The electric potential of some configuration is given by the expressionV(r)=Ae-λrrWhere A and λ are constants. Find the electric field E(r), the charge density ρ(r), and the total charge Q.

Answer The electric field is E=Ae-λr(1+rλ)^ rr2.The charge density is Aε0[4πδ3r-λ2re-λr].The total charge Q is 0.

Q2.50P - Introduction to Electrodynamics

Q2.50P

Question

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 $Q$ .

Step-by-Step Solution

Verified

Answer

Answer

The electric field is $E = A e^{- λ r} (1 + r λ) \frac{\hat{} r}{r^{2}}$ .

The charge density is $A ε_{0} [4 π δ^{3} (r) - \frac{λ^{2}}{r} e^{- λ r}]$ .

The total charge Q is 0.

1Step 1: Define functions

Consider the given expression.,

$V (r) = A \frac{e^{λ r}}{r}$ …… (1)

Here, A and $λ$ are constant.

2Step 2: Determine electric filed

Write the representation of electric filed is gradient of a scalar potential.,

$E = - \nabla V$ ……(2)

Hence, the electric filed is calculated as follows:,

$E = - \nabla V = - \frac{\partial}{\partial r} (A \frac{e^{- λ r}}{r}) = - A [\frac{- r e^{- λ r} (- λ) - e^{- λ r} (1)}{r^{2}}] r = - A [\frac{- r λ e^{- λ r} - e^{- λ r} (1)}{r^{2}}] r$

Solve further.

$E = A (r λ e^{- λ r} + e^{- λ r}) \frac{\hat{} r}{r^{2}} = A e^{- λ r} (1 + r λ) \frac{\hat{} r}{r^{2}}$ …… (3)

Thus, the electric field is $E = A e^{- λr} (1 + rλ) \frac{\hat{} r}{r^{2}}$ .

3Step 3: Determine charge density

The Gauss’s law in different way

$\nabla \cdot E = \frac{1}{ε_{0}} ρ ρ = ε_{0} \nabla \cdot E$ …… (4)

Substitute $E = A e^{- λr} (1 + rλ) \frac{\hat{} r}{r^{2}}$ in equation (4).

$ρ = ε \nabla \cdot (A e^{- λr} (1 + rλ) \frac{\hat{} r}{r^{2}}) = ε_{0} [(A e^{- λr} (1 + rλ)) (\nabla \cdot \frac{\hat{} r}{r^{2}}) + (\frac{r}{r^{2}}) \cdot \nabla (A e^{- λ r}) (1 + r λ)]$

But $\nabla \cdot \frac{\hat{} r}{r^{2}} = 4 {πδ}^{3} (r)$ ,

Therefore,

$ρ = ε_{0} [(A e^{- λr} (1 + rλ)) (4 π δ^{3} (r)) + (\frac{r}{r^{2}}) \nabla \cdot (A e^{- λr} (1 + rλ))] = ε_{0} [A 4 π δ^{3} (r) + (\frac{r}{r^{2}}) \nabla \cdot (A e^{- λr} (1 + rλ))]$

Since $(e^{- λ r} (1 + r λ)) (4 {πδ}^{3} (r)) = 4 π δ^{3} (r)$ ,

$\nabla (A e^{- λ r} (1 + r λ)) = \hat{} r \frac{\partial}{\partial r} [A e^{- λ r} (1 + r λ)] = \hat{} r A \frac{\partial}{\partial r} [e^{- λ r} (1 + r λ)] = \hat{} r A [(1 + r λ) \frac{\partial}{\partial r} (e^{- λ r}) + e^{- λ r} \frac{\partial}{\partial r} (1 + r λ)] = \hat{} r A [(1 + r λ) (- λ) + e^{- λ r} + e^{- λ r} (λ)]$

Solve further. as,

$\nabla (A e^{- λ r} (1 + r λ)) = \hat{} r A [λ e^{- λ r} - λ (1 + r λ) e^{- λ r}] = \hat{} r A [λ e^{- λ r} - λ e^{- λ r} (1 + λ r)] = \hat{} r A [λ e^{- λ r} - ((1 + λ r))] = \hat{} r A [λ e^{- λ r} - (1 - 1 - λ r)]$

Solve further. as,

$\nabla ({Ae}^{- λr} (1 + rλ)) = \hat{} r A [{λe}^{- λr} (- rλ)] = \hat{} r A [- λ^{2} {re}^{- λr}]$

Multiply by $(\frac{r}{r^{2}})$ on both sides.,

$(\frac{r}{r^{2}}) \cdot \nabla (A e^{- λ r} (1 + r λ)) = (\frac{r}{r^{2}}) \cdot \nabla (A e^{- λ r} (1 + r λ)) = (\frac{r}{r^{2}}) \cdot A (λ e^{- λ r} + (1 + r λ) e^{- λ r} (- r λ)) = (\frac{r}{r^{2}}) \cdot [A (λ e^{- λ r} - λ e^{- λ r} - r λ^{2} e^{- λ r}) r] = (\frac{r}{r^{2}}) \cdot A [- r λ^{2} e^{- λ r}]$

Thus,

$(\frac{r}{r^{2}}) \cdot \nabla (A e^{- λ r} (1 + r λ)) = - \frac{A λ^{2}}{r} e^{- λ r}$

Hence,

$ρ = ε_{0} [A 4 π δ^{3} (r) + (\frac{r}{r^{2}}) \nabla \cdot (A^{- λ r}) (1 + λ)] = ε_{0} [A 4 π δ^{3} (r) - \frac{A λ^{2}}{r} e^{- λ r}] = A ε_{0} [A 4 π δ^{3} (r) - \frac{λ^{2}}{r} e^{- λ r}]$

Thus, the charge density is $A ε_{0} [A 4 {πδ}^{3} (r) - \frac{λ^{2}}{r} e^{- λr}]$ .

4Step 4: Determine the total charge

Write the expression for the total charge.

$Q = \int ρ d ζ = \int A ε_{0} [4 {πδ}^{3} (r) - \frac{λ^{2}}{r} e^{- λr}] d ζ = \int A ε_{0} 4 π δ^{3} (r) d ζ - \int A ε_{0} \frac{λ^{2}}{r} e^{- λ r} d ζ = 4 π ε_{0} A \int δ^{3} (r) d ζ - A ε_{0} λ^{2} \int \frac{e^{- λ r}}{r} d ζ$

Simplify further,

$Q = 4 π ε_{0} A (1) - A ε_{0} λ^{2} \int \frac{e^{- λ r}}{r} (4 π r^{2} d r) = 4 π ε_{0} A - 4 π A ε_{0} λ^{2} \int r e^{2} d r = 4 π ε_{0} A - 4 π A ε_{0} λ^{2} \int r e^{2} d r = 4 π ε_{0} A - 4 π A ε_{0} λ^{2} {[- \frac{r e^{- λ r}}{λ} - \frac{e^{- λ r}}{λ^{2}}]}_{0}^{\infty}$

Also,

$Q = 4 π ε_{0} A - 4 π A ε_{0} λ^{2} (\frac{1}{λ^{2}}) = 4 π ε_{0} A - 4 π A ε_{0} = 0$

Thus, the total charge Q is 0.

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