题目
一无限大平面将相对介电常数分别为varepsilon _(1)和varepsilon _(2)两种介质隔开.在介质2中放一点电荷q,该点电荷与界面垂直距离为h,如图所示.求界面极化电荷的分布,以及界面上极化电荷总量.A n-|||-ε1 r-|||-ε2 A-|||-h-|||-r-|||-q
一无限大平面将相对介电常数分别为$\varepsilon _{1}$和$\varepsilon _{2}$两种介质隔开.在介质$2$中放一点电荷$q$,该点电荷与界面垂直距离为$h$,如图所示.求界面极化电荷的分布,以及界面上极化电荷总量.
题目解答
答案
如图,设两介质界面法向是由$2$指向$1$的方向为正向,$\widehat{n}$表示其单位矢量.设界面上某点$A$的极化电荷面密度为$\sigma '\left ( r\right )$,$A$点的电场包含三部分场源的贡献:(ⅰ)点电荷$q$;(ⅱ)点电荷$q$周围的极化电荷$q''$;(ⅲ)介质界面上的极化面电荷密度$\sigma'(r)$.为求界面上极化电荷面密度,只需求出$A$点处垂直于界面的电场分量.
显然,介质界面上的极化面电荷(ⅲ)对这个电场分量的贡献仅仅来自于$A$点附近极化面电荷$\sigma'$,它对$1$侧和$2$侧的贡献分别为
$E'_{ 1n}=\dfrac{ \sigma' }{2 {\varepsilon_{0} }}$,$E'_{2n}=-\dfrac{ \sigma'}{2 {\varepsilon_{0}}}$
再利用介质表面的极化电荷与自由电荷的关系式$\left ( 1.64\right )$和式$\left ( 1.65\right )$,立即得点电荷$q$周围的极化电荷$q''$以及$q + q''$,有
$q''=-\dfrac{\varepsilon_{2}-1}{\varepsilon_{2}}q$和$q+q''= \dfrac{q}{\varepsilon_{2}}$
以致$q$和$q''$在$A$点产生的电场法向分量为
$E''_{1n}=E''_{2n}=\dfrac{\dfrac{q}{\varepsilon_{2}}}{4\pi\varepsilon_{0}r^{2} }\dfrac{h}{r}$
全部场源对$A$点两侧电场法向分量的总贡献为
$E_{1n} = E'_{1n} + E''_{1n} =\dfrac{\sigma'}{2{\varepsilon_{0}}} + \dfrac{qh}{4\pi\varepsilon_{0}\varepsilon_{2}r^{3}}$,$E_{2n} = E'_{2n} + E''_{2n} = -\dfrac{\sigma'}{2\varepsilon_{0}} + \dfrac{qh}{4\pi\varepsilon_{0}\varepsilon_{2}r^{3}}$
由于界面上无自由电荷,利用电位移矢量沿法向分量连续的边界条件式$\left ( 1.59\right )$,得
$D_{1n} = D_{2n} \Rightarrow \varepsilon_{0}\varepsilon_{1}E_{1n} = \varepsilon_{0}\varepsilon_{2}E_{2n}$
代入$E_{ 1n }$和$E_{2n}$,从中解出界面上极化电荷面密度表达式
$\sigma'\left ( r\right ) = \dfrac{qh\left ( \varepsilon_{2} - \varepsilon_{1}\right )}{2\pi\varepsilon_{2}\left ( \varepsilon_{2} + \varepsilon_{1}\right )r^{3}}$
界面上极化电荷总量
$q'= \sum \limits_{0}^{\infty}\sigma'\left ( r\right ) \cdot 2\pi r'dr' = \dfrac{qh\left ( \varepsilon_{2} - \varepsilon_{1}\right )}{ \varepsilon_{2} \left ( \varepsilon_{2} +\varepsilon_{1}\right )}\sum \limits_{0}^{\infty}\dfrac{r'dr'}{r^{3}}$
$= \dfrac{qh\left ( \varepsilon_{2} - \varepsilon_{1}\right )}{\varepsilon_{2} \left ( \varepsilon_{2} +\varepsilon_{1}\right )}\sum \limits_{0}^{\infty}\dfrac{r'dr'}{\sqrt{\left ( r'^{2} + h^{2}\right )^{3}}} = \dfrac{qh\left ( \varepsilon_{2} - \varepsilon_{1}\right )}{ \varepsilon_{2} \left ( \varepsilon_{2} +\varepsilon_{1}\right )}\sum \limits_{0}^{\infty}\dfrac{r'dr'}{\left ( r'^{2} + h^{2}\right )^{3/2}} $
$= \dfrac{\left ( \varepsilon_{2} - \varepsilon_{1}\right )q}{\varepsilon_{2}\left ( \varepsilon_{2} + \varepsilon_{1}\right )}$
显然,介质界面上的极化面电荷(ⅲ)对这个电场分量的贡献仅仅来自于$A$点附近极化面电荷$\sigma'$,它对$1$侧和$2$侧的贡献分别为
$E'_{ 1n}=\dfrac{ \sigma' }{2 {\varepsilon_{0} }}$,$E'_{2n}=-\dfrac{ \sigma'}{2 {\varepsilon_{0}}}$
再利用介质表面的极化电荷与自由电荷的关系式$\left ( 1.64\right )$和式$\left ( 1.65\right )$,立即得点电荷$q$周围的极化电荷$q''$以及$q + q''$,有
$q''=-\dfrac{\varepsilon_{2}-1}{\varepsilon_{2}}q$和$q+q''= \dfrac{q}{\varepsilon_{2}}$
以致$q$和$q''$在$A$点产生的电场法向分量为
$E''_{1n}=E''_{2n}=\dfrac{\dfrac{q}{\varepsilon_{2}}}{4\pi\varepsilon_{0}r^{2} }\dfrac{h}{r}$
全部场源对$A$点两侧电场法向分量的总贡献为
$E_{1n} = E'_{1n} + E''_{1n} =\dfrac{\sigma'}{2{\varepsilon_{0}}} + \dfrac{qh}{4\pi\varepsilon_{0}\varepsilon_{2}r^{3}}$,$E_{2n} = E'_{2n} + E''_{2n} = -\dfrac{\sigma'}{2\varepsilon_{0}} + \dfrac{qh}{4\pi\varepsilon_{0}\varepsilon_{2}r^{3}}$
由于界面上无自由电荷,利用电位移矢量沿法向分量连续的边界条件式$\left ( 1.59\right )$,得
$D_{1n} = D_{2n} \Rightarrow \varepsilon_{0}\varepsilon_{1}E_{1n} = \varepsilon_{0}\varepsilon_{2}E_{2n}$
代入$E_{ 1n }$和$E_{2n}$,从中解出界面上极化电荷面密度表达式
$\sigma'\left ( r\right ) = \dfrac{qh\left ( \varepsilon_{2} - \varepsilon_{1}\right )}{2\pi\varepsilon_{2}\left ( \varepsilon_{2} + \varepsilon_{1}\right )r^{3}}$
界面上极化电荷总量
$q'= \sum \limits_{0}^{\infty}\sigma'\left ( r\right ) \cdot 2\pi r'dr' = \dfrac{qh\left ( \varepsilon_{2} - \varepsilon_{1}\right )}{ \varepsilon_{2} \left ( \varepsilon_{2} +\varepsilon_{1}\right )}\sum \limits_{0}^{\infty}\dfrac{r'dr'}{r^{3}}$
$= \dfrac{qh\left ( \varepsilon_{2} - \varepsilon_{1}\right )}{\varepsilon_{2} \left ( \varepsilon_{2} +\varepsilon_{1}\right )}\sum \limits_{0}^{\infty}\dfrac{r'dr'}{\sqrt{\left ( r'^{2} + h^{2}\right )^{3}}} = \dfrac{qh\left ( \varepsilon_{2} - \varepsilon_{1}\right )}{ \varepsilon_{2} \left ( \varepsilon_{2} +\varepsilon_{1}\right )}\sum \limits_{0}^{\infty}\dfrac{r'dr'}{\left ( r'^{2} + h^{2}\right )^{3/2}} $
$= \dfrac{\left ( \varepsilon_{2} - \varepsilon_{1}\right )q}{\varepsilon_{2}\left ( \varepsilon_{2} + \varepsilon_{1}\right )}$