工程電磁場部分課后習題答案?第一章矢量分析習題11已知矢量\(\vec{A}=3\vec{e}_x+4\vec{e}_y5\vec{e}_z\)，\(\vec{B}=2\vec{e}_x+\vec{e}_y+3\vec{e}_z\)，求\(\vec{A}+\vec{B}\)，\(\vec{A}\vec{B}\)，\(\vec{A}\cdot\vec{B}\)，\(\vec{A}\times\vec{B}\)以及\(\vec{A}\)與\(\vec{B}\)之間的夾角。

解：1.\(\vec{A}+\vec{B}\)\(\vec{A}+\vec{B}=(32)\vec{e}_x+(4+1)\vec{e}_y+(5+3)\vec{e}_z=\vec{e}_x+5\vec{e}_y2\vec{e}_z\)2.\(\vec{A}\vec{B}\)\(\vec{A}\vec{B}=(3+2)\vec{e}_x+(41)\vec{e}_y+(53)\vec{e}_z=5\vec{e}_x+3\vec{e}_y8\vec{e}_z\)3.\(\vec{A}\cdot\vec{B}\)\(\vec{A}\cdot\vec{B}=3\times(2)+4\times1+(5)\times3=6+415=17\)4.\(\vec{A}\times\vec{B}\)\(\vec{A}\times\vec{B}=\begin{vmatrix}\vec{e}_x&\vec{e}_y&\vec{e}_z\\3&4&5\\2&1&3\end{vmatrix}\)\(=\vec{e}_x(4\times3(5)\times1)\vec{e}_y(3\times3(5)\times(2))+\vec{e}_z(3\times14\times(2))\)\(=\vec{e}_x(12+5)\vec{e}_y(910)+\vec{e}_z(3+8)\)\(=17\vec{e}_x+\vec{e}_y+11\vec{e}_z\)5.\(\vec{A}\)與\(\vec{B}\)之間的夾角\(\theta\)根據\(\vec{A}\cdot\vec{B}=|\vec{A}||\vec{B}|\cos\theta\)\(|\vec{A}|=\sqrt{3^2+4^2+(5)^2}=\sqrt{9+16+25}=\sqrt{50}\)\(|\vec{B}|=\sqrt{(2)^2+1^2+3^2}=\sqrt{4+1+9}=\sqrt{14}\)則\(\cos\theta=\frac{\vec{A}\cdot\vec{B}}{|\vec{A}||\vec{B}|}=\frac{17}{\sqrt{50}\times\sqrt{14}}\)\(\theta=\arccos(\frac{17}{\sqrt{50}\times\sqrt{14}})\)

習題12證明：\(\nabla\cdot(\vec{A}\times\vec{B})=\vec{B}\cdot(\nabla\times\vec{A})\vec{A}\cdot(\nabla\times\vec{B})\)

證明：設\(\vec{A}=A_x\vec{e}_x+A_y\vec{e}_y+A_z\vec{e}_z\)，\(\vec{B}=B_x\vec{e}_x+B_y\vec{e}_y+B_z\vec{e}_z\)1.先計算\(\vec{A}\times\vec{B}\)\(\vec{A}\times\vec{B}=\begin{vmatrix}\vec{e}_x&\vec{e}_y&\vec{e}_z\\A_x&A_y&A_z\\B_x&B_y&B_z\end{vmatrix}\)\(=(A_yB_zA_zB_y)\vec{e}_x+(A_zB_xA_xB_z)\vec{e}_y+(A_xB_yA_yB_x)\vec{e}_z\)2.再計算\(\nabla\cdot(\vec{A}\times\vec{B})\)\(\nabla\cdot(\vec{A}\times\vec{B})=\frac{\partial(A_yB_zA_zB_y)}{\partialx}+\frac{\partial(A_zB_xA_xB_z)}{\partialy}+\frac{\partial(A_xB_yA_yB_x)}{\partialz}\)\(=\left(\frac{\partialA_yB_z}{\partialx}\frac{\partialA_zB_y}{\partialx}\right)+\left(\frac{\partialA_zB_x}{\partialy}\frac{\partialA_xB_z}{\partialy}\right)+\left(\frac{\partialA_xB_y}{\partialz}\frac{\partialA_yB_x}{\partialz}\right)\)\(=\left(B_z\frac{\partialA_y}{\partialx}+A_y\frac{\partialB_z}{\partialx}B_y\frac{\partialA_z}{\partialx}A_z\frac{\partialB_y}{\partialx}\right)+\left(B_x\frac{\partialA_z}{\partialy}+A_z\frac{\partialB_x}{\partialy}B_z\frac{\partialA_x}{\partialy}A_x\frac{\partialB_z}{\partialy}\right)+\left(B_y\frac{\partialA_x}{\partialz}+A_x\frac{\partialB_y}{\partialz}B_x\frac{\partialA_y}{\partialz}A_y\frac{\partialB_x}{\partialz}\right)\)3.計算\(\vec{B}\cdot(\nabla\times\vec{A})\vec{A}\cdot(\nabla\times\vec{B})\)\(\nabla\times\vec{A}=\begin{vmatrix}\vec{e}_x&\vec{e}_y&\vec{e}_z\\\frac{\partial}{\partialx}&\frac{\partial}{\partialy}&\frac{\partial}{\partialz}\\A_x&A_y&A_z\end{vmatrix}\)\(=\left(\frac{\partialA_z}{\partialy}\frac{\partialA_y}{\partialz}\right)\vec{e}_x+\left(\frac{\partialA_x}{\partialz}\frac{\partialA_z}{\partialx}\right)\vec{e}_y+\left(\frac{\partialA_y}{\partialx}\frac{\partialA_x}{\partialy}\right)\vec{e}_z\)\(\nabla\times\vec{B}=\begin{vmatrix}\vec{e}_x&\vec{e}_y&\vec{e}_z\\\frac{\partial}{\partialx}&\frac{\partial}{\partialy}&\frac{\partial}{\partialz}\\B_x&B_y&B_z\end{vmatrix}\)\(=\left(\frac{\partialB_z}{\partialy}\frac{\partialB_y}{\partialz}\right)\vec{e}_x+\left(\frac{\partialB_x}{\partialz}\frac{\partialB_z}{\partialx}\right)\vec{e}_y+\left(\frac{\partialB_y}{\partialx}\frac{\partialB_x}{\partialy}\right)\vec{e}_z\)\(\vec{B}\cdot(\nabla\times\vec{A})=B_x\left(\frac{\partialA_z}{\partialy}\frac{\partialA_y}{\partialz}\right)+B_y\left(\frac{\partialA_x}{\partialz}\frac{\partialA_z}{\partialx}\right)+B_z\left(\frac{\partialA_y}{\partialx}\frac{\partialA_x}{\partialy}\right)\)\(\vec{A}\cdot(\nabla\times\vec{B})=A_x\left(\frac{\partialB_z}{\partialy}\frac{\partialB_y}{\partialz}\right)+A_y\left(\frac{\partialB_x}{\partialz}\frac{\partialB_z}{\partialx}\right)+A_z\left(\frac{\partialB_y}{\partialx}\frac{\partialB_x}{\partialy}\right)\)經整理可得\(\vec{B}\cdot(\nabla\times\vec{A})\vec{A}\cdot(\nabla\times\vec{B})=\nabla\cdot(\vec{A}\times\vec{B})\)，證畢。

習題13已知標量場\(u=x^2y+y^2z+z^2x\)，求\(\nablau\)在點\((1,1,1)\)處的值。

解：\(\nablau=\frac{\partialu}{\partialx}\vec{e}_x+\frac{\partialu}{\partialy}\vec{e}_y+\frac{\partialu}{\partialz}\vec{e}_z\)1.求\(\frac{\partialu}{\partialx}\)\(\frac{\partialu}{\partialx}=2xy+z^2\)2.求\(\frac{\partialu}{\partialy}\)\(\frac{\partialu}{\partialy}=x^2+2yz\)3.求\(\frac{\partialu}{\partialz}\)\(\frac{\partialu}{\partialz}=y^2+2zx\)4.在點\((1,1,1)\)處\(\frac{\partialu}{\partialx}\big|_{(1,1,1)}=2\times1\times1+1^2=3\)\(\frac{\partialu}{\partialy}\big|_{(1,1,1)}=1^2+2\times1\times1=3\)\(\frac{\partialu}{\partialz}\big|_{(1,1,1)}=1^2+2\times1\times1=3\)所以\(\nablau\big|_{(1,1,1)}=3\vec{e}_x+3\vec{e}_y+3\vec{e}_z\)

第二章靜電場習題21真空中有兩個點電荷，\(q_1=2\times10^{6}C\)，\(q_2=3\times10^{6}C\)，它們相距\(r=0.5m\)，求兩電荷之間的相互作用力。

解：根據庫侖定律\(\vec{F}=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\vec{e}_{r12}\)\(\frac{1}{4\pi\epsilon_0}=9\times10^9N\cdotm^2/C^2\)\(\vec{F}=9\times10^9\times\frac{2\times10^{6}\times(3\times10^{6})}{(0.5)^2}\vec{e}_{r12}\)\(=0.216\vec{e}_{r12}N\)兩電荷之間的相互作用力大小為\(0.216N\)，方向沿兩電荷連線方向指向\(q_2\)。

習題22在均勻電場\(\vec{E}=3\vec{e}_x+4\vec{e}_y\)中，有一個邊長為\(a\)的正方形平面，其法向單位矢量\(\vec{n}=\vec{e}_z\)，求通過該平面的電通量\(\varPhi_E\)。

解：根據電通量公式\(\varPhi_E=\vec{E}\cdot\vec{S}=\vec{E}\cdot\vec{n}S\)\(S=a^2\)\(\varPhi_E=(3\vec{e}_x+4\vec{e}_y)\cdot\vec{e}_z\timesa^2=0\)

習題23真空中有一半徑為\(R\)的均勻帶電球面，總電量為\(Q\)，求球內外的電場強度分布。

解：1.球外（\(r\gtR\)）應用高斯定理\(\oint_{S}\vec{E}\cdotd\vec{S}=\frac{Q}{\epsilon_0}\)作半徑為\(r\)（\(r\gtR\)）的高斯球面\(S\)，\(\vec{E}\)與\(d\vec{S}\)同向\(\oint_{S}\vec{E}\cdotd\vec{S}=E\cdot4\pir^2\)則\(E\cdot4\pir^2=\frac{Q}{\epsilon_0}\)解得\(E=\frac{Q}{4\pi\epsilon_0r^2}\)2.球內（\(r\ltR\)）作半徑為\(r\)（\(r\ltR\)）的高斯球面\(S\)，高斯面內包含的電量\(q=0\)由高斯定理\(\oint_{S}\vec{E}\cdotd\vec{S}=\frac{q}{\epsilon_0}\)可得\(E\cdot4\pir^2=0\)所以\(E=0\)

習題24真空中有一均勻帶電球體，半徑為\(R\)，體電荷密度為\(\rho\)，求球內外的電場強度分布。

解：1.球外（\(r\gtR\)）應用高斯定理\(\oint_{S}\vec{E}\cdotd\vec{S}=\frac{Q}{\epsilon_0}\)球體總電量\(Q=\frac{4}{3}\piR^3\rho\)作半徑為\(r\)（\(r\gtR\)）的高斯球面\(S\)，\(\vec{E}\)與\(d\vec{S}\)同向\(\oint_{S}\vec{E}\cdotd\vec{S}=E\cdot4\pir^2\)則\(E\cdot4\pir^2=\frac{\frac{4}{3}\piR^3\rho}{\epsilon_0}\)解得\(E=\frac{\rhoR^3}{3\epsilon_0r^2}\)2.球內（\(r\ltR\)）作半徑為\(r\)（\(r\ltR\)）的高斯球面\(S\)高斯面內包含的電量\(q=\frac{4}{3}\pir^3\rho\)由高斯定理\(\oint_{S}\vec{E}\cdotd\vec{S}=\frac{q}{\epsilon_0}\)即\(E\cdot4\pir^2=\frac{\frac{4}{3}\pir^3\rho}{\epsilon_0}\)解得\(E=\frac{\rhor}{3\epsilon_0}\)

第三章靜電場中的導體和電介質習題31一導體球半徑為\(R_1\)，帶電量為\(Q\)，球外有一同心的導體球殼，內半徑為\(R_2\)，外半徑為\(R_3\)，球殼原來不帶電，求：1.各區域的電場強度分布；2