ 不变，而 $C$ 相应变大，所以 $W_e$ 变小。
(b) 因为 $W_e=\frac{1}{2}CU^2$，$U$ 不变，而 $C$ 相应变大，所以 $W_e$ 变大。

Chapter 24 Problem 10

In a dynamic random access memory (DRAM) computer chip, each memory cell chiefly consists of a capacitor for charge storage. Each of these cells represents a single binary-bit value of $1$ when its $35$-$\mathrm{fF}$ capacitor ($1 \mathrm{fF} = 10^{-15} \mathrm{F}$) is charged at $1.5$ $\mathrm{V}$, or $0$ when uncharged at $0$ $\mathrm{V}$.
(a) When it is fully charged, how many excess electrons are on a cell capacitor's negative plate?
(b) After charge has been placed on a cell capacitor's plate, it slowly "leaks" off (through a variety of me- chanisms) at a constant rate of $0.30$ $\mathrm{fC/s}$. How long does it take for the potential difference across this capacitor to decrease by $1.0\%$ from its fully charged value? (Because of this leakage effect, the charge on a DRAM capacitor is "refreshed" many times per second.)

(a)

$$n=\frac{Q}{e}=\frac{CU}{e}=3.3 \times 10^5$$
(b) $$t=\frac{\triangle Q}{v_Q}=\frac{0.01CV}{v_Q}=1.8\ \mathrm{s}$$

Chapter 24 Problem 15

Dry air will break down if the electric field exceeds about $3.0 \times 10^6\ \mathrm{V/m}$. What amount of charge can be placed on a capacitor if the area of each plate is $6.8\ \mathrm{cm^2}$ ?

$$Q=CU=\frac{\epsilon_0S}{d}Ed=\epsilon_0ES=1.8 \times 10^{-8}\ \mathrm{C}$$

Chapter 24 Problem 20

In an electrostatic air cleaner ("precipitator"), the strong nonuniform electric field in the central region of a cylindrical capacitor (with outer and inner cylindrical radii $R_a$ and $R_b$) is used to create ionized air molecules for use in charging dust and soot particles (Fig. 24-21). Under standard atmospheric conditions, if air is subjected to an electric field magnitude that exceeds its dielectric strength $E_s=2.7 \times 10^6\ \mathrm{N/C}$, air molecules will dissociate into positively charged ions and free electrons. In a precipitator, the region within which air is ionized (the corona discharge region) occupies a cylindrical volume of radius $R$ that is typically five times that of the inner cylinder. Assume a particular precipitator is constructed with $R_b = 0.10\ \mathrm{mm}$ and $R_a = 10.0\ \mathrm{cm}$. In order to create a corona discharge region with radius $R = 5.0\ R_b$, what potential difference $V$ should be applied between the precipitator's inner and outer conducting cylinders? [Besides dissociating air, the charged inner cylinder repels the resulting positive ions from the corona discharge region, where they are put to use in charging dust particles, which are then "collected" on the negatively charged outer cylinder.]

$$U=ER\ln\frac{R_a}{R_b}=9.3\times 10^3\ \mathrm{V}$$

Chapter 24 Problem 39

Suppose one plate of a parallel-plate capacitor is tilted so it makes a small angle $\theta$ with the other plate, as shown in Fig. 24-28. Determine a formula for the capacitance $C$ in terms of $A$, $d$, and $\theta$, where $A$ is the area of each plate and $\theta$ is small. Assume the plates are square. [Hint: Imagine the capacitor as many infinitesimal capacitors in parallel. ]

$$C=\int_0^{\sqrt{A}}\frac{\epsilon_0\sqrt{A}\ \mathrm{d}x}{d+x\sin\theta}=\frac{\epsilon_0\sqrt{A}\ln\left(1+\frac{\sqrt{A}}{d}\sin\theta\right)}{\sin\theta}$$
由于 $\theta$ 非常小，所以我们近似认为 $\sin\theta\approx\theta$，$\ln(1+x)\approx x-\frac{1}{2}x^2$，可得：
$$C=\frac{\epsilon_0\sqrt{A}}{\theta}\ln\left({1+\frac{\theta\sqrt{A}}{d}}\right)=\frac{\epsilon_0A}{d}\left(1-\frac{\theta\sqrt{A}}{2d}\right)$$

Chapter 27 Question 1

A compass needle is not always balanced parallel to the Earth's surface, but one end may dip downward. Explain.

一枚钢镚抛抛抛，其正面朝上的概率不可能完美对半分，何况磁场线也不和地面保持平行，地面还坑坑洼洼。 @v@

Chapter 27 Question 14

The force on a particle in a magnetic field is the idea behind electromagnetic pumping. It is used to pump metallic fluids (such as sodium) and to pump blood in artificial heart machines. The basic design is shown in Fig. 27-36. An electric field is applied perpendicular to a blood vessel and to a magnetic field. Explain how ions are caused to move. Do positive and negative ions feel a force in the same direction?

就 图27-36 而言，磁场方向向右，电流方向向下，根据左手定则，则可知，力的方向垂直纸面向外。
正负离子受力方向相同。由于在电路中，正离子向下运动，负离子向上运动，都相当于向下运动的电流，所以受力方向相同。

Chapter 27 Question 19

How can you make a compass without using iron or other ferromagnetic material?

制作一个电路，电路中绕一个可以自由旋转的线圈，通上电，这个线圈就相当于是一个“磁铁”。然后再根据电流方向与安培右手定则，确定“磁极”。指南针即制作完成。

Chapter 27 Problem 5

The force on a wire is a maximum of $7.50 \times 10^{-2}\ \mathrm{N}$ when placed between the pole faces of a magnet. The current flows horizontally to the right and the magnetic field is vertical. The wire is observed to "jump" toward the observer when the current is turned on.
(a) What type of magnetic pole is the top pole face?
(b) If the pole faces have a diameter of $10.0\ \mathrm{cm}$, estimate the current in the wire if the field is $0.220\ \mathrm{T}$.
(c) If the wire is tipped so that it makes an angle of $10.0^{\circ}$ with the horizontal, what force will it now feel?

(a) 上端是 S 极。
(b)

$$I=\frac{F}{Bl}=3.4\ \mathrm{A}$$
(c) 其受力方向不会改变，受力大小将会变成原来的 $\cos10^{\circ}$，即 $$F=F_0\cos10^{\circ}=7.39\times 10^{-2}\ \mathrm{N}$$

Chapter 27 Problem 15

Alpha particles of charge $q=+2e$ and mass $m = 6.6 \times 10^{-27}\ \mathrm{kg}$ are emitted from a radioactive source at a speed of $1.6 \times 10^7\ \mathrm{m/s}$. What magnetic field strength would be required to bend them into a circular path of radius $r = 0.18\ \mathrm{m}$?

$$F=qvB=m\frac{v^2}{r}\ \Rightarrow\ B=\frac{mv}{qr}=1.8\ \mathrm{T}$$

Chapter 27 Problem 31

Suppose the Earth's magnetic field at the equator has magnitude $0.50 \times 10^{-4}\ \mathrm{T}$ and a northerly direction at all points. Estimate the speed a singly ionized uranium ion ($m = 238\ \mathrm{u}$, $q = e$) would need to circle the Earth $5.0\ \mathrm{km}$ above the equator. Can you ignore gravity? [Ignore relativity.]

计入重力，其中取 $g=0.96\ \mathrm{m/s^2}$：

$$qv_gB+mg=m\frac{v_g^2}{r}\ \Rightarrow\ v_g=1.3 \times 10^8\ \mathrm{m/s}$$
不计入重力：
$$qvB=m\frac{v^2}{r}\ \Rightarrow\ v=1.3 \times 10^8\ \mathrm{m/s}$$
所以，无需计入重力。