查看“︁學院物理/電位”︁的源代码

{{wikipedia|電位}}
==定義==
<math>\begin{align}U_r=\frac{kqQ}r\qquad V_r=\frac{U_r}q=\frac{kQ}r\qquad V_p=k\int\frac{dq}r\qquad\Delta V=-\int_1^2\vec Ed\vec S\\\vec E=\frac{-\partial V}{\partial x}\hat i\end{align}</math>
==性質==
{{see also|學院物理/連續電荷分布與高斯定律#六{{!}}圖23.4}}
<span style=display:inline-block;width:30px></span>[[w:絕緣體|絕緣體]][[w:球面#包围的体积|球]]<span style=display:inline-block;width:40px></span>[[w:導體|導體]]球
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<div style="margin-left:44px"><math>a\qquad\qquad\quad\;a</math></div>
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<div style="left:20px;position:absolute;top:20px"><math>\propto r\,\propto\frac1{r^2}\qquad\quad\,\propto\frac1{r^2}</math></div>
<div style="left:48px;position:absolute;top:110px"><math>\begin{align}&\begin{smallmatrix}\propto3a^2-\end{smallmatrix}r^2\\&\propto\frac1 r\qquad\quad\;\,\propto\frac1 r\end{align}</math></div>
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===絕緣體球===
:<math>\begin{align}V(r<a)&=\frac{kQ}{2a}\left(3-\frac{r^2}{a^2}\right)=\frac{kQ}{2a^3}\left(3a^2-r^2\right)\\V(r>a)&=\frac{kQ}r\end{align}</math>
===導體球===
:<math>\begin{align}V(r<a)=\frac{kQ}a\\V(r>a)=\frac{kQ}r\end{align}</math>

==例題==
===五===
<div style=float:right>{{see also|學院物理/連續電荷分布與高斯定律#二{{!}}圖23.3}}圖24.14</div>
<math>\begin{align}V=k\int\frac{dq}r=\frac{kQ}\sqrt{x^2+a^2}\\E_x=\frac{-dV}{dx}=\frac{kxQ}{\sqrt{x^2+a^2}^3}\end{align}</math>
===六===
<div style=float:right>{{see also|學院物理/連續電荷分布與高斯定律#三{{!}}圖23.4}}圖24.15</div>
<math>\begin{align}V=\int dV=\int\frac{kdq}\sqrt{x^2+r^2}=\int\frac{k2\pi\sigma rdr}\sqrt{x^2+r^2}=k2\pi\sigma\int_0^R\frac{rdr}\sqrt{x^2+r^2}\\=k2\pi\sigma\int_0^R\frac 1{2r}r\left(x^2+r^2\right)^{\frac{-1}2}d\left(x^2+r^2\right)=k\pi\sigma2\left[(x^2+r^2)^{\frac1 2}\right]_0^R\\=2k\pi\sigma(\sqrt{x^2+R^2}-x)\\E_x=\frac{-dV}{dx}=-2k\pi\sigma\frac{d\left[(x^2+R^2)^{\frac1 2}-x\right]}{dx}=-2k\pi\sigma\left(\frac{2x}2\frac1\sqrt{x^2+R^2}-1\right)\\=2k\pi\sigma\left(1-\frac x\sqrt{x^2+R^2}\right)\end{align}</math>
===七===
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<div style=left:0;position:absolute;top:0><math>\begin{array}{l}P\,\bullet\nwarrow\qquad\\\quad\;\,a\quad\;r\\\quad\;\gets x\to\searrow \dashv dx\vdash\\\quad\;\longleftarrow\quad l\quad\longrightarrow\end{array}</math></div>
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圖24.16</div>
{{see also|微积分学/不定积分/三角代換法{{!}}三角代換法|學院物理/連續電荷分布與高斯定律#一{{!}}例題23.1}}
<math>\begin{align}V_p=k\int\frac{dq}r=k\int_0^l\frac{\lambda dx}\sqrt{x^2+a^2}=k\lambda\int_0^l\frac{dx}\sqrt{x^2+a^2}\\\because\int\frac{dx}\sqrt{x^2+a^2}=\ln\left(\frac{x+\sqrt{x^2+a^2}}a\right)+\text{C}\\\therefore V_p=k\lambda\left[\ln\left(\frac{x+\sqrt{x^2+a^2}}a\right)\right]_0^l\\=k\lambda\left[\ln\left(\frac{l+\sqrt{l^2+a^2}}a\right)-\ln\frac\sqrt{a^2}a\right]\\=k\lambda\ln\left(\frac{l+\sqrt{l^2+a^2}}a\right)\end{align}</math>

==靜電平衡==
導體的靜電平衡有以下性質：
#導體內的電場為零
#導體的電荷在表面
#導體外的電場為<math>\sigma/\epsilon</math>
#導體的電荷分布在曲率半徑越大時密度越大
==練習==
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<div style=left:120px;position:absolute;top:70px><math>\theta_1</math></div>
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<math>
\begin{align}&
\lambda=a\sin\theta
\\&
dE=\frac{kdq}{r^2}
\\&
dq=\lambda ds=a\sin\theta rd\theta
\\&
dE_x=dE\cos\theta=\frac{kdq\cos\theta}{r^2}=\frac{ka\sin\theta\cos\theta d\theta}r
\\&
dE_y=dE\sin\theta=\frac{kdq\sin\theta}{r^2}=\frac{ka\sin^2\theta d\theta}r
\\&
E_x=\frac{ka}r\int_{\theta_1}^{\theta_2}\sin\theta\cos\theta d\theta=\frac{ka}{4r}\int_{\theta_1}^{\theta_2}\sin2\theta d2\theta
\\&
=\frac{-ka}{4r}(\cos2\theta_2-\cos2\theta_1)
\\&
E_y=\frac{ka}r\int_{\theta_1}^{\theta_2}\sin^2\theta d\theta=\frac{ka}{4r}\int_{\theta_1}^{\theta_2}\left(1-\cos2\theta\right)d2\theta
\\&
=\frac{ka}{4r}\bigg[2\theta-\sin2\theta\bigg]_{\theta_1}^{\theta_2}=\frac{ka}{4r}\left(2\theta_2-2\theta_1-\sin2\theta_2+\sin2\theta_1\right)
\\&
\vec E=E_x\hat i-E_y\hat j
\\&
=\frac{-ka}{4r}\left(\cos2\theta_2-\cos2\theta_1\right)\hat i-\frac{ka}{4r}\left(2\theta_2-2\theta_1-\sin2\theta_2+\sin2\theta_1\right)\hat j
\\&
V_x=-\int E_xdS=-\int E_xrd\theta
\\&
=\frac{ka}8\int_{\theta_2}^{\theta_1}\cos2\theta d2\theta

=\frac{ka}8\bigg[\sin2\theta\bigg]_{\theta_2}^{\theta_1}
\\&
=\frac{ka}8\left(\sin2\theta_2-\sin2\theta_1\right)
\\&
V_y=-\int E_ydS=-\int E_yrd\theta
\\&
=\frac{-ka}{4r}\int_{\theta_1}{\theta_2}2\theta-\sin2\theta d\theta
\\&
=\frac{-ka}{4r}\bigg[\theta^2+\frac{\cos2\theta}2\bigg]_{\theta_1}^{\theta_2}
\end{align}
</math>