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青青子衿 发表于 2018-11-3 10:49

与矩形均匀带电薄片电场强度有关的二重积分

[i=s] 本帖最后由 青青子衿 于 2018-11-17 21:22 编辑 [/i]

2.25  矩形带电薄片(带正电荷)位于\(z=-z_0\)(其中\(z_0>0\))的平面上,且带电薄片限定于\(-a\le x\le a\)与\(-b\le y\le b\)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)
\[ \left|\overrightarrow{\boldsymbol{E}}\,\right|\,=\int_{-b}^b\int_{-a}^a \dfrac{\sigma\,\,{\rm\,d}x{\rm\,d}y}{4\pi\varepsilon_0\left(x^2+y^2+{z_{\overset{\,}{0}}}^2\right)}\dfrac{z_0}{\sqrt{x^2+y^2+{z_{\overset{\,}{0}}}^2}} \]
\begin{align*}
\big|\overrightarrow{\boldsymbol{E}}\big|\,&=\int_{-b}^b\int_{-a}^a \dfrac{\sigma z_0}{4\pi\varepsilon_0\left(x^2+y^2+{z_{\overset{\,}{0}}}^2\right)^{\frac{3}{2}}}{\rm\,d}x{\rm\,d}y\\
\,\\
&=\dfrac{\sigma z_{\overset{\,}{0}}}{4\pi\varepsilon_0}\int_{-b}^b\int_{-a}^a \dfrac{1}{\left(x^2+y^2+{z_{\overset{\,}{0}}}^2\right)^{\frac{3}{2}}}{\rm\,d}x{\rm\,d}y\\
&=\dfrac{\sigma z_{\overset{\,}{0}}}{\pi\varepsilon_0}\int_0^b\int_0^a \dfrac{1}{\left(x^2+y^2+{z_{\overset{\,}{0}}}^2\right)^{\frac{3}{2}}}{\rm\,d}x{\rm\,d}y
\end{align*}
...[code]Desmos code
\int_{-b}^b\int_{-a}^a\frac{1}{\left(x^2+y^2+z_0^2\right)^{\frac{3}{2}}}dxdy[/code]

青青子衿 发表于 2018-11-17 16:33

[i=s] 本帖最后由 青青子衿 于 2019-7-5 16:31 编辑 [/i]

[b]回复 [url=http://kuing.orzweb.net/redirect.php?goto=findpost&pid=28732&ptid=5720]1#[/url] [i]青青子衿[/i] [/b]
\begin{align*}
\overrightarrow{\boldsymbol{E}}\,
&=\int_{-b}^b\int_{-a}^a \dfrac{\sigma z_{\overset{\,}{0}}}{4\pi\varepsilon_{\overset{\,}{0}}\left(x^2+y^2+{z_{\overset{\,}{0}}}^2\right)^{\frac{3}{2}}}{\rm\,d}x{\rm\,d}y\,\vv{\boldsymbol{n}}\\
&=\dfrac{\sigma{\color{red}{z_{\overset{\,}{0}}}}\vv{\boldsymbol{n}}}{\pi\varepsilon_0}\int_0^b\int_0^a \dfrac{1}{\left(x^2+y^2+{z_{\overset{\,}{0}}}^2\right)^{\frac{3}{2}}}{\rm\,d}x{\rm\,d}y
\,\\
&=\dfrac{\sigma\vv{\boldsymbol{n}}}{\pi\varepsilon_0}\arctan\left(\dfrac{ab}{z_{\overset{\,}{0}}\sqrt{a^2+b^2+{z_{\overset{\,}{0}}}^2}}\right)\\
&=\dfrac{\sigma}{\pi\varepsilon_0}\arctan\left(\dfrac{ab}{z_{\overset{\,}{0}}\sqrt{a^2+b^2+{z_{\overset{\,}{0}}}^2}}\right)\Big(\,0,\,0,\,1\,\Big)\\
\end{align*}

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