题目
一半径为R的带电球体,其电荷体密度分布为r≤R(q为一正的常量)(I>R试求:(1)带电球体的总电荷:2)球内、外各点的电场强度;(3)球内、外各点的电势
一半径为R的带电球体,其电荷体密度分布为r≤R(q为一正的常量)(I>R试求:(1)带电球体的总电荷:2)球内、外各点的电场强度;(3)球内、外各点的电势
题目解答
答案
解析
步骤 1:计算带电球体的总电荷
在球内取半径为r、厚为dr的薄球壳,该壳内所包含的电荷为
$$
dq = \rho dV = \frac{q}{\pi R^4} 4\pi r^2 dr = \frac{4qr^3}{R^4} dr
$$
则球体所带的总电荷为
$$
Q = \int dq = \int_0^R \frac{4qr^3}{R^4} dr = \frac{4q}{R^4} \int_0^R r^3 dr = \frac{4q}{R^4} \left[\frac{r^4}{4}\right]_0^R = q
$$
步骤 2:计算球内、外各点的电场强度
在球内作一半径为r1的高斯球面,按高斯定理有
$$
4\pi r_1^2 E_1 = \frac{1}{\epsilon_0} \int_{r_1} q(r) dV = \frac{1}{\epsilon_0} \int_0^{r_1} \frac{4qr^3}{R^4} 4\pi r^2 dr = \frac{4q}{\epsilon_0 R^4} \int_0^{r_1} r^5 dr = \frac{4q}{\epsilon_0 R^4} \left[\frac{r^6}{6}\right]_0^{r_1} = \frac{q r_1^2}{3 \epsilon_0 R^4}
$$
得
$$
E_1 = \frac{q r_1^2}{12 \pi \epsilon_0 R^4}
$$
方向沿半径向外。
在球体外作半径为r2的高斯球面,按高斯定理有
$$
4\pi r_2^2 E_2 = \frac{1}{\epsilon_0} \int_{r_2} q(r) dV = \frac{1}{\epsilon_0} \int_0^R \frac{4qr^3}{R^4} 4\pi r^2 dr = \frac{4q}{\epsilon_0 R^4} \int_0^R r^5 dr = \frac{4q}{\epsilon_0 R^4} \left[\frac{r^6}{6}\right]_0^R = \frac{q}{3 \epsilon_0}
$$
得
$$
E_2 = \frac{q}{4 \pi \epsilon_0 r_2^2}
$$
(r2>R),E2方向沿半径向外。
步骤 3:计算球内、外各点的电势
球内电势
$$
U_1 = \int_{r_1}^R E_1 dr + \int_R^\infty E_2 dr = \int_{r_1}^R \frac{q r_1^2}{12 \pi \epsilon_0 R^4} dr + \int_R^\infty \frac{q}{4 \pi \epsilon_0 r_2^2} dr = \frac{q}{12 \pi \epsilon_0 R^4} \left[\frac{r_1^3}{3}\right]_{r_1}^R + \frac{q}{4 \pi \epsilon_0} \left[-\frac{1}{r_2}\right]_R^\infty = \frac{q}{36 \pi \epsilon_0 R^4} (R^3 - r_1^3) + \frac{q}{4 \pi \epsilon_0 R}
$$
球外电势
$$
U_2 = \int_R^\infty E_2 dr = \int_R^\infty \frac{q}{4 \pi \epsilon_0 r_2^2} dr = \frac{q}{4 \pi \epsilon_0} \left[-\frac{1}{r_2}\right]_R^\infty = \frac{q}{4 \pi \epsilon_0 R}
$$
在球内取半径为r、厚为dr的薄球壳,该壳内所包含的电荷为
$$
dq = \rho dV = \frac{q}{\pi R^4} 4\pi r^2 dr = \frac{4qr^3}{R^4} dr
$$
则球体所带的总电荷为
$$
Q = \int dq = \int_0^R \frac{4qr^3}{R^4} dr = \frac{4q}{R^4} \int_0^R r^3 dr = \frac{4q}{R^4} \left[\frac{r^4}{4}\right]_0^R = q
$$
步骤 2:计算球内、外各点的电场强度
在球内作一半径为r1的高斯球面,按高斯定理有
$$
4\pi r_1^2 E_1 = \frac{1}{\epsilon_0} \int_{r_1} q(r) dV = \frac{1}{\epsilon_0} \int_0^{r_1} \frac{4qr^3}{R^4} 4\pi r^2 dr = \frac{4q}{\epsilon_0 R^4} \int_0^{r_1} r^5 dr = \frac{4q}{\epsilon_0 R^4} \left[\frac{r^6}{6}\right]_0^{r_1} = \frac{q r_1^2}{3 \epsilon_0 R^4}
$$
得
$$
E_1 = \frac{q r_1^2}{12 \pi \epsilon_0 R^4}
$$
方向沿半径向外。
在球体外作半径为r2的高斯球面,按高斯定理有
$$
4\pi r_2^2 E_2 = \frac{1}{\epsilon_0} \int_{r_2} q(r) dV = \frac{1}{\epsilon_0} \int_0^R \frac{4qr^3}{R^4} 4\pi r^2 dr = \frac{4q}{\epsilon_0 R^4} \int_0^R r^5 dr = \frac{4q}{\epsilon_0 R^4} \left[\frac{r^6}{6}\right]_0^R = \frac{q}{3 \epsilon_0}
$$
得
$$
E_2 = \frac{q}{4 \pi \epsilon_0 r_2^2}
$$
(r2>R),E2方向沿半径向外。
步骤 3:计算球内、外各点的电势
球内电势
$$
U_1 = \int_{r_1}^R E_1 dr + \int_R^\infty E_2 dr = \int_{r_1}^R \frac{q r_1^2}{12 \pi \epsilon_0 R^4} dr + \int_R^\infty \frac{q}{4 \pi \epsilon_0 r_2^2} dr = \frac{q}{12 \pi \epsilon_0 R^4} \left[\frac{r_1^3}{3}\right]_{r_1}^R + \frac{q}{4 \pi \epsilon_0} \left[-\frac{1}{r_2}\right]_R^\infty = \frac{q}{36 \pi \epsilon_0 R^4} (R^3 - r_1^3) + \frac{q}{4 \pi \epsilon_0 R}
$$
球外电势
$$
U_2 = \int_R^\infty E_2 dr = \int_R^\infty \frac{q}{4 \pi \epsilon_0 r_2^2} dr = \frac{q}{4 \pi \epsilon_0} \left[-\frac{1}{r_2}\right]_R^\infty = \frac{q}{4 \pi \epsilon_0 R}
$$