|
|
本帖最后由 青青子衿 于 2019-7-5 17:35 编辑 2.32 均匀带电立方体（带正电荷）位于三维区域\(\,\,\Omega\colon\,
\left\{\begin{split}
-a\leqslant\,\!x&\leqslant\,a\\
-b\leqslant\,\!y&\leqslant\,b\\
-z_{\overset{\,}2}\leqslant\,\!z&\leqslant-z_{\overset{\,}1}\end{split}\right.\,\)里（其中\(\,a>0\),\(\,b>0\),\(\,c>0\)）,
其电荷体密度为\(\,\rho\)，试求出点\(\,A(0,0,0)\,\)处的电场强度\(\boldsymbol{E}\)（矢量）
\[ \left|\overrightarrow{\boldsymbol{E}}\,\right|\,=\displaystyle\iiint\limits_{\Omega}\dfrac{\rho\,\,{\rm\,d}V}{4\pi\varepsilon_0\left(x^2+y^2+z^2\right)}\dfrac{\left|z\right|}{\sqrt{x^2+y^2+z^2}} \]
\begin{align*}
\left|\overrightarrow{\boldsymbol{E}}\,\right|\,&=\iiint\limits_{\Omega}\dfrac{\rho\,\,\mathrm{d}x\mathrm{d}y\mathrm{d}z}{4\pi\varepsilon_0\left(x^2+y^2+z^2\right)}\dfrac{\left|z\right|}{\sqrt{x^2+y^2+z^2}}\\
&=\int_{z_{\overset{\,}{1}}\,}^{z_{\overset{\,}{2}}\,}\int_{-b}^{b}\int_{-a}^{a}\dfrac{\rho\,\,\mathrm{d}x\mathrm{d}y\mathrm{d}z}{4\pi\varepsilon_0\left(x^2+y^2+z^2\right)}\dfrac{z}{\sqrt{x^2+y^2+z^2}}\\
&=\int_{z_{\overset{\,}{1}}\,}^{z_{\overset{\,}{2}}\,}\int_{0}^{b}\int_{0}^{a}\dfrac{\rho\,\,\mathrm{d}x\mathrm{d}y\mathrm{d}z}{\pi\varepsilon_0\left(x^2+y^2+z^2\right)}\dfrac{z}{\sqrt{x^2+y^2+z^2}}\\
&=\int_{z_{\overset{\,}{1}}\,}^{z_{\overset{\,}{2}}}\int_{0}^{b}\int_{0}^{a}\dfrac{\rho\,z\,\mathrm{d}x\mathrm{d}y\mathrm{d}z}{\pi\varepsilon_0\left(x^2+y^2+z^2\right)^{\frac{3}{2}}}\\
&=\dfrac{\rho}{\pi\varepsilon_0}\int_{z_{\overset{\,}{1}}\,}^{z_{\overset{\,}{2}}}\arctan\left(\dfrac{ab}{z\sqrt{a^2+b^2+z^2}}\right)\mathrm{d}z\\
\end{align*}
另见：kuing.cjhb.site/forum.php?mod=viewthread&tid=5720 |
|
