6-21 电荷Q均匀分布在半径为R的球体内，试证明离球心r(r<R)处的电势为 Gamma =dfrac (Q(3{R)^2-(r)^2)}(8{pi r)^0(R)^3}

步骤 1：计算球内电荷体密度
球内电荷体密度为：
$\rho =\dfrac {Q}{\dfrac {4}{3}\pi {R}^{3}}$

步骤 2：计算球心到 r 的小球的电势
该小球的体积为 $\dfrac {4}{3}\pi {r}^{3}$，电荷为 $q=\rho \times \dfrac {4}{3}\pi {r}^{3}$。
电势 由这部分电荷产生的在距离 r 的点为：
${U}_{1}=\dfrac {q}{4\pi {\varepsilon }_{0}r}$
${U}_{1}=\dfrac {\rho \times \dfrac {4}{3}\pi {r}^{3}}{4\pi {\varepsilon }_{0}r}$
${U}_{1}=\dfrac {Q{r}^{2}}{4\pi {\varepsilon }_{0}\times \dfrac {4}{3}{R}^{3}}$

步骤 3：计算从r到R的球壳的电势
对于球壳上的一个微小电荷，其半径为 r'，厚度为 dr'，产生的电势为:
${U}_{2}=\dfrac {dq}{4\pi {\varepsilon }_{0}r}$
其中:
$dq=\rho \times 4\pi {r'}^{2}dr'$
代入上式得到：
${U}_{2}=\dfrac {\rho \times 4\pi {r'}^{2}dr'}{4\pi {\varepsilon }_{0}r}$
对 dU2从r 到 R 积分得到：
${U}_{2}=\dfrac {Q}{4\pi {\varepsilon }_{0}\times \dfrac {4}{3}{R}^{3}}{\int }_{r}^{R}r'dr'$
${U}_{2}=\dfrac {Q}{4\pi {\varepsilon }_{0}\times \dfrac {4}{3}{R}^{3}}\times \dfrac {1}{2}({R}^{2}-{r}^{2})$

步骤 4：计算总电势
总电势为：
$J={U}_{1}+{U}_{2}$
将 ${U}_{1}$ 和 ${U}_{2}$ 的表达式代入，得到：
$J=\dfrac {Q{r}^{2}}{4\pi {\varepsilon }_{0}\times \dfrac {4}{3}{R}^{3}}+\dfrac {Q}{4\pi {\varepsilon }_{0}\times \dfrac {4}{3}{R}^{3}}\times \dfrac {1}{2}({R}^{2}-{r}^{2})$
$J=\dfrac {Q(3{R}^{2}-{r}^{2})}{8\pi {\varepsilon }_{0}{R}^{3}}$