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期末|大学物理
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期末|大学物理

December 16, 2025
7 min read
笔记-大学物理

Page 1: 静电场基础

库仑定律
F=kq1q2r2er,k=14πε0F = k \frac{q_1 q_2}{r^2} e_r, \quad k = \frac{1}{4\pi\varepsilon_0}
F=14πε0q1q2r2erF = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2} e_r

场强
E=Fq=Q4πε0r2erE = \frac{F}{q} = \frac{Q}{4\pi\varepsilon_0 r^2} e_r
=dE=14πε0r2dqer(er为单位向量)= \int dE = \int \frac{1}{4\pi\varepsilon_0 r^2} dq \cdot e_r \quad (e_r \text{为单位向量})
电荷元:dq=λdl,σds,ρdvdq = \lambda dl, \sigma ds, \rho dv

无限长电棒周围电场
E=λ2πε0a向外E = \frac{\lambda}{2\pi\varepsilon_0 a} \quad \text{向外}

例:圆环轴线电场
总电量 QQ
dq=λdl=Q2πRdldq = \lambda dl = \frac{Q}{2\pi R} dl
dE=14πε0dqr2dE = \frac{1}{4\pi\varepsilon_0} \frac{dq}{r^2}
Ep=14πε0Qx(x2+R2)3/2E_p = \frac{1}{4\pi\varepsilon_0} \frac{Qx}{(x^2+R^2)^{3/2}}

例:圆盘轴线电场 (面密度 σ\sigma)
dq=σ2πrdrdq = \sigma \cdot 2\pi r dr
dE=xdq4πε0(x2+r2)3/2dE = \frac{x dq}{4\pi\varepsilon_0 (x^2+r^2)^{3/2}}
E=0Rxσ2πrdr4πε0(x2+r2)3/2=σ2ε0(1xx2+R2)E = \int_0^R \frac{x \cdot \sigma 2\pi r dr}{4\pi\varepsilon_0 (x^2+r^2)^{3/2}} = \frac{\sigma}{2\varepsilon_0} (1 - \frac{x}{\sqrt{x^2+R^2}})

高斯定理
Φe=SEdS=1ε0qi=1ε0Vρdv\Phi_e = \oiint_S \vec{E} \cdot d\vec{S} = \frac{1}{\varepsilon_0} \sum q_i = \frac{1}{\varepsilon_0} \int_V \rho dv

特殊几何体电场

  • 球体
    • 内 (r<Rr<R): E=qr4πε0R3E = \frac{qr}{4\pi\varepsilon_0 R^3}
    • 外 (r>Rr>R): E=q4πε0r2E = \frac{q}{4\pi\varepsilon_0 r^2}
  • 柱体 (线密度 λ\lambda):
    • 内: E=λr2πε0R2E = \frac{\lambda r}{2\pi\varepsilon_0 R^2}
    • 外: E=λ2πε0rE = \frac{\lambda}{2\pi\varepsilon_0 r}
  • 无限大平面: E=σ2ε0E = \frac{\sigma}{2\varepsilon_0}

电势
U=Edl=dq4πε0rU = \int \vec{E} \cdot d\vec{l} = \int \frac{dq}{4\pi\varepsilon_0 r}
做功:W=qΔUW = q \Delta U

  • 球体电势
    • 内: U=q4πε0RU = \frac{q}{4\pi\varepsilon_0 R} (?) 注:此处笔记可能简写,实心球体内部电势公式较复杂,此处可能指球壳内部等势
    • 外: U=q4πε0rU = \frac{q}{4\pi\varepsilon_0 r}
  • 圆环轴线: U=Q4πε0x2+R2U = \frac{Q}{4\pi\varepsilon_0 \sqrt{x^2+R^2}}
  • 圆盘轴线: U=σ2ε0(R2+x2x)U = \frac{\sigma}{2\varepsilon_0} (\sqrt{R^2+x^2} - x)

Page 2: 电介质与电场能量

E=grad UE = -\text{grad } U
P=limΔV0piΔV(电极化强度)\vec{P} = \lim_{\Delta V \to 0} \frac{\sum \vec{p_i}}{\Delta V} \quad (\text{电极化强度})

极化电荷

  • 面极化电荷: q=PdSq' = \oiint \vec{P} \cdot d\vec{S}
  • 介质表面极化电荷密度: σ=Pn\sigma' = \vec{P} \cdot \vec{n} (法线方向)

总场
E=E0+E(外场+退极化场)\vec{E} = \vec{E_0} + \vec{E'} \quad (\text{外场} + \text{退极化场})

介质中高斯定理
SDdS=q0(Sε0εrEdS=Q0i)\oiint_S \vec{D} \cdot d\vec{S} = \sum q_0 \quad (\oint_S \varepsilon_0 \varepsilon_r \vec{E} \cdot d\vec{S} = \sum Q_{0i})
D=ε0E+P\vec{D} = \varepsilon_0 \vec{E} + \vec{P} (电位移矢量)
对各向同性介质: D=εE\vec{D} = \varepsilon \vec{E}, ε=ε0εr\varepsilon = \varepsilon_0 \varepsilon_r (介质电容率)

电容
C=qUC = \frac{q}{U}

  • 串联: 1C=1C1+1C2C=C1C2C1+C2\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} \Rightarrow C = \frac{C_1 C_2}{C_1+C_2} (注意笔记中距离 d1,d2d_1, d_2 的示意)
  • 并联: C=C1+C2C = C_1 + C_2 (面积 S1,S2S_1, S_2)

能量
做功 A=0QUdq=0QqCdq=12Q2CA = \int_0^Q U dq = \int_0^Q \frac{q}{C} dq = \frac{1}{2} \frac{Q^2}{C}
任意电容器储能 W=12Q2C=12CU2=12UQW = \frac{1}{2} \frac{Q^2}{C} = \frac{1}{2} CU^2 = \frac{1}{2} UQ

电场能量密度
we=dWdV=12εE2w_e = \frac{dW}{dV} = \frac{1}{2} \varepsilon E^2
W=VwedV=V12εE2dVW_{\text{总}} = \int_V w_e dV = \int_V \frac{1}{2} \varepsilon E^2 dV

平板电容
E=σεE = \frac{\sigma}{\varepsilon}
UaUb=Ed=qεSdU_a - U_b = Ed = \frac{q}{\varepsilon S} d
C=qUab=εSdC = \frac{q}{U_{ab}} = \frac{\varepsilon S}{d}
即:计算 EU(Edl)C(qU)E \to U (\int \vec{E} \cdot d\vec{l}) \to C (\frac{q}{U})


Page 3: 稳恒磁场

毕奥-萨伐尔定律 (Biot-Savart Law)
dB=μ4πIdl×rr3dB=μ4πIdlsinαr2dB = \frac{\mu}{4\pi} \frac{I d\vec{l} \times \vec{r}}{r^3} \quad \text{或} \quad dB = \frac{\mu}{4\pi} \frac{I dl \sin\alpha}{r^2}
μ=μ0μr\mu = \mu_0 \mu_r

长直导线磁场
dB=μ04πIdxsinαr2dB = \frac{\mu_0}{4\pi} \frac{I dx \sin\alpha}{r^2}
换元: x=atanβ,dx=adβcos2β,r=acosβ,α=π2+βx = a \tan\beta, dx = \frac{a d\beta}{\cos^2\beta}, r = \frac{a}{\cos\beta}, \alpha = \frac{\pi}{2} + \beta
B=μ0I4πaβ1β2cosβdβ无限长μ0I2πaB = \frac{\mu_0 I}{4\pi a} \int_{\beta_1}^{\beta_2} \cos\beta d\beta \xrightarrow{\text{无限长}} \frac{\mu_0 I}{2\pi a}

圆环中心轴线磁场
dB=μ04πIdlsinαr2(α=90)dB = \frac{\mu_0}{4\pi} \frac{I dl \sin\alpha}{r^2} \quad (\alpha=90^\circ)
dB=μ0I4πr2dlsinθdB_{\text{轴}} = \frac{\mu_0 I}{4\pi r^2} \oint dl \cdot \sin\theta
B=μ0IR22r3=μ0IR22(R2+x2)3/2B = \frac{\mu_0 I R^2}{2 r^3} = \frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}

运动电荷
I=qnSvIdl=qnSvdt=(nSdl)qv=dNqvI = q n S v \Rightarrow I d\vec{l} = q n S v d\vec{t} = (n S dl) q \vec{v} = dN q \vec{v}
磁矩: Pm=ISn\vec{P_m} = I S \vec{n}
大小: Pm=ISP_m = IS
单电荷B: B=dBdN=μ4πdNqv×rr3dN=μ4πqv×rr3\vec{B} = \frac{d\vec{B}}{dN} = \frac{\mu}{4\pi} \frac{dN q \vec{v} \times \vec{r}}{r^3 dN} = \frac{\mu}{4\pi} \frac{q \vec{v} \times \vec{r}}{r^3}
单电荷I: I=ΔqΔt=fq=qTI = \frac{\Delta q}{\Delta t} = f q = \frac{q}{T}

磁通量
Φm=SBdS=SBcosθds\Phi_m = \iint_S \vec{B} \cdot d\vec{S} = \iint_S B \cos\theta ds

安培环路定理
LBdl=μ0Ii(包围内)\oint_L \vec{B} \cdot d\vec{l} = \mu_0 \sum I_i \quad (\text{包围内})

同轴电缆 (Cylindrical Shell)

  • 外 (r>Rr>R): B=μ0I2πrB = \frac{\mu_0 I}{2\pi r}
  • 内 (r<Rr<R): B=μ0Ir2πR2B = \frac{\mu_0 I r}{2\pi R^2}
    推导: Bdl=μ0Ii\oint \vec{B} \cdot d\vec{l} = \mu_0 \sum I_i
    B2πr=μ0Iπr2πR2B=μ0Ir2πR2B \cdot 2\pi r = \mu_0 I \frac{\pi r^2}{\pi R^2} \Rightarrow B = \frac{\mu_0 I r}{2\pi R^2}

Page 4: 磁场力与电磁感应

螺线管

  • 内: B=μ0nIB = \mu_0 n I
  • 外: B0B \approx 0
    螺绕环: B=μ0NI2πrB = \frac{\mu_0 N I}{2\pi r}

磁对电 (洛伦兹力)
F=qv×B\vec{F} = q \vec{v} \times \vec{B}
匀速圆周运动: qvB=mv2RR=mvqB,T=2πRv=2πmqBqvB = m \frac{v^2}{R} \Rightarrow R = \frac{mv}{qB}, T = \frac{2\pi R}{v} = \frac{2\pi m}{qB}
一般情况: F=qE+qv×B\vec{F} = q\vec{E} + q\vec{v} \times \vec{B}

霍尔效应: UH=KHIBd,KH=1nqU_H = K_H \frac{IB}{d}, K_H = \frac{1}{nq} (霍尔系数,q正负则正负)

安培力:
dF=Idl×Bd\vec{F} = I d\vec{l} \times \vec{B}
F=IL×B\vec{F} = I \vec{L} \times \vec{B} (任意形状导线首尾相连直线电流受力,均匀磁场)

磁做功: A=IΔΦA = I \Delta \Phi (IdΦI \int d\Phi)

感应电动势:
E=dΦdt\mathcal{E} = - \frac{d\Phi}{dt}
Ekdl=SBtdS\int \vec{E_k} d\vec{l} = - \int_S \frac{\partial \vec{B}}{\partial t} d\vec{S}
动生电动势: E=Fmq=(v×B)dl\mathcal{E} = \frac{F_m}{q} = \int (\vec{v} \times \vec{B}) \cdot d\vec{l}

自感与互感:

  • 自感: L=ΦIL = \frac{\Phi}{I}, EL=LdIdt\mathcal{E}_L = -L \frac{dI}{dt} (磁链 Ψ=NΦ\Psi = N\Phi)
  • 互感: M=Ψ21I1=Ψ12I2M = \frac{\Psi_{21}}{I_1} = \frac{\Psi_{12}}{I_2}
    E21=MdI1dt\mathcal{E}_{21} = -M \frac{dI_1}{dt}, E12=MdI2dt\mathcal{E}_{12} = -M \frac{dI_2}{dt}

磁能:
WL=12LI2W_L = \frac{1}{2} L I^2
W12=E12I1dt=MI1I2=W21W_{12} = \int \mathcal{E}_{12} I_1 dt = M I_1 I_2 = W_{21}
磁能密度: wm=12B2μ=12BH=12μH2w_m = \frac{1}{2} \frac{B^2}{\mu} = \frac{1}{2} BH = \frac{1}{2} \mu H^2 (B=μHB=\mu H)
磁场总能: Wm=wmdVW_m = \int w_m dV
(对比电场能量密度 we=εE22w_e = \frac{\varepsilon E^2}{2})


Page 5: 麦克斯韦方程组

介质中的安培环路定理
LBdl=μ0(I+I)\oint_L \vec{B} \cdot d\vec{l} = \mu_0 (\sum I + I')
L(Bμ0M)dl=I\Rightarrow \oint_L (\frac{\vec{B}}{\mu_0} - \vec{M}) d\vec{l} = \sum I
LHdl=I\Rightarrow \oint_L \vec{H} d\vec{l} = \sum I (H\vec{H}: 磁场强度矢量)
各向同性介质: M=χmH\vec{M} = \chi_m \vec{H} (χm\chi_m: 磁化率)
B=μ0(1+χm)HμH\vec{B} = \mu_0 (1 + \chi_m) \vec{H} \Rightarrow \mu \vec{H} (μ=μ0μr\mu = \mu_0 \mu_r)

位移电流: Id=dΦDdtI_d = \frac{d\Phi_D}{dt}
电流密度: jd=Dt\vec{j_d} = \frac{\partial \vec{D}}{\partial t}
全电流定律: LHdl=I=I+dΦDdt=S(j+jd)dS\oint_L \vec{H} \cdot d\vec{l} = I_{\text{全}} = I + \frac{d\Phi_D}{dt} = \iint_S (\vec{j} + \vec{j_d}) \cdot d\vec{S}
(+SdDdtε0dEdt+ S \frac{dD}{dt} \sim \varepsilon_0 \frac{dE}{dt})

电磁场总结 (麦克斯韦方程组)

  1. LEdl=SBtdS\oint_L \vec{E} \cdot d\vec{l} = -\iint_S \frac{\partial \vec{B}}{\partial t} \cdot d\vec{S} (法拉第电磁感应,变磁生电)
  2. LHdl=Ii+SDtdS\oint_L \vec{H} \cdot d\vec{l} = \sum I_i + \iint_S \frac{\partial \vec{D}}{\partial t} \cdot d\vec{S} (安培-麦克斯韦,变电生磁)
  3. SDdS=qi\oiint_S \vec{D} \cdot d\vec{S} = \sum q_i (高斯定理,有源)
  4. SBdS=0\oiint_S \vec{B} \cdot d\vec{S} = 0 (磁通连续,无源)

辅助方程:
D=εE\vec{D} = \varepsilon \vec{E}
B=μH\vec{B} = \mu \vec{H}
J=σE\vec{J} = \sigma \vec{E} (欧姆定律微分形式)

微分形式
D=ρ\nabla \cdot \vec{D} = \rho
B=0\nabla \cdot \vec{B} = 0
×E=Bt\nabla \times \vec{E} = - \frac{\partial \vec{B}}{\partial t}
×H=J+Dt\nabla \times \vec{H} = \vec{J} + \frac{\partial \vec{D}}{\partial t}


Page 6: 电磁波

E=E0cosω(txv)E = E_0 \cos \omega (t - \frac{x}{v})
H=H0cosω(txv)H = H_0 \cos \omega (t - \frac{x}{v})
振幅关系: εE0=μH0\sqrt{\varepsilon} E_0 = \sqrt{\mu} H_0
波速: v=1εμv = \frac{1}{\sqrt{\varepsilon \mu}}, 真空中 c=1ε0μ0c = \frac{1}{\sqrt{\varepsilon_0 \mu_0}}
v=cεrμrv = \frac{c}{\sqrt{\varepsilon_r \mu_r}}

能量密度: w=we+wm=(DE+BH)12=εE2w = w_e + w_m = (\vec{D}\cdot\vec{E} + \vec{B}\cdot\vec{H}) \frac{1}{2} = \varepsilon E^2 (或 B2μ\frac{B^2}{\mu})
坡印廷矢量 (能流密度):
S=E×H\vec{S} = \vec{E} \times \vec{H}
方向:波传播方向。


Page 7: 相对论与量子物理基础

相对论 (Relativity)
SS' 系相对 SS 系沿 xx 轴以 vv 运动:
洛伦兹变换:
x=xvt1v2/c2,y=y,z=zx' = \frac{x-vt}{\sqrt{1-v^2/c^2}}, \quad y'=y, \quad z'=z
t=tvc2x1v2/c2t' = \frac{t - \frac{v}{c^2}x}{\sqrt{1-v^2/c^2}}
速度变换:
ux=uxv1uxv/c2u_x' = \frac{u_x - v}{1 - u_x v / c^2}
uy=uy1v2/c21uxv/c2u_y' = \frac{u_y \sqrt{1-v^2/c^2}}{1 - u_x v / c^2}

效应

  • 同时的相对性
  • 时间膨胀: Δt=Δt01β2=γΔt0\Delta t = \frac{\Delta t_0}{\sqrt{1-\beta^2}} = \gamma \Delta t_0 (ΔtΔt0\Delta t \ge \Delta t_0)
  • 长度收缩: L=L01β2L' = L_0 \sqrt{1-\beta^2} (LL0L \le L_0)
  • 动质量: m=m01v2/c2m = \frac{m_0}{\sqrt{1-v^2/c^2}}
  • 能量: E=Ek+E0=(mc2m0c2)+m0c2=mc2E = E_k + E_0 = (m c^2 - m_0 c^2) + m_0 c^2 = m c^2
  • 动量能量关系: E2=E02+(pc)2=(m0c2)2+p2c2E^2 = E_0^2 + (pc)^2 = (m_0 c^2)^2 + p^2 c^2
  • 力: F=d(mv)dt=mdvdt+vdmdt\vec{F} = \frac{d(m\vec{v})}{dt} = m \frac{d\vec{v}}{dt} + \vec{v} \frac{dm}{dt}

量子物理基础
λmT=b\lambda_m T = b (维恩位移定律, b=2.898×103mKb=2.898 \times 10^{-3} mK)
En=nε,n=1,2...E_n = n \varepsilon, n=1,2... (普朗克量子假说)
ε=hν\varepsilon = h \nu (能量子,普朗克常数 h=6.63×1034Jsh=6.63 \times 10^{-34} J \cdot s)

光电效应:
ε=hν=12mvm2+W\varepsilon = h \nu = \frac{1}{2} m v_m^2 + W (逸出功)
红限 ν0=Wh\nu_0 = \frac{W}{h}
遏止电压 eVa=12mvm2Va=heνWee V_a = \frac{1}{2} m v_m^2 \Rightarrow V_a = \frac{h}{e} \nu - \frac{W}{e}

光子说:
E=hν=mc2E = h \nu = m c^2
质量 m=hνc2=hλcm = \frac{h \nu}{c^2} = \frac{h}{\lambda c}, 静质量 m0=0m_0 = 0
动量 p=mc=hλp = mc = \frac{h}{\lambda}

康普顿散射:
λλ0=2hm0csin2φ2=2Λsin2φ2\lambda - \lambda_0 = \frac{2h}{m_0 c} \sin^2 \frac{\varphi}{2} = 2 \Lambda \sin^2 \frac{\varphi}{2}
Λ=2.41×1012m\Lambda = 2.41 \times 10^{-12} m

常数: e=1.602×1019Ce=1.602 \times 10^{-19} C, me=9.1×1031kgm_e = 9.1 \times 10^{-31} kg, 1eV=1.602×1019J1 eV = 1.602 \times 10^{-19} J


Page 8: 量子力学初步

玻尔模型:
能级 En=E1n2=13.6n2eVE_n = \frac{E_1}{n^2} = \frac{-13.6}{n^2} eV
半径 rn=n2r1r_n = n^2 r_1
角动量量子化: L=mvr=nL = mvr = n \hbar
德布罗意波: λ=hp=hmv\lambda = \frac{h}{p} = \frac{h}{mv}
不确定度关系: ΔxΔp\Delta x \Delta p \ge \hbar (或 hh, h2\frac{h}{2}, 2\frac{\hbar}{2})

量子数 (原子)

  1. 主量子数 nn
  2. 角量子数 ll: l=0,1,...n1l = 0, 1, ... n-1. L=l(l+1)L = \sqrt{l(l+1)} \hbar
  3. 磁量子数 mlm_l: ml=0,±1,...±lm_l = 0, \pm 1, ... \pm l. Lz=mlL_z = m_l \hbar
  4. 自旋量子数 msm_s: ms=±12m_s = \pm \frac{1}{2}. Sz=msS_z = m_s \hbar
    电子可能状态数 2n22n^2

一维无限深势阱:
能量 En=n2h28ma2E_n = \frac{n^2 h^2}{8 m a^2}
波函数 ψn(x)=2asin(nπxa)(0xa)\psi_n(x) = \sqrt{\frac{2}{a}} \sin(\frac{n\pi x}{a}) \quad (0 \le x \le a)

常数: c2.9979×108m/sc \approx 2.9979 \times 10^8 m/s

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