Fernando Garcia

PHYS405 - Electricity & Magnetism I

The electric potential of some configuration is given by the expression

V (r) = A \frac{e^{- λ r}}{r}

Where $A$ and $λ$ are constants. Find the electric field $E (r)$ , the charge density $ρ (r)$ , and the total charge $Q$ .

Start by recalling that

E = - \nabla V

Using the spherical gradient, it is clear that

\begin{aligned} E & = - A \frac{\partial}{\partial r} (\frac{e^{- λ r}}{r}) \hat{r} \\ = A \frac{e^{- λ r}}{r^{2}} (r λ + 1) \hat{r} \end{aligned}

To find $ρ$ , we recall

ρ = ϵ_{0} \nabla \cdot E

To calculate such divergence, we need to recall two things:

1. Product rule.

\nabla \cdot (f v) = \nabla (f) \cdot v + f (\nabla \cdot v)

2. The special case of the divergence of $1 / r^{2}$ .

\nabla \cdot (\frac{\hat{r}}{r^{2}}) = 4 π δ^{3} (r)

With that in mind, we proceed:

\begin{aligned} ρ & = ϵ_{0} \nabla \cdot (A \frac{e^{- λ r}}{r^{2}} (r λ + 1) \hat{r}) \\ = ϵ A (e^{- λ r} (r λ + 1) \nabla \cdot (\frac{\hat{r}}{r^{2}}) + (\frac{\hat{r}}{r^{2}}) \cdot \nabla (e^{- λ r} (r λ + 1))) \\ = ϵ_{0} A (4 π δ^{3} (r) - \frac{λ^{2}}{r} e^{- λ r}) \end{aligned}

To get the charge, we integrate

\begin{aligned} Q & = \int ρ d τ \\ = ϵ_{0} A \int 4 π δ^{3} (r) d τ - ϵ_{0} A λ^{2} \int \frac{e^{- λ r}}{r} d τ \\ = ϵ_{0} A 4 π - ϵ_{0} A λ^{2} \int_{ϕ = 0}^{2 π} \int_{θ = 0}^{π} \int_{r = 0}^{\infty} \frac{e^{- λ r}}{r} r^{2} \sin (θ) d r d θ d ϕ \\ = ϵ_{0} A 4 π - ϵ_{0} A λ^{2} 4 π \int_{0}^{\infty} r e^{- λ r} d r \\ = ϵ_{0} A 4 π - ϵ_{0} A λ^{2} 4 π \frac{1}{λ^{2}} \\ = 0 \end{aligned}