题目
9.18 如题9.18图所示的绝缘细线上均匀分布着线密度为λ的正电荷,两直导线的长度和半-|||-圆环的半径都等于R.试求环中心O点处的场强和电势.-|||-↑y-|||-dl-|||-dx θ R-|||-x-|||-A B 矛 C R D-|||-dE-|||-题9.18图
题目解答
答案
解析
步骤 1：计算O点处的场强
由于电荷均匀分布与对称性，A B和CD段电荷在O点产生的场强互相抵消。取 $dl=Rd\theta$，则 $dq=\lambda Rd\theta$。产生O点的场强dE如图，由于对称性，O点场强沿y轴负方向。因此，我们有：
$E=\int d{E}_{y}={\int }_{\dfrac {\pi }{2}}^{\dfrac {\pi }{2}}\dfrac {\lambda Rd\theta }{4\pi {\varepsilon }_{0}{R}^{2}}\cos \theta $ $=\dfrac {\lambda }{4\pi {\varepsilon }_{0}R}[ \sin (-\dfrac {\pi }{2})-\sin \dfrac {\pi }{2}] $ $=\dfrac {-\lambda }{2\pi {\varepsilon }_{0}R}$
步骤 2：计算O点处的电势
AB电荷在O点产生电势，以 ${U}_{总}=0$，则 ${U}_{1}={\int }_{B}^{A}\dfrac {\lambda dx}{4\pi {e}_{0}x}={\int }_{R}^{2R}\dfrac {\lambda dx}{4\pi {e}_{0}x}=\dfrac {\lambda }{4\pi {e}_{0}}\ln 2$。同理，CD产生 ${U}_{2}=\dfrac {\lambda }{4\pi {\varepsilon }_{0}}\ln 2$。半圆环产生 ${U}_{3}=\dfrac {\pi R\lambda }{4\pi {\varepsilon }_{0}R}=\dfrac {\lambda }{4{\varepsilon }_{0}}$。因此，O点处的电势为：
${U}_{0}={U}_{1}+{U}_{2}+{U}_{3}=\dfrac {\lambda }{2\pi {\varepsilon }_{0}}\ln 2+\dfrac {\lambda }{4{\varepsilon }_{0}}$
由于电荷均匀分布与对称性，A B和CD段电荷在O点产生的场强互相抵消。取 $dl=Rd\theta$，则 $dq=\lambda Rd\theta$。产生O点的场强dE如图，由于对称性，O点场强沿y轴负方向。因此，我们有：
$E=\int d{E}_{y}={\int }_{\dfrac {\pi }{2}}^{\dfrac {\pi }{2}}\dfrac {\lambda Rd\theta }{4\pi {\varepsilon }_{0}{R}^{2}}\cos \theta $ $=\dfrac {\lambda }{4\pi {\varepsilon }_{0}R}[ \sin (-\dfrac {\pi }{2})-\sin \dfrac {\pi }{2}] $ $=\dfrac {-\lambda }{2\pi {\varepsilon }_{0}R}$
步骤 2：计算O点处的电势
AB电荷在O点产生电势，以 ${U}_{总}=0$，则 ${U}_{1}={\int }_{B}^{A}\dfrac {\lambda dx}{4\pi {e}_{0}x}={\int }_{R}^{2R}\dfrac {\lambda dx}{4\pi {e}_{0}x}=\dfrac {\lambda }{4\pi {e}_{0}}\ln 2$。同理，CD产生 ${U}_{2}=\dfrac {\lambda }{4\pi {\varepsilon }_{0}}\ln 2$。半圆环产生 ${U}_{3}=\dfrac {\pi R\lambda }{4\pi {\varepsilon }_{0}R}=\dfrac {\lambda }{4{\varepsilon }_{0}}$。因此，O点处的电势为：
${U}_{0}={U}_{1}+{U}_{2}+{U}_{3}=\dfrac {\lambda }{2\pi {\varepsilon }_{0}}\ln 2+\dfrac {\lambda }{4{\varepsilon }_{0}}$