平面几何问题

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}\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*}

\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*}

 

如图,分别以$\triangle ABC$的边$AB,AC$为一边在三角形外作正方形$ABEF$和$ACGH$, $M$为$FH$上的中点,求证: $MA\perp BC$.

如图,分别以$\triangle ABC$的边$AB,AC$为一边在三角形外作正方形$ABEF$和$ACGH$, $MA\perp BC$,求证: $M$为$FH$上的中点.

 

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转载自www.cnblogs.com/Eufisky/p/10080420.html
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