与正三角形均匀带电薄片电场强度有关的曲面积分

青青子衿

2.28  正三角形带电薄片（带正电荷）位于\(\,\Sigma\colon\,x+y+z=-a\)（其中\(\,a>0\)）的平面上，且带电薄片限定于\(-a\le x\le 0\)与\(-a\le y\le 0\)之间，
        其电荷面密度为\(\sigma\)，试求出原点处的电场强度\(\boldsymbol{E}\)（矢量）
\[ \left|\overrightarrow{\boldsymbol{E}}\,\right|\,=\sqrt{3}\displaystyle\iint\limits_{\Sigma\colon\,x+y+z=-a}\dfrac{\sigma\,\,{\rm\,d}S}{4\pi\varepsilon_0\left(x^2+y^2+z^2\right)}\dfrac{\left|x\right|}{\sqrt{x^2+y^2+z^2}} \]

\[
\big|\overrightarrow{\boldsymbol{E}}\big|\,\cos\langle\overrightarrow{\boldsymbol{E}},\overrightarrow{\boldsymbol{i}}\rangle=\displaystyle\iint\limits_{\Sigma\colon\,x+y+z=-a}\dfrac{\sigma\,\left|x\right|\,{\rm\,d}S}{4\pi\varepsilon_0\left(x^2+y^2+z^2\right)^{\frac{3}{2}}}
\]
\[
\big|\overrightarrow{\boldsymbol{E}}\big|\,\cos\langle\overrightarrow{\boldsymbol{E}},\overrightarrow{\boldsymbol{j}}\rangle=\displaystyle\iint\limits_{\Sigma\colon\,x+y+z=-a}\dfrac{\sigma\,\left|y\right|\,{\rm\,d}S}{4\pi\varepsilon_0\left(x^2+y^2+z^2\right)^{\frac{3}{2}}}
\]
\[
\big|\overrightarrow{\boldsymbol{E}}\big|\,\cos\langle\overrightarrow{\boldsymbol{E}},\overrightarrow{\boldsymbol{k}}\rangle=\displaystyle\iint\limits_{\Sigma\colon\,x+y+z=-a}\dfrac{\sigma\,\left|z\right|\,{\rm\,d}S}{4\pi\varepsilon_0\left(x^2+y^2+z^2\right)^{\frac{3}{2}}}
\]

\begin{align*}  
\big|\overrightarrow{\boldsymbol{E}}\big|\,&=\sqrt{3\,}\left(\big|\overrightarrow{\boldsymbol{E}}\big|\,\cos\langle\overrightarrow{\boldsymbol{E}},\overrightarrow{\boldsymbol{i}}\rangle\right)\\
&=\sqrt{3}\displaystyle\iint\limits_{\Sigma\colon\,x+y+z=-a}\dfrac{\sigma\,\left|x\right|\,{\rm\,d}S}{4\pi\varepsilon_0\left(x^2+y^2+z^2\right)^{\frac{3}{2}}}\\  
&=\frac{\sqrt{3}\,\sigma}{4\pi\varepsilon_0}
\displaystyle\iint\limits_{\Sigma\colon\,x+y+z=-a}\dfrac{\,\left|x\right|\,}{\left(x^2+y^2+z^2\right)^{\frac{3}{2}}}{\rm\,d}S\\
&=\frac{\sqrt{3}\,\sigma}{4\pi\varepsilon_0}
\int_{-a}^0\int_{-a-y}^0 \dfrac{\sqrt{3}\,\left|x\right|}{\left(x^2+y^2+\left(-a-x-y\right)^2\right)^{\frac{3}{2}}}{\rm\,d}x{\rm\,d}y\\
&=\frac{\sqrt{3}\,\sigma}{4\pi\varepsilon_0}
\int_{-a}^0\int_{-a-y}^0 \dfrac{\sqrt{3}\,\left|x\right|}{\left(x^2+y^2+\left(a+x+y\right)^2\right)^{\frac{3}{2}}}{\rm\,d}x{\rm\,d}y\\
&=\frac{\sqrt{3}\,\sigma}{4\pi\varepsilon_0}
\int_0^a\int_0^{a-y} \dfrac{\sqrt{3}\,x}{\left(x^2+y^2+\left(a-x-y\right)^2\right)^{\frac{3}{2}}}{\rm\,d}x{\rm\,d}y\\
&=\frac{3\,\sigma}{4\pi\varepsilon_0}
\int_0^a\int_0^{a-y} \dfrac{\,x}{\left(x^2+y^2+\left(a-x-y\right)^2\right)^{\frac{3}{2}}}{\rm\,d}x{\rm\,d}y\\
\end{align*}

青青子衿

回复 1# 青青子衿
\begin{align*}
\overrightarrow{\boldsymbol{E}}\,  
&=\left(\big|\overrightarrow{\boldsymbol{E}}\big|\,\cos\langle\overrightarrow{\boldsymbol{E}},\overrightarrow{\boldsymbol{i}}\rangle\right)\,\vv{\boldsymbol{i}}
+\left(\big|\overrightarrow{\boldsymbol{E}}\big|\,\cos\langle\overrightarrow{\boldsymbol{E}},\overrightarrow{\boldsymbol{j}}\rangle\right)\,\vv{\boldsymbol{j}}
+\left(\big|\overrightarrow{\boldsymbol{E}}\big|\,\cos\langle\overrightarrow{\boldsymbol{E}},\overrightarrow{\boldsymbol{k}}\rangle\right)\,\vv{\boldsymbol{k}}
\\
  
&=\iint\limits_{\Sigma\colon\,x+y+z=-a}\dfrac{\sigma\,\left|x\right|\,{\rm\,d}S}{4\pi\varepsilon_0\left(x^2+y^2+z^2\right)^{\frac{3}{2}}}\,\left(\vv{\boldsymbol{i}}+\vv{\boldsymbol{j}}+\vv{\boldsymbol{k}} \right)
\,\\   
&=\frac{\sqrt{3\,}\,\sigma}{4\pi\varepsilon_0}   
\int_0^a\int_0^{a-y} \dfrac{\,x}{\left(x^2+y^2+\left(a-x-y\right)^2\right)^{\frac{3}{2}}}{\rm\,d}x{\rm\,d}y\,\left(\vv{\boldsymbol{i}}+\vv{\boldsymbol{j}}+\vv{\boldsymbol{k}} \right)\\  
&=\frac{\sqrt{3\,}\,\sigma}{4\pi\varepsilon_0}\cdot
\frac{\pi}{6}\,\left(\vv{\boldsymbol{i}}+\vv{\boldsymbol{j}}+\vv{\boldsymbol{k}} \right)\\
&=\frac{\sqrt{3\,}\,\sigma}{\,\,24\,\varepsilon_0}\,\left(\vv{\boldsymbol{i}}+\vv{\boldsymbol{j}}+\vv{\boldsymbol{k}} \right)\\
\,\\
\big|\overrightarrow{\boldsymbol{E}}\big|\,&=\sqrt{3\,}\left(\big|\overrightarrow{\boldsymbol{E}}\big|\,\cos\langle\overrightarrow{\boldsymbol{E}},\overrightarrow{\boldsymbol{i}}\rangle\right)\\  
&=\sqrt{3}\displaystyle\iint\limits_{\Sigma\colon\,x+y+z=-a}\dfrac{\sigma\,\left|x\right|\,{\rm\,d}S}{4\pi\varepsilon_0\left(x^2+y^2+z^2\right)^{\frac{3}{2}}}\\  
&=\frac{3\,\sigma}{4\pi\varepsilon_0}   
\int_0^a\int_0^{a-y} \dfrac{\,x}{\left(x^2+y^2+\left(a-x-y\right)^2\right)^{\frac{3}{2}}}{\rm\,d}x{\rm\,d}y\\
&=\frac{3\,\sigma}{4\pi\varepsilon_0}\cdot\frac{\pi}{6}=\frac{\,\sigma}{8\,\varepsilon_0}
\end{align*}

\begin{align*}
\color{red}{\boxed{\quad\color{black}{\int_0^a\int_0^{a-y} \dfrac{\,x}{\left(x^2+y^2+\left(a-x-y\right)^2\right)^{\frac{3}{2}}}{\rm\,d}x{\rm\,d}y\,=\,\frac{\pi}{6}}\quad}}
\end{align*}

青青子衿

\begin{align*}
\color{red}{\boxed{\quad\color{black}{\int_0^a\int_0^{a-y} \dfrac{\,x}{\left(x^2+y^2+\left(a-x-y\right)^2\right)^{\frac{3}{2}}}{\rm\,d}x{\rm\,d}y\,=\,\frac{\pi}{6}}\quad}}
\end{align*}
青青子衿 发表于 2018-12-6 20:08

回复 2# 青青子衿

\begin{align*}
&&&\int_0^a\int_0^{a-y} \dfrac{\,x}{\left(x^2+y^2+\left(a-x-y\right)^2\right)^{\frac{3}{2}}}{\rm\,d}x{\rm\,d}y\\
&&\overset{\begin{cases}  
x\,=\,a\,\cdot\,u\\  
y\,=\,a\,\cdot\,v\\   
\end{cases}}{\overline{\overline{\hspace{3cm}}}}&\int_0^a\int_0^{a-av} \dfrac{\,a\cdot u}{\left(a^2u^2+a^2v^2+\left(a-au-av\right)^2\right)^{\frac{3}{2}}}{\rm\,d}\left(au\right){\rm\,d}\left(av\right)\\
&&=&\int_0^1\int_0^{1-v} \dfrac{u}{\left(u^2+v^2+\left(1-u-v\right)^2\right)^{\frac{3}{2}}}{\rm\,d}u{\rm\,d}v\\
&&=&\int_0^1\int_0^{1-y} \dfrac{\,x}{\left(x^2+y^2+\left(1-x-y\right)^2\right)^{\frac{3}{2}}}{\rm\,d}x{\rm\,d}y\\
\end{align*}
\begin{align*}
&&&\int_0^{1-y} \dfrac{x}{\left(x^2+y^2+\left(1-x-y\right)^2\right)^{\frac{3}{2}}}{\rm\,d}x\\  
&&\overset{
b=\frac{1-y}{2}
}{\overline{\overline{\hspace{2cm}}}}&\int_0^{2b} \dfrac{x}{\left(x^2+y^2+\left(-x+1-y\right)^2\right)^{\frac{3}{2}}}{\rm\,d}x\\
&&=&\int_0^{2b} \dfrac{x}{\left(x^2+y^2+x^2-2\left(1-y\right)x+\left(1-y\right)^2\right)^{\frac{3}{2}}}{\rm\,d}x\\
&&=&\int_0^{2b} \dfrac{x}{\left(2x^2-2\left(1-y\right)x+2\left(\frac{1-y}{2}\right)^2+y^2+\frac{\left(1-y\right)^2}{2}\right)^{\frac{3}{2}}}{\rm\,d}x\\
&&=&\int_0^{2b} \dfrac{x}{\left(2\left(x-\frac{1-y}{2}\right)^2+y^2+\frac{\left(1-y\right)^2}{2}\right)^{\frac{3}{2}}}{\rm\,d}x\\
&&=&\int_0^{2b} \dfrac{x}{\left(2\left(x-b\right)^2+y^2+2b^2\right)^{\frac{3}{2}}}{\rm\,d}x\\
&&=&\int_{-b}^b \dfrac{t+b}{\left(2t^2+y^2+2b^2\right)^{\frac{3}{2}}}{\rm\,d}t=\int_0^b \dfrac{2b}{\left(2t^2+y^2+2b^2\right)^{\frac{3}{2}}}{\rm\,d}t\\
&&=&\left.\dfrac{2bt}{\left(y^2+2b^2\right)\sqrt{2t^2+y^2+2b^2}}\right|_0^b=\dfrac{2b^2}{\left(y^2+2b^2\right)\sqrt{2b^2+y^2+2b^2}}\\
&&=&\dfrac{2b^2}{\left(y^2+2b^2\right)\sqrt{y^2+4b^2}}=\dfrac{2\left(\frac{1-y}{2}\right)^2}{\left(y^2+2\left(\frac{1-y}{2}\right)^2\right)\sqrt{y^2+4\left(\frac{1-y}{2}\right)^2}}\\
&&=&\dfrac{\left(1-y\right)^2}{\left(2y^2+\left(1-y\right)^2\right)\sqrt{y^2+\left(1-y\right)^2}}\\
\end{align*}
\begin{align*}
&&&\int_0^1\dfrac{\left(1-y\right)^2}{\left(2y^2+\left(1-y\right)^2\right)\sqrt{y^2+\left(1-y\right)^2}}{\rm\,d}y\\  
&&=&\int_0^1\dfrac{y^2}{\left(2\left(1-y\right)^2+y^2\right)\sqrt{\left(1-y\right)^2+y^2}}{\rm\,d}y\\
&&=&\int_0^1\dfrac{y^2}{\left(2\left(y^2-2y+1\right)+y^2\right)\sqrt{2y^2-2y+1}}{\rm\,d}y\\
&&=&\int_0^1\dfrac{y^2}{\left(3y^2-4y+2\right)\sqrt{2\left(y-\frac{1}{2}\right)^2+\frac{1}{2}}}{\rm\,d}y\\
&&\overset{
y-\frac{1}{2}=\frac{1}{2}\sinh t
}{\overline{\overline{\hspace{3cm}}}}&\int_{\ln\left(\sqrt{2}-1\right)}^{\ln\left(\sqrt{2}+1\right)}\dfrac{\left(\frac{1}{2}\sinh t
+\frac{1}{2}\right)^2}{\left(3\left(\frac{1}{2}\sinh t
+\frac{1}{2}\right)^2-4\left(\frac{1}{2}\sinh t
+\frac{1}{2}\right)+2\right)\sqrt{\frac{1}{2}\sinh^2t
+\frac{1}{2}}}{\rm\,d}\left(\frac{1}{2}\sinh t
\right)\\
&&=&\frac{1}{\sqrt{2\,}\,}\int_{\ln\left(\sqrt{2}-1\right)}^{\ln\left(\sqrt{2}+1\right)}\dfrac{\left(\frac{1}{2}\sinh t
+\frac{1}{2}\right)^2}{3\left(\frac{1}{2}\sinh t
+\frac{1}{2}\right)^2-4\left(\frac{1}{2}\sinh t
+\frac{1}{2}\right)+2}{\rm\,d}t\\
&&=&\frac{1}{\sqrt{2\,}\,}\int_{\ln\left(\sqrt{2}-1\right)}^{\ln\left(\sqrt{2}+1\right)}\dfrac{\left(e^{2t}+2e^t-1\right)^2}{3e^{4t}-4e^{3t}+6e^{2t}+4e^t+3}{\rm\,d}t\\
&&=&\frac{1}{\sqrt{2\,}\,}\int_{\ln\left(\sqrt{2}-1\right)}^{\ln\left(\sqrt{2}+1\right)}\dfrac{\left(e^{2t}+2e^t-1\right)^2}{\left(3e^{4t}-4e^{3t}+6e^{2t}+4e^t+3\right)e^t}{\rm\,d}\left(e^t\right)\\
&&\overset{e^t=w}{\overline{\overline{\hspace{3cm}}}}&\,\,\frac{1}{\sqrt{2\,}\,}\int_{\sqrt{2}-1}^{\sqrt{2}+1}\dfrac{\left(w^2+2w-1\right)^2}{\left(w^2-2w+3\right)\left(3w^2+2w+1\right)w}{\rm\,d}w\\
\end{align*}

\begin{align*}
&&&\frac{1}{\sqrt{2\,}\,}\int_{\sqrt{2}-1}^{\sqrt{2}+1}\dfrac{\left(w^2+2w-1\right)^2}{\left(w^2-2w+3\right)\left(3w^2+2w+1\right)w}{\rm\,d}w\\
&&=&\frac{1}{\sqrt{2\,}\,}\int_{\sqrt{2}-1}^{\sqrt{2}+1}\dfrac{\left(w^2+2w-1\right)^2}{\left(w^2-2w+3\right)\left(3w^2+2w+1\right)w}{\rm\,d}w\\
&&=&\frac{1}{\sqrt{2\,}\,}\int_{\sqrt{2}-1}^{\sqrt{2}+1}\dfrac{1}{3w}{\rm\,d}w+\frac{1}{\sqrt{2\,}\,}\int_{\sqrt{2}-1}^{\sqrt{2}+1}\dfrac{2\left(1+w\right)}{3\left(w^2-2w+3\right)}{\rm\,d}w-\frac{1}{\sqrt{2\,}\,}\int_{\sqrt{2}-1}^{\sqrt{2}+1}\dfrac{2\left(1+w\right)}{3w^2+2w+1}{\rm\,d}w\\
&&=&\,\boxed{\frac{\pi}{6}}
\end{align*}