Q47P - Introduction to Electrodynamics

Question

(a) Write an expression for the volume charge density p(r) of a point charge q at r'. Make sure that the volume integral of p equals q.

(b) What is the volume charge density of an electric dipole, consisting of a point? charge -q at the origin and a point charge +q at a?

(c) What is the volume charge density (in spherical coordinates) of a uniform, in-finitesimally thin spherical shell of radius Rand total charge Q, centered at the origin? [Beware: the integral over all space must equal Q.]

Step-by-Step Solution

Answer

(a) The volume charge density of a point charge is defined as$\rho \left(r\right)=q{\delta }^2(r-r^{\text{'}})$.

(b) The volume charge density of an electric dipole is defined as$\rho \left(r\right)=q{\delta }^3(r-a)-q{\delta }^3\left(r\right)$.

(c) The volume charge density within the spherical shell is defined as$\rho \left(r\right)=\frac{Q}{4\mathrm{\pi }{R}^2}\delta \left(r-R\right)$.

Step 1: Describe the given information

It is given that the volume charge density of a point charge q at r' is p(r), provided that the volume integral of p equals q.

Step 2: Define the Dirac delta function

The Dirac delta function, which is represented as $\mathit{\delta }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$, is defined as $\mathit{\delta }\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{a}\mathbf{)}\mathbf{=}\mathbf{\{}\begin{array}{ll}\mathbf{0}& \mathbf{x}\mathbf{\ne }\mathbf{0}\\ \mathbf{\infty }& \mathbf{x}\mathbf{=}\mathbf{0}\end{array}$ , the Dirac delta function has the property ${\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathit{\delta }\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{a}\mathbf{)}\mathit{d}\mathit{X}\mathbf{=}\mathit{f}\mathbf{\left(}\mathit{a}\mathbf{\right)}$, where $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$ is a continuous containing ,$\mathit{x}\mathbf{=}\mathbf{0}$ and ${\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}\mathit{\delta }\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{a}\mathbf{)}\mathit{d}\mathit{X}\mathbf{=}\mathbf{1}$.

Step: 3 Find the volume charge density of a charge

$\int {\delta }^2\left(r-r\text{'}\right)d\tau =1$(a)

Let us define the volume charge density $\rho \left(r\right)$ using the Dirac delta function as follows:

${\int }_{all space}\rho \left(r\right){\delta }^2\left(r-a\right)d\tau =\rho \left(a\right)$

 

Integrating the volume charge density gives the total charge $q$. Thus, the above definition of $\rho \left(r\right)$ can be written as follows:

 

Compute the value of $\int {\delta }^2\left(r-r\text{'}\right)d\tau$ as follows:

 

$\int {\delta }^2\left(r-r\text{'}\right)d\tau =1$ 

 

Substitute 1 for $\int {\delta }^2\left(r-r\text{'}\right)d\tau =1$ into the equation .

${\int }_{all space}\rho \left(r\right)d\tau =q\int {\delta }^2\left(r-r\text{'}\right)d\tau$.

  $$\begin{gathered} {\int }_{all space}\rho \left(r\right)d\tau =q\left(1\right) \\ =q \end{gathered}$$

Therefore, the volume charge density of a point charge is defined as $\rho \left(r\right)=q{\delta }^2\left(r-r\text{'}\right)$.

Step: 4 Find the volume charge density of an electric dipole

(b)

Let us define the dipole moment $p\left(r\right)$ using the Dirac delta function for continuous charge distribution as follows:

 

$$\begin{gathered} p\left(a\right)={\int }_{all space}\rho \left(r\right)\delta {}^2(r-a)d\tau \\ =\rho \left(a\right) \end{gathered}$$

 

Two equal and opposite charges are separated by a distance constitute an electric dipole. Let the position of the positive charge be , and the position of the negative charge be $r_{-}$, such that $r_{+}-r=a$.

 

The dipole moment can be written as follows:

 

$$\begin{gathered} p\left(a\right)={\int }_{all space}\rho \left(r\right)\delta {}^2(r-a)d\tau \\ =\rho \left(a\right) \end{gathered}$$

 

Compute the value of volume charge density using the Dirac delta function as follows:

 

  $$\begin{gathered} p\left(r\right)=q\left[{\delta }^3\left(r-a\right)-{\delta }^3 \left(r\right)\right] \\ =q{\delta }^3(r-a)-q{\delta }^3 \left(r\right) \end{gathered}$$

 

Therefore, the volume charge density $\rho \left(r\right)$ of an electric dipole is defined as $\rho \left(r\right)=q{\delta }^3\left(r-a\right)-q{\delta }^3\left(r\right)$.

Step: 5 Find the volume charge density within the spherical shell

(c)

It is known that the charge inside a spherical shell of the radius is zero, but there is a finite amount of charge present on the surface of the sphere.  Thus, the charge is non-zero only on the surface where$r=R$. Hence, the volume density on the surface of the spherical shell can be defined using the Dirac delta function:

$\rho \left(r\right)=A\delta (r-R)$

 

Here, A is a constant.

 

The integral of volume charge density $\rho \left(r\right)$ gives total charge $Q$ as follows:

 $Q=\int \rho \left(r\right)d\tau$

 

Substitute $A\delta \left(r-R\right)$ for $p\left(r\right)$ into $Q=\int p\left(r\right)d\tau$ as follows:

$Q=\int A\delta \left(r-R\right)d\tau$

 

The infinitesimal volume element $d\tau =4\pi R^{2}dr$. 

 

Substitute $4{\mathrm{\pi R}}^{2}dr$ for $d\tau$ into the equation $Q=\int A\delta \left(r-R\right)d\tau$.

$$\begin{gathered} Q=\int A\delta \left(r-R\right)4{\mathrm{\pi R}}^{2}dr \\ =A\left(4\pi R^2\right)\int \delta \left(r-R\right) dr \end{gathered}$$

 

The value of $\int \delta \left(r-R\right)dr$ is 1. 

 

Substitute 1 for $\int \delta \left(r-R\right)dr$ into $Q=A\left(4\pi R\right)\int \delta (r-R)dr$.

$$\begin{gathered} Q=A\left(4\pi R^2\right)\left(1\right) \\ A=\frac{Q}{4\pi R^2} \end{gathered}$$

 

Substitute $\frac{Q}{4\mathrm{\pi }{R}^2}$ for A into $\rho \left(r\right)=A\delta (r-R)$.

$\rho \left(r\right)=\frac{Q}{4\pi R^2}\delta (r-R)$

 

Therefore, the volume charge density within the spherical shell is defined as$\rho \left(r\right)=\frac{Q}{4\pi R^2}\delta (r-R)$.