91影视

Consider a time-dependent point charge at the origin,$蚁\left(\text{r,t}\right)=\text{q}\left(\text{t}\right){未}^{\text{3}}\left(\text{r}\right)$, is fed by a current $\text{J}\left(\text{r,t}\right)=-\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4\mathrm{蟺}$}\right.\right)\left(\raisebox{1ex}{$q$}\!\left/ \!\raisebox{-1ex}{$r^2$}\right.\right)\hat{r}$$\mathrm{J}(\mathrm{r},\mathrm{t})=-\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\mathrm{蟺}\right)\left(\raisebox{1ex}{$\mathrm{q}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{r}}^2$}\right.\right)\hat{\mathrm{r}}$.

Write the formula of .

$鈭�路\mathbf{J}=鈭�路\left[-\left(\frac{\mathbf{1}}{\mathbf{4}蟺}\right)\left(\frac{\stackrel{\mathbf{藱}}{\mathbf{q}}}{{\mathbf{r}}^{\mathbf{2}}}\right)\hat{\mathbf{r}}\right]$ 鈥︹�� (1)

Here, q is time derivative of q and r is radius.

Write the formula of scalar in the Coulomb gauge.

$\mathbf{V}\left(\mathbf{r}\mathbf{,}\mathbf{t}\right)=\frac{\mathbf{q}\left(\mathbf{t}\right)}{\mathbf{4}蟺{蔚}_{\mathbf{0}}}鈭�\frac{{未}^{\mathbf{3}}\left(\mathbf{r}-\mathbf{r}\mathbf{\text{'}}\right)}{\mathbf{r}\mathbf{\text{'}}}\mathbf{d}蟿\text{'}$ 鈥︹�� (2)

Here, q is time derivative of q and r is radius, is relative permittivity, is the position vector of the point from the origin and is the position vector of the volume element and is element.

Write the formula ofscalar and magnetic vector potential of the electric field.

$\mathbf{E}\mathbf{=}\mathbf{-}\mathbf{鈭�}\mathbf{V}\mathbf{-}\frac{\mathbf{鈭�}\mathbf{A}}{\mathbf{鈭�}\mathbf{t}}$ 鈥︹�� (3)

Here, is derivative, V is voltage and A is vector potential.

Write the formula ofMaxwell equation.

$鈭�路\mathbf{E}=\frac{蚁}{{\mathbf{蔚}}_{\mathbf{0}}}$ 鈥︹�� (4)

Here, is the charge density and is relative permittivity

Equation of continuity: It claims that the change in free charge density over time inside a volume is equal to the divergence of electric current density. The conservation of charge concept is adhered to. The phrase is given as,

$鈭�.J=-\frac{鈭�蚁}{鈭�t}$ 鈥︹�� (5)

Here, is current density and is the charge density enclosed by the volume. The negative sign tells that charge is flowing out of the volume.

It is given that,

$$\begin{gathered} 蚁\left(r,t\right)=q\left(t\right){未}^3\left(r\right) \\ =-\left(\frac{1}{4蟺}\right)\left(\frac{\stackrel{藱}{q}}{r^2}\right)\hat{r} \end{gathered}$$

Here, the value of is the time derivative of that is $\frac{dq}{dt}$, r is the position vector and the delta function at the position .

Confirm the conservation of charge by using the equation of continuity.

Determine the value of is

.$鈭�.J=\left(-\frac{\stackrel{藱}{q}}{4蟺}\right)\left[鈭�.\left(\frac{\hat{r}}{r^2}\right)\right]$

Substitute ${鈭�.\left(\frac{\hat{r}}{r^2}\right)=4蟺{未}^3\left(r\right)}_{}$ and for r into equation (1).

$$\begin{gathered} 鈭�.J=\left(-\frac{\stackrel{藱}{q}}{4蟺}\right){\left[鈭�.\left(\frac{\hat{r}}{r^2}\right)\right]}^3 \\ =\left(-\frac{\stackrel{藱}{q}}{4蟺}\right)4蟺{未}^3\left(r\right) \\ =\left(-\frac{dq}{dt}\right){未}^3\left(r\right) \\ =\left(\frac{d\left(q{未}^3\left(r\right)\right)}{dt}\right) \end{gathered}$$

Substitute$蚁\left(r,t\right)$ for $q\left(t\right){未}^3\left(r\right)$ into above equation.

$鈭�.J=-\frac{d蚁}{dt}$ 鈥︹�� (6)

Equation (6) is the equation of continuity. Hence, it confirms the conservation of charge.

In general, potential in coulomb gauge is defined as:

$$\begin{gathered} V\left(r,t\right)=\frac{1}{4蟺{蔚}_0}鈭�\frac{蚁\left(r\text{'},t\right)}{r\text{'}\text{'}}d蟿\text{'} \\ =\frac{q\left(t\right)}{4蟺{蔚}_0}鈭�\frac{{未}^3\left(r\right)}{r\text{'}\text{'}}d蟿\text{'} \end{gathered}$$ 鈥︹�� (7)

Draw the circuit diagram of showing the arrangement for the above potential is given as,

Figure 1

Here, is the position vector of the point from the origin where potential is to be calculated, is the position vector of the volume element from the origin and is the position vector of the point from the volume element.

From the figure, the relation between, and is given as follows,

$\$r\text{'}+r\text{'}\text{'}=r\$$

Rearrange the above equation, the expression is found at:

$\$r\text{'}=r-r\text{'}\text{'}\$$

Substitute the value of from the above equation and $鈭�f\left(x\right){未}^3\left(x-a\right)d蟿=f\left(a\right)$(property of delta function) in equation (7).

Determine the scalar in the Coulomb gauge.

$$\begin{gathered} V\left(r,t\right)=\frac{q\left(t\right)}{4蟺{蔚}_0}鈭�\frac{{未}^3\left(r-r\text{'}\text{'}\right)}{r\text{'}}d蟿\text{'} \\ =\frac{q\left(t\right)}{4蟺{蔚}_0}\left(\frac{1}{r}\right) \end{gathered}$$ 鈥︹�� (8)

After examining equation (8), it is clear that a static charge distribution is to blame for the scalar potential, and that a charge at rest won't generate a magnetic field.

B = 0 鈥︹�� (9)

Therefore, its divergence will also be zero.

$鈭�.B=0$ 鈥︹�� (10)

Further can be expressed as,

$B=鈭�脳A$ 鈥︹�� (11)

Here, is the vector potential.

Using equation (9) and (10)

$鈭�.\left(鈭�脳A\right)=0$

Therefore, it is clear that

In terms of scalar and magnetic vector potential, determine the electric field is written as follows:

Substitute A = 0 and $V=\frac{q\left(t\right)}{4蟺{蔚}_0}\left(\frac{1}{r}\right)$ into equation (3).

$$\begin{gathered} E=-鈭�V \\ =-鈭�\left[\frac{q\left(t\right)}{4蟺{蔚}_0}\left(\frac{1}{r}\right)\right] \\ =\frac{1}{4蟺{蔚}_0}\left(\frac{q\left(t\right)}{r^2}\right)\hat{r} \end{gathered}$$

Maxwell first equation says that the divergence of the electric field is equal to the times the total charge density.

Substitute $鈭�路\left(\frac{\hat{r}}{r^2}\right)=4蟺{未}^3\left(r\right)$ and $鈭�路\left(\frac{\hat{r}}{r^2}\right)=4蟺{未}^3\left(r\right)$into equation (4).

$$\begin{gathered} 鈭�路E=\frac{1}{4蟺{蔚}_0}鈭�路\left(\frac{q\left(t\right)}{r^2}\right)\hat{r} \\ =\frac{q\left(t\right)}{4蟺{蔚}_0}鈭�路\left(\frac{\hat{r}}{r^2}\right) \\ =\frac{q\left(t\right)\left[4蟺{未}^3\left(r\right)\right]}{4蟺{蔚}_0} \\ =\frac{q\left(t\right){未}^3\left(r\right)}{{蔚}_0} \end{gathered}$$

Substitute $q\left(t\right){未}^3\left(r\right)$ for$蚁\left(r,t\right)$.

$鈭�路E=\frac{蚁}{{蔚}_0}$

Hence, the value of first Maxwell equation is .

Recall equation (10).

$蚁\left(r,t\right)=q\left(t\right){未}^3\left(r\right)$ 鈥︹�� (12)

Hence, the value of second Maxwell equation is.

According to the third equation of Maxwell, an electric field that is conserved has no curl and only an induced electric field has curl.

$鈭�路B=0$ 鈥︹�� (13)

Substitute the value of from equation (8) in the above equation.

$鈭�脳E=-\frac{鈭�B}{鈭�t}$ 鈥︹�� (14)

Hence, the value of third Maxwell equation is .

According to Maxwell's fourth equation, the magnetic field's curvature may be expressed in terms of current density and displacement current.

$鈭�脳B={渭}_{0}J+{渭}_0{e}_0\frac{鈭�E}{鈭�t}$ 鈥︹�� (15)

From equation (9) , so the left-hand side of the above equation will also be zero.

Determine the right hand side of the equation (15).

$$\begin{gathered} {渭}_{0}J+{渭}_0{蔚}_0\frac{鈭�E}{鈭�t}={渭}_0\left[-\frac{1}{4蟺}\left(\frac{\stackrel{藱}{q}}{r^3}\right)\hat{r}\right]+{渭}_0{蔚}_0\left(\frac{\stackrel{藱}{q}}{4蟺{蔚}_0{r}^2}\hat{r}\right) \\ =0 \end{gathered}$$

According to Maxwell's fourth equation, the magnetic field's curvature may be expressed in terms of current density and displacement current.

Hence, all four Maxwell鈥檚 equations are satisfied.