The electric potential of some configuration is given by the expression
V( r )=A \frac{e^{-\lambda r}}{r} ,
where A and λ are constants. Find the electric field E(r), the charge density ρ(r ), and the total charge Q. \left[\text { Answer: } \rho=\epsilon_{0} A\left(4 \pi \delta^{3}( r )-\lambda^{2} e^{-\lambda r} / r\right)\right]
E =- \nabla V=-A \frac{\partial}{\partial r}\left(\frac{e^{-\lambda r}}{r}\right) \hat{ r }=-A\left\{\frac{r(-\lambda) e^{-\lambda r}-e^{-\lambda r}}{r^{2}}\right\} \hat{ r }=A e^{-\lambda r}(1+\lambda r) \frac{\hat{ r }}{r^{2}} .
\rho=\epsilon_{0} \nabla \cdot E =\epsilon_{0} A\left\{e^{-\lambda r}(1+\lambda r) \nabla \cdot\left(\frac{\hat{r}}{r^{2}}\right)+\frac{\hat{r}}{r^{2}} \cdot \nabla\left(e^{-\lambda r}(1+\lambda r)\right)\right\} . \quad \text { But } \nabla \cdot\left(\frac{\hat{r}}{r^{2}}\right)=4 \pi \delta^{3}( r ) (Eq. 1.99), and e^{-\lambda r}(1+\lambda r) \delta^{3}( r )=\delta^{3}( r ) (Eq. 1.88). Meanwhile,
f(x) \delta(x)=f(0) \delta(x) (1.88)
\nabla \left(e^{-\lambda r}(1+\lambda r)\right)=\hat{ r } \frac{\partial}{\partial r}\left(e^{-\lambda r}(1+\lambda r)\right)=\hat{ r }\left\{-\lambda e^{-\lambda r}(1+\lambda r)+e^{-\lambda r} \lambda\right\}=\hat{ r }\left(-\lambda^{2} r e^{-\lambda r}\right) .
\text { So } \frac{\hat{ r }}{r^{2}} \cdot \nabla\left(e^{-\lambda r}(1+\lambda r)\right)=-\frac{\lambda^{2}}{r} e^{-\lambda r} , and \rho=\epsilon_{0} A\left[4 \pi \delta^{3}( r )-\frac{\lambda^{2}}{r} e^{-\lambda r}\right]
Q=\int \rho d \tau=\epsilon_{0} A\left\{4 \pi \int \delta^{3}( r ) d \tau-\lambda^{2} \int \frac{e^{-\lambda r}}{r} 4 \pi r^{2} d r\right\}=\epsilon_{0} A\left(4 \pi-\lambda^{2} 4 \pi \int_{0}^{\infty} r e^{-\lambda r} d r\right) .
\text { But } \int_{0}^{\infty} r e^{-\lambda r} d r=\frac{1}{\lambda^{2}}, \text { so } Q=4 \pi \epsilon_{0} A\left(1-\frac{\lambda^{2}}{\lambda^{2}}\right)=\text { zero. }