与矩形均匀带电薄片电场强度有关的二重积分

青青子衿

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*}
...

1. Desmos code
2. \int_{-b}^b\int_{-a}^a\frac{1}{\left(x^2+y^2+z_0^2\right)^{\frac{3}{2}}}dxdy

复制代码

青青子衿

回复 1# 青青子衿
\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*}