91Ó°ÊÓ

It is given that the volume charge density of a point charge qat r' isp(r),provided thatthe volume integral of pequals q.

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

$∫{\mathrm{δ}}^2\left(r-r\text{'}\right)d\mathrm{τ}=1$(a)

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

${∫}_{allspace}\mathrm{ÒÏ}\left(r\right){\mathrm{δ}}^2\left(r-a\right)d\mathrm{Ï„}=\mathrm{ÒÏ}\left(a\right)$

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

Compute the value of$∫{\mathrm{δ}}^2\left(r-r\text{'}\right)d\mathrm{τ}$ as follows:

$∫{\mathrm{δ}}^2\left(r-r\text{'}\right)d\mathrm{τ}=1$

Substitute 1 for$∫{\mathrm{δ}}^2\left(r-r\text{'}\right)d\mathrm{τ}=1$ into the equation .

role="math" localid="1657368114501" ${∫}_{allspace}\mathrm{ÒÏ}\left(r\right)d\mathrm{Ï„}=q∫{\mathrm{δ}}^2\left(r-r\text{'}\right)d\mathrm{Ï„}$.

role="math" localid="1657368168462" $$\begin{gathered} {∫}_{allspace}\mathrm{ÒÏ}\left(r\right)d\mathrm{Ï„}=q\left(1\right) \\ =q \end{gathered}$$

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

(b)

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

role="math" localid="1657368463939" $$\begin{gathered} p\left(a\right)={∫}_{allspace}\mathrm{ÒÏ}\left(r\right)\mathrm{δ}{}^2(r-a)d\mathrm{Ï„} \\ =\mathrm{ÒÏ}\left(a\right) \end{gathered}$$

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:

$$\begin{gathered} p\left(a\right)={∫}_{allspace}\mathrm{ÒÏ}\left(r\right)\mathrm{δ}{}^2(r-a)d\mathrm{Ï„} \\ =\mathrm{ÒÏ}\left(a\right) \end{gathered}$$

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

role="math" localid="1657368891826" $$\begin{gathered} p\left(r\right)=q\left[{\mathrm{δ}}^3\left(r-a\right)-{\mathrm{δ}}^3\left(r\right)\right] \\ =q{\mathrm{δ}}^3(r-a)-q{\mathrm{δ}}^3\left(r\right) \end{gathered}$$

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

(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:

role="math" localid="1657369143005" $\mathrm{ÒÏ}\left(r\right)=A\mathrm{δ}(r-R)$

Here, A is a constant.

The integral of volume charge density$\mathrm{ÒÏ}\left(r\right)$ gives total charge$Q$ as follows:

$Q=∫\mathrm{ÒÏ}\left(r\right)d\mathrm{Ï„}$

Substitute$A\mathrm{δ}\left(r-R\right)$ for$p\left(r\right)$ into$Q=∫p\left(r\right)d\mathrm{τ}$ as follows:

$Q=∫A\mathrm{δ}\left(r-R\right)d\mathrm{τ}$

The infinitesimal volume element $d\mathrm{Ï„}=4\mathrm{Ï€}{R}^{2}dr$.

Substitute$4{\mathrm{Ï€¸é}}^{2}dr$ for$d\mathrm{Ï„}$ into the equation role="math" localid="1657369605513" $Q=∫A\mathrm{δ}\left(r-R\right)d\mathrm{Ï„}$.

$$\begin{gathered} Q=∫A\mathrm{δ}\left(r-R\right)4{\mathrm{Ï€¸é}}^{2}dr \\ =A\left(4\mathrm{Ï€}{R}^2\right)∫\mathrm{δ}\left(r-R\right)dr \end{gathered}$$

The value of$∫\mathrm{δ}\left(r-R\right)dr$ is 1.

Substitute 1 for$∫\mathrm{δ}\left(r-R\right)dr$ into $Q=A\left(4\mathrm{π}R\right)∫\mathrm{δ}(r-R)dr$.

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

Substitute $\frac{Q}{4\mathrm{Ï€}{R}^2}$for A into $\mathrm{ÒÏ}\left(r\right)=A\mathrm{δ}(r-R)$.

$\mathrm{ÒÏ}\left(r\right)=\frac{Q}{4\mathrm{Ï€}{R}^2}\mathrm{δ}(r-R)$

Therefore, the volume charge density within the spherical shell is defined as$\mathrm{ÒÏ}\left(r\right)=\frac{Q}{4\mathrm{Ï€}{R}^2}\mathrm{δ}(r-R)$.