# 物理學 Physics (U1559)
###### tags: `Physics 2020 (U1559)`
Lecturer: 江振宇 副教授

---

## Chapter 23: Electric Fields 積分技巧補充
######
本章在計算continuous electric charge distribution的時候，會用到一些基本的積分技巧，會使用到以下兩種基本的積分形式：
1. $\int_a^bx^{-2}dx$
2. $\int_a^b\frac{x}{(x^2+y^2)^{1.5}}dx$

請各位同學詳讀以下解法。

### 1. 簡單積分式
$\int_a^bx^{-2}dx＝(-1)x^{-1}\Big|_a^b=\frac{-1}{b}-\frac{-1}{a}=\frac{1}{a}-\frac{1}{b}$

### 2. 需要變數代換的積分式 $\int_a^b\frac{x}{(x^2+y^2)^{1.5}}dx$

在這裡我們要利用變數代換的技巧($u$ substitution)，從數學式可以觀察到分子$x$是分母的$(x^2+y^2)$以$x$微分後再乘以$0.5$，也就是：

$\frac{d(x^2+y^2)}{dx}=2x$

所以我們可以令 $u=x^2+y^2$，也就是：

$\frac{du}{dx}=2x$

或是$xdx=\frac{1}{2}du$

接下來我們可以改寫原本的積分式：

$\int\frac{x}{(x^2+y^2)^{1.5}}dx＝\int\frac{\frac{1}{2}du}{(u)^{\frac{3}{2}}}=\int\frac{1}{2}u^{\frac{-3}{2}}du=-u^{\frac{-1}{2}}+K$

其中$K$是一個constant

因為$u=x^2+y^2$且上式的積分是對$u$這個變數做處理，所以原本對於$x$的上下限要換成對於$u$的上下限：
下限：因為 $x=a$ 所以 $u=x^2+y^2=a^2+y^2$
上限：因為 $x=b$ 所以 $u=x^2+y^2=b^2+y^2$

最後：

$\int_a^b\frac{x}{(x^2+y^2)^{1.5}}dx=-u^{\frac{-1}{2}}\Big|_{a^2+y^2}^{b^2+y^2}=(a^2+y^2)^{\frac{-1}{2}}-(b^2+y^2)^{\frac{-1}{2}}$

###### <div style="text-align: right"> Last updated: March 08, 2020, 08:41 GMT </div>
---

## Mini Project 1
請寫一個C語言程式，給予一些point source charge的位置（不只一個）以及test point的位置（一個位置），便可以得到test point的electric field的相量。

### 程式要求
寫一個C語言程式 comelefield_xxxxxxxxx.c，xxxxxxxxx代表你的學號，這個程式可以用以下方法計算出電場向量，列印到螢幕上：

`comelefield.exe x y z < source_charge.txt`

其中 comelefield 為執行檔的名稱，x、y、z分別是test point的空間位置，source_charge.txt這個檔案裡面記錄了point source charge的電量以及空間位置。比如：

D:\comelefield.exe 0.2 0.1 0.9 < source_charge.txt
Ex=0.123456
Ey=1.234456
Ez=-7.234423

也就是計算點於 $(x,y,z)=(0.2,0.1,0.9)$ 因為point source charge(資訊儲存於source_charge.txt)而貢獻出的電場向量<Ex,Ey,Ez>，此source_charge.txt裡面儲存資訊如下：

N
x1 y1 z1 q1
x2 y2 z2 q2
...
...
...
xN yN zN qN

N: number of the point source charges
xi, yi, and zi represent position of i-th point charge on x-axis, y-axis, and z-axis, respectively.
qi represents charge of i-th point source charge.

###### <div style="text-align: right"> Last updated: March 30, 2020, 12:16 GMT </div>
---

## Mini Project 2
請手寫計算下圖原點(紅色點的地方）的**電場**以及**電位**，source charges 分佈在藍色粗線圓弧上，其電荷密度為：
$\lambda(\theta)=asin(\theta)\tag{1}$
其中 $a$ 為一個常數， $\theta$ 為角度(單位：radian)，$\lambda(\theta)$ 的單位為 $C/m$，$r$ 為圓弧的半徑(單位: $m$)，此圓弧是由 $\theta_1$ 一直彎到 $\theta_2$。


請將手寫過程掃描成pdf檔，並繳交到數位學苑，檔名為：
pj2-xxxxxxxx.pdf 其中 xxxxxxxx 為你的學號。


###### <div style="text-align: right"> Last updated: May 4, 2020, 06:07 GMT </div>

---

## Mini Project 3
請寫一個C語言程式，產生Mini Project 2裡面的source charges，讓使用mini project 1程式計算的電場以及電位以科學記號表示後：
$c\times 10^n\tag{2}$
$c$ 的小數點在四捨五入後到小數點6為的精確度。其中 $a=0.01$ ，$r=0.1$ ，$\theta_1=0$，$\theta_2=1.5\pi$。請假設這些source charges都在xy plane上！
### 程式要求
#### 1. Generate source charges
寫一個C語言程式 gen_charges_xxxxxxxxx.c，xxxxxxxxx代表你的學號，這個程式可以產生 source_charge.txt。

`gen_charges.exe > source_charge.txt`

source_charge.txt內容：

N
x1 y1 z1 q1
x2 y2 z2 q2
...
...
...
xN yN zN qN

N: number of the point source charges
xi, yi, and zi represent position of i-th point charge on x-axis, y-axis, and z-axis, respectively.
qi represents charge of i-th point source charge.

#### 2. Calculate electric field and electric potential
寫另一個C語言程式 comfieldpot_xxxxxxxxx.c ，計算原點的electric field以及potential。比如：

D:\comfieldpot_xxxxxxxxx.exe 0.0 0.0 0.0 < source_charge.txt
Ex=0.123456
Ey=1.234456
Ez=-7.234423
V=0.123456

###### <div style="text-align: right"> Last updated: April 20, 2020, 07:12 GMT </div>

---

## Midterm Examination 解答
### Problem 1
Three spherical conductors of diameters (直徑) $d_1$, $d_2$ ,and $d_3$ are separated by a distance much greater than the radius of each sphere. The spheres are connected by a conducting wire as shown in Figure 1. Assume each of the three spheres are uniformly charged in electrostatic equilibrium. The total charges of the three spheres are $Q$ Coulombs. Find the charge density (6 points), electric potential (3 points), and magnitude of electric field (6 points) on the surface of each of the spherical conductors. (15 points)
###### <div style="text-align: center"></div>
Ans:
#### Find the charge densities
Assume the charges on the spherical conductors with diameters $d_1$, $d_2$, and $d_3$ are $q_1$, $q_2$, and $q_3$, respectively. Since the three spherical condoctors are in electrostatic equilibrium, the electric potentials on the three conductors are the same, i.e.,
$$V_1=V_2=V_3$$
where
$$V_1=k_eq_1/(d_1/2)$$

$$V_2=k_eq_2/(d_2/2)$$

$$V_3=k_eq_3/(d_3/2)$$
So we know that
$$q_1:q_2:q_3=d_1:d_2:d_3$$

$$Q=q_1+q_2+q_3 \to$$

$$q_1=Qd_1/(d_1+d_2+d_3)$$

$$q_2=Qd_2/(d_1+d_2+d_3)$$

$$q_3=Qd_3/(d_1+d_2+d_3)$$

The charge densities $\sigma_1$, $\sigma_2$, and $\sigma_3$ for the conductors with diameters $d_1$, $d_2$ ,and $d_3$, are respectively:
$$\sigma_1=q_1/(4\pi (d_1/2)^2)=q_1/(\pi d_1^2)\\
=Qd_1/(\pi d_1^2(d_1+d_2+d_3))\\
=Q/(\pi d_1(d_1+d_2+d_3))$$

$$\sigma_2=Q/(\pi d_2(d_1+d_2+d_3))$$

$$\sigma_3=Q/(\pi d_3(d_1+d_2+d_3))$$

#### Find the electric potentials
$$V_1=V_2=V_3=k_eq_1/(d_1/2)=2k_eq_1/d_1\\
=2k_e(Qd_1/(d_1+d_2+d_3))/d_1\\
=2k_eQ/(d_1+d_2+d_3)$$

#### Find the magnitudes of electric fields
$$E_1=k_eq_1/(d_1/2)^2=4k_eq_1/d_1^2\\
=4k_e(Qd_1/(d_1+d_2+d_3))/d_1^2\\
=4k_eQ/(d_1(d_1+d_2+d_3))$$

$$E_2
=4k_eQ/(d_2(d_1+d_2+d_3))$$

$$E_3
=4k_eQ/(d_2(d_1+d_2+d_3))$$

---

### Problem 2
Suppose two infinite planes of charge are parallel to each other (Figure 2), one positively charged and the other negatively charged. The positively and negatively charged planes have charge densities of $\sigma_L$ and $-\sigma_R$, respectively. What are the directions and magnitudes of electric field at point P1 (to the left of the positively charged plane), P3 (to the right of the negatively charged plane) and P2 (between the two planes). (6 answers with 12 points)
###### <div style="text-align: center"></div>
Ans:
P1: $E_{P1}=\sigma_{L}/(2\epsilon_0)-\sigma_R/(2\epsilon_0)$, point left

P2: $E_{P2}=\sigma_{L}/(2\epsilon_0)+\sigma_R/(2\epsilon_0)$, point right

P3: $E_{P3}=\sigma_{L}/(2\epsilon_0)-\sigma_R/(2\epsilon_0)$, point right

---

### Problem 3
A solid cylindrical conductor of radius a and charge $Q$ is coaxial with a cylindrical shell of negligible thickness, radius $b > a$, and charge $–Q$ (Figure 3).
a) Find the capacitance of this cylindrical capacitor if its length is $l$. (5 points)
b) Calculate the energy stroed in the capacitance. (5 points)
c) Find the resistance between the solid cylindrical conductor and the cylindrical shell. Assume the resistivity between the solid cylindrical conductor and the cylindrical shell is $\sigma$. (5 points)
###### <div style="text-align: center"></div>
Ans:
#### capacitance
$$Q=CV$$
$$V=-\int_{r=a}^{b}E(r)dr$$
$$E(r)A=\frac{Q}{\epsilon_0}$$
where $A=2\pi rl$
$$E(r)=\frac{Q}{2\pi l\epsilon_0r}$$
$$V=-\int_{r=a}^{b}\frac{Q}{2\pi l\epsilon_0r}dr=-\frac{Q}{2\pi l\epsilon_0}\int_{r=a}^{b}\frac{1}{r}dr\\
=\frac{Q}{2\pi l\epsilon_0}ln(b/a)$$
$$Q=C\frac{Q}{2\pi l\epsilon_0}ln(b/a)$$
$$C=\frac{2\pi l\epsilon_0}{ln(b/a)}$$

#### stored energy
$$energy=\frac{Q^2}{2C}=\frac{1}{2}\frac{ln(b/a)}{2\pi l\epsilon_0}Q^2=\frac{ln(b/a)}{4\pi l\epsilon_0}Q^2$$

#### resistance
$$dR=\frac{\sigma dr}{2\pi rl}$$
$$R=\int_{r=a}^{b}dR=\int_{r=a}^{b}\frac{\sigma dr}{2\pi rl}=\frac{\sigma ln(b/a)}{2\pi l}$$

---

### Problem 4
As shown in Figure 4, a solid insulating sphere of radius a carries a net positive charge Q uniformly distributed throughout its volume. A conducting spherical shell of inner radius b and outer radius c is concentric with the solid sphere and carries a net charge –5Q. Note that the whole system is in electrostatic equilibrium. Please find: (a) the electric field in the regions labeled 1, 2, 3, and 4 (8 points), (b) the charge distribution in the conducting spherical shell (region 3) (4 points), and (c) the capacitance of whole system (3 points)
###### <div style="text-align: center"> </div>
Ans:
#### (a) the electric field in the regions labeled 1, 2, 3, and 4
##### region 1:
$$4\pi r^2E(r)=\frac{q_{in}}{\epsilon_0}$$

$$q_{in}=\frac{r^3}{a^3}Q$$

$$E(r)=\frac{\frac{r^3}{a^3}Q}{4\pi r^2\epsilon_0}=\frac{rQ}{4\pi a^3\epsilon_0}$$

##### region 2:
$$4\pi r^2E(r)=\frac{q_{in}}{\epsilon_0}$$

$$q_{in}=Q$$

$$E(r)=\frac{Q}{4\pi r^2\epsilon_0}$$

##### region 3:
$$4\pi r^2E(r)=\frac{q_{in}}{\epsilon_0}$$

$$q_{in}=0$$

$$E(r)=0$$

##### region 4:
$$4\pi r^2E(r)=\frac{q_{in}}{\epsilon_0}$$

$$q_{in}=-4Q$$

$$E(r)=\frac{-Q}{\pi r^2\epsilon_0}$$

#### (b) the charge distribution in the conducting spherical shell (region 3)
The charges on the inner shell ($r=b$) are uniformly distibuted with the total charges of $-Q$.
The charges on the outer shell ($r=c$) are uniformly distibuted with the total charges of $-4Q$.

#### \(c\) the capacitance of whole system
$$q=CV$$

$$q=-4Q$$

$$V=\frac{k_e(-4Q)}{c}$$

$$-4Q=C\frac{k_e(-4Q)}{c}$$

$$C=\frac{c}{k_e}$$

---

### Problem 5
The electric potential at the position of a point $P$ in three dimensions be given by the vector $r = (x, y, z)$ in Cartesian Coordinates is $(x^2+y^2+z^2)^{0.5}$. Please find the electric field at the point $(1, 2, 3)$.
Ans:
$$V(x,y,z)=(x^2+y^2+z^2)^{0.5}$$

$$E_x(x,y,z)=-\frac{\partial V(x,y,z)}{\partial x}=-0.5(x^2+y^2+z^2)^{-0.5}(2x)=-x(x^2+y^2+z^2)^{-0.5}\\
=(-1)(1^2+2^2+3^2)^{-0.5}=\frac{-1}{\sqrt{14}}$$

$$E_y(x,y,z)=-\frac{\partial V(x,y,z)}{\partial y}=-y(x^2+y^2+z^2)^{-0.5}=\frac{-2}{\sqrt{14}}$$

$$E_z(x,y,z)=-\frac{\partial V(x,y,z)}{\partial z}=-z(x^2+y^2+z^2)^{-0.5}=\frac{-3}{\sqrt{14}}$$

---

### Problem 6
Please see [Mini-Project-2](#Mini-Project-2)

---

### Problems 7-11
Please read the textbook

###### <div style="text-align: right"> Last updated: May 4, 2020, 15:07 GMT </div>