Question

Find the (Lorenz gauge) potentials and fields of a time-dependent ideal electric dipole $\mathbf{p}\left(\mathbf{t}\right)$ at the origin. (It is stationary, but its magnitude and/or direction are changing with time.) Don't bother with the contact term. [Answer:

$\begin{array}{l}\mathbf{V}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_0}\frac{\widehat{r}}{{\mathbf{r}}^2}\mathbf{\cdot }\mathbf{[}\mathbf{p}\mathbf{+}\mathbf{(}\mathbf{r}\mathbf{/}\mathbf{c}\mathbf{)}\dot{\mathbf{p}}\mathbf{]}\\ \mathbf{A}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\frac{{\mathbf{\mu }}_0}{\mathbf{4}\mathbf{\pi }}\left[\frac{}{}\right]\\ \mathbf{E}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\mathbf{-}\frac{{\mathbf{\mu }}_0}{\mathbf{4}\mathbf{\pi }}\left\{\frac{\ddot{P}-\widehat{r}(\widehat{r}\cdot \ddot{p})}{}+c^2\frac{[p+(r/c)\dot{p}]-3\widehat{r}(\widehat{r}\cdot [p+(r/c)\dot{p}])}{r^3}\right\}\\ \mathbf{B}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\mathbf{-}\frac{{\mathbf{\mu }}_0}{\mathbf{4}\mathbf{\pi }}\left\{\frac{\widehat{r}\times [\dot{p}+(r/c)\ddot{p}]}{r^2}\right\}\end{array}$

Where all the derivatives of $\mathbf{p}$ are evaluated at the retarded time.]

Short answer

The value of (Lorenz gauge) potentials of a time-dependent ideal electric dipole $\text{p}\left(\text{t}\right)$at the origin is $\mathrm{V}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\widehat{\mathrm{r}}}{{\mathrm{r}}^2}\cdot (\overrightarrow{\mathrm{p}}\left({\mathrm{t}}_0\right)+(\mathrm{r}/\mathrm{c})\dot{\overrightarrow{\mathrm{p}}}\left({\mathrm{t}}_0\right))$.

The value of vector potentials of a time-dependent ideal electric dipole$\text{p}\left(\text{t}\right)$ at the origin is$\overrightarrow{\mathrm{A}}\approx \frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\dot{\overrightarrow{\mathrm{p}}}\left({\mathrm{t}}_0\right)}{\mathrm{r}}$.

The value of magnetic field of a time-dependent ideal electric dipole $\text{p}\left(\text{t}\right)$ at the origin is $\overrightarrow{\mathrm{B}}=-\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{1}{{\mathrm{r}}^2}(\widehat{\mathrm{r}}\times \dot{\overrightarrow{\mathrm{p}}}\left({\mathrm{t}}_0\right)+(\mathrm{r}/\mathrm{c})\widehat{\mathrm{r}}\times \ddot{\overrightarrow{\mathrm{p}}}\left({\mathrm{t}}_0\right))$.

The value of electric field of a time-dependent ideal electric dipole $\text{p}\left(\text{t}\right)$ at the origin is $\overrightarrow{E}=-\frac{{\mu }_0}{4\pi }\left\{\frac{\overrightarrow{p}\left(t_0\right)-\widehat{r}(\widehat{r}\cdot \ddot{\overrightarrow{p}}\left(t_0\right))}{r}+\frac{c^2}{r^3}\left[(\overrightarrow{p}+\frac{r}{c}\dot{\overrightarrow{p}}\left(t_0\right))-3\widehat{r}(\overrightarrow{r}\cdot (\overrightarrow{p}-\frac{r}{c}\dot{\overrightarrow{p}}\left(t_0\right)))\right]\right\}$

Step-by-step solution

Write the given data from the question.

Consider the distance between the source and the point of interest will be measured using the letter $l$ due to the site's limitations $I$.

Determine the formula of scalar and vector potentials of a time-dependent ideal electric dipole p(t)  at the origin, magnetic and electrical field of a time-dependent ideal electric dipole p(t) at the origin.

Write the formula of (Lorenz gauge) potentials of a time-dependent ideal electric dipole at the origin.

$\mathbf{V}\mathbf{=}\frac{1}{\mathbf{4}{\mathbf{\pi \epsilon }}_0}{\int }_V\frac{\rho ({\overrightarrow{r}}^{\text{'}},t_r)}{\iota }{\mathbf{dV}}^{\mathbf{\text{'}}}$ …… (1)

Here, role="math" localid="1658842513366" $\mathit{\rho }$ is charge density, ${\mathbf{\epsilon }}_{\mathbf{0}}$ is permittivity, role="math" localid="1658842442385" ${\overrightarrow{\mathbf{r}}}^{\mathbf{\text{'}}}$is resistance, role="math" localid="1658842619960" ${\mathit{t}}_{\mathbf{r}}$ is retarded time and role="math" localid="1658842641287" $\mathit{\iota }$ is letter at the distance from the source to the point of interest.

Write the formula of vector potentials of a time-dependent ideal electric dipole at the origin.

$\overrightarrow{\mathbf{A}}\mathbf{=}\frac{{\mathbf{\mu }}_0}{\mathbf{4}\mathbf{\pi }}{\int }_V\frac{\overrightarrow{J}({\overrightarrow{r}}^{\text{'}},t_r)}{\iota }{\mathbf{dV}}^{\mathbf{\text{'}}}$ …… (2)

Here, ${\mathbf{\mu }}_{\mathbf{0}}$is permeability, ${\overrightarrow{\mathbf{r}}}^{\mathbf{\text{'}}}$is resistance, ${\mathit{t}}_{\mathbf{r}}$ is retarded time and $\mathit{\iota }$ is letter at the distance from the source to the point of interest.

Write the formula of magnetic field of a time-dependent ideal electric dipole at the origin.

$\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{\nabla }\mathbf{\times }\overrightarrow{\mathbf{A}}$ …… (3)

Here, $\mathbf{\nabla }$ is derivative, $\stackrel{}{\mathbf{A}}$ is vector potential.

Write the formula of electric field of a time-dependent ideal electric dipole at the origin.

$\overrightarrow{\mathbf{E}}\mathbf{=}\mathbf{-}\mathbf{\nabla }\mathbf{V}\mathbf{-}\frac{\mathbf{\partial }\overrightarrow{\mathbf{A}}}{\mathbf{\partial }\mathbf{t}}$ …… (4)

Here, $\mathbf{\nabla }$ is derivative, $\mathbf{V}$ is velocity and $\mathit{A}$ is vector potential.

Determine the value of scalar and vector potentials of a time-dependent ideal electric dipole p(t) at the origin, magnetic and electrical field of a time-dependent ideal electric dipole p(t) at the origin.

Determine the (Lorenz gauge) potentials of a time-dependent ideal electric dipole$\text{p}\left(\text{t}\right)$at the origin.

Substitute $\frac{1}{r}\left(1+\frac{\overrightarrow{r}\cdot r^{\text{'}}}{r^2}\right)$for $\iota$and $t_0+\frac{\overrightarrow{r}\cdot {\overrightarrow{r}}^{\text{'}}}{r^2}$ for $t_r$ into equation (1).

Where, $t_0=t-r/c$ with this we can expand the potentials:

$\begin{array}{l}V=\frac{1}{4\pi {\epsilon }_0}{\int }_V\frac{\rho ({\overrightarrow{r}}^{\text{'}},t_r)}{\iota }d{V}^{\text{'}}\\ \approx \frac{1}{4\pi {\epsilon }_0}{\int }_V(\rho ({\overrightarrow{r}}^{\text{'}},t_0)+\frac{\overrightarrow{r}\cdot {\overrightarrow{r}}^{\text{'}}}{rc}\dot{\rho }({\overrightarrow{r}}^{\text{'}},t_0))(1+\frac{\overrightarrow{r}\cdot {\overrightarrow{r}}^{\text{'}}}{rc})\frac{1}{r}d{V}^{\text{'}}\\ \approx \frac{1}{4\pi {\epsilon }_0}{\int }_V(\rho ({\overrightarrow{r}}^{\text{'}},t_0)+\frac{\overrightarrow{r}}{r^2}\cdot \left(\rho ({\overrightarrow{r}}^{\text{'}},t_0){\overrightarrow{r}}^{\text{'}}\right)+\frac{\overrightarrow{r}}{rc}\cdot \left(\dot{\rho }({\overrightarrow{r}}^{\text{'}},t_0){\overrightarrow{r}}^{\text{'}}\right))d{V}^{\text{'}}\end{array}$

Using the definition of a dipole moment, ignoring the fact that the first term is the total charge (zero), and leaving out the $o\left({r^{\text{'}}}^2\right)$terms:

Therefore, the value of (Lorenz gauge) potentials of a time-dependent ideal electric dipole at the origin $p\left(t\right)$is $V=\frac{1}{4\pi {\epsilon }_0}\frac{\widehat{r}}{r^2}\cdot (\overrightarrow{p}\left(t_0\right)+(r/c)\dot{\overrightarrow{p}}\left(t_0\right))$.

Determine the vector potentials of a time-dependent ideal electric dipole $p\left(t\right)$at the origin.

Substitute $\frac{1}{r}\left(1+\frac{\overrightarrow{r}\cdot r^{\text{'}}}{r^2}\right)$ for $\iota$and$t_0+\frac{\overrightarrow{r}\cdot {\overrightarrow{r}}^{\text{'}}}{r^2}$ for $t_r$ into equation (1).

role="math" localid="1658843272323" $\overrightarrow{A}=\frac{{\mu }_0}{4\pi }{\int }_V(\overrightarrow{J}({\overrightarrow{r}}^{\text{'}},t_0)+\frac{\overrightarrow{r}\cdot {\overrightarrow{r}}^{\text{'}}}{rc}\overrightarrow{J}({\overrightarrow{r}}^{\text{'}},t_0))\left(1+\frac{\overrightarrow{r}\cdot {\overrightarrow{r}}^{\text{'}}}{rc}\right)\frac{1}{r}d{V}^{\text{'}}$

However, according to the (eq. 5.31) we have:

${\int }_V\overrightarrow{J}d{V}^{\text{'}}=\dot{\overrightarrow{p}}$

And

$\begin{array}{c}\overrightarrow{\mathrm{J}}\left({\overrightarrow{\mathrm{r}}}^{\text{'}}\right)=\mathrm{\rho }{\overrightarrow{\mathrm{\upsilon }}}^{\text{'}}\\ =\mathrm{\rho }\dot{\overrightarrow{\mathrm{r}}}\end{array}$

Therefore, all terms other than the first will have an $o\left({r^{\text{'}}}^2\right)$ or higher and may be eliminated. Due to this, just the initial phrase is preserved.

$\begin{array}{c}A\approx \frac{{\mu }_0}{4\pi r}{\int }_V\overrightarrow{J}({\overrightarrow{r}}^{\text{'}},t_0)d{V}^{\text{'}}\\ =\frac{{\mu }_0}{4\pi }\frac{\dot{\overrightarrow{\rho }}\left(t_0\right)}{r}\end{array}$

Therefore, the value of vector potentials of a time-dependent ideal electric dipole $\text{p}\left(\text{t}\right)$at the origin is $\overrightarrow{A}\approx \frac{{\mu }_0}{4\pi }\frac{\dot{\overrightarrow{p}}\left(t_0\right)}{r}$.

Determine the magnetic field of a time-dependent ideal electric dipole $\text{p}\left(\text{t}\right)$ at the origin.

Substitute$\frac{{\mu }_0}{4\pi }\nabla$ for $\nabla$ and $\frac{\dot{\overrightarrow{p}}\left(t_0\right)}{r}$ for $\overrightarrow{A}$into equation (3).

$\begin{array}{c}\overrightarrow{\mathrm{B}}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\nabla \times \frac{\dot{\overrightarrow{\mathrm{p}}}\left({\mathrm{t}}_0\right)}{\mathrm{r}}\\ =\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\left(\frac{1}{\mathrm{r}}\nabla \times \dot{\overrightarrow{\mathrm{p}}}\left({\mathrm{t}}_0\right)-\dot{\overrightarrow{\mathrm{p}}}\left({\mathrm{t}}_0\right)\times \nabla \frac{1}{\mathrm{r}}\right)\\ =\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\left(\frac{1}{\mathrm{r}}\nabla {\mathrm{t}}_0\times \ddot{\overrightarrow{\mathrm{p}}}\left({\mathrm{t}}_0\right)+\frac{1}{{\mathrm{r}}^2}\dot{\overrightarrow{\mathrm{p}}}\left({\mathrm{t}}_0\right)\times \widehat{\mathrm{r}}\right)\\ =-\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\left(\frac{\widehat{\mathrm{r}}}{\mathrm{rc}}\times \ddot{\overrightarrow{\mathrm{p}}}\left({\mathrm{t}}_0\right)+\frac{1}{{\mathrm{r}}^2}\dot{\overrightarrow{\mathrm{p}}}\left({\mathrm{t}}_0\right)\times \widehat{\mathrm{r}}\right)\end{array}$

Solve further as

$\overrightarrow{B}=-\frac{{\mu }_0}{4\pi }\frac{1}{r^2}(\widehat{r}\times \dot{\overrightarrow{p}}\left(t_0\right)+(r/c)\widehat{r}\times \ddot{\overrightarrow{p}}\left(t_0\right))$

Therefore, the value of magnetic field of a time-dependent ideal electric dipole $\text{p}\left(\text{t}\right)$ at the origin is $\overrightarrow{B}=-\frac{{\mu }_0}{4\pi }\frac{1}{r^2}(\widehat{r}\times \dot{\overrightarrow{p}}\left(t_0\right)+(r/c)\widehat{r}\times \ddot{\overrightarrow{p}}\left(t_0\right))$.

Determine the electric field of a time-dependent ideal electric dipole $p\left(t\right)$at the origin.

Substitute $\frac{{\mu }_0}{4\pi }\frac{\ddot{\overrightarrow{p}}\left(t_0\right)}{r}$ for $\frac{\partial \overrightarrow{A}}{\partial t}$ and $\frac{1}{4\pi {\epsilon }_0}\nabla \left(\frac{\widehat{r}\cdot \overrightarrow{p}\left(t_0\right)}{r^2}+\frac{\widehat{r}\cdot \dot{\overrightarrow{p}}\left(t_0\right)}{r}\right)$for $V$ into equation (4).

$\overrightarrow{E}=-\frac{1}{4\pi {\epsilon }_0}\nabla \left(\frac{\widehat{r}\cdot \overrightarrow{p}\left(t_0\right)}{r^2}+\frac{\widehat{r}\cdot \dot{\overrightarrow{p}}\left(t_0\right)}{r}\right)-\frac{{\mu }_0}{4\pi }\frac{\ddot{\overrightarrow{p}}\left(t_0\right)}{r}$

We may broaden the gradients of the scalar products using identity (ii) on page 21:

$\begin{array}{l}\nabla \frac{\widehat{r}\cdot \overrightarrow{p}\left(t_0\right)}{r^2}=\frac{\widehat{r}}{r^2}\times (\nabla \times \overrightarrow{p}\left(t_0\right))+\left(\frac{\widehat{r}}{r^2}\cdot \nabla \right)\overrightarrow{p}\left(t_0\right)+(\overrightarrow{p}\left(t_0\right)\cdot \nabla )\frac{\widehat{r}}{r^2}\\ \nabla \frac{\widehat{r}\cdot \overrightarrow{p}\left(t_0\right)}{r^2}=\frac{\widehat{r}}{rc}\times (\nabla \times \overrightarrow{p}\left(t_0\right))+\left(\frac{\widehat{r}}{r^2}\cdot \nabla \right)\overrightarrow{p}\left(t_0\right)+(\overrightarrow{p}\left(t_0\right)\cdot \nabla )\frac{\widehat{r}}{rc}\end{array}$

Where we used the fact that $\nabla \times (\widehat{r}/r^2)=0$. The terms are:

$\begin{array}{c}\nabla \times \overrightarrow{\mathrm{p}}\left({\mathrm{t}}_0\right)=-\frac{\widehat{\mathrm{r}}}{\mathrm{c}}\times \dot{\overrightarrow{\mathrm{p}}}\left({\mathrm{t}}_0\right)\\ \nabla \times \dot{\overrightarrow{\mathrm{p}}}\left({\mathrm{t}}_0\right)=-\frac{\widehat{\mathrm{r}}}{\mathrm{c}}\times \ddot{\overrightarrow{\mathrm{p}}}\left({\mathrm{t}}_0\right)\\ \left(\frac{\widehat{\mathrm{r}}}{{\mathrm{r}}^2}\cdot \nabla \right)\overrightarrow{\mathrm{p}}\left({\mathrm{t}}_0\right)=\left(\frac{\mathrm{x}}{{\mathrm{r}}^3}\frac{\partial }{\partial \mathrm{x}}+\frac{\mathrm{y}}{{\mathrm{r}}^3}\frac{\partial }{\partial \mathrm{y}}+\frac{\mathrm{z}}{{\mathrm{r}}^3}\right)\overrightarrow{\mathrm{p}}\left({\mathrm{t}}_0\right)\\ =\frac{1}{{\mathrm{r}}^3}\left(\mathrm{x}\dot{\overrightarrow{\mathrm{p}}}\left({\mathrm{t}}_0\right)\frac{\partial {\mathrm{t}}_0}{\partial \mathrm{x}}+\mathrm{y}\dot{\overrightarrow{\mathrm{p}}}\left({\mathrm{t}}_0\right)\frac{\partial {\mathrm{t}}_0}{\partial \mathrm{y}}+\mathrm{z}\dot{\overrightarrow{\mathrm{p}}}\left({\mathrm{t}}_0\right)\frac{\partial {\mathrm{t}}_0}{\partial \mathrm{z}}\right)\end{array}$

Solve further as

$\begin{array}{c}\left(\frac{\widehat{r}}{r^2}\cdot \nabla \right)\overrightarrow{p}\left(t_0\right)=\frac{\dot{\overrightarrow{p}}\left(t_0\right)}{r^3}\overrightarrow{r}\cdot \nabla t_0\\ =-\frac{\dot{\overrightarrow{p}}\left(t_0\right)}{r^{2}c}\end{array}$

Similarly:

$\left(\frac{\widehat{r}}{rc}\cdot \nabla \right)\dot{\overrightarrow{p}}\left(t_0\right)=-\frac{\ddot{\overrightarrow{p}}\left(t_0\right)}{c^{2}r}$

The final two terms are:

$\begin{array}{c}(\overrightarrow{\mathrm{p}}\left({\mathrm{t}}_0\right)\cdot \nabla )\frac{\widehat{\mathrm{r}}}{{\mathrm{r}}^2}=\left({\mathrm{p}}_{\mathrm{x}}\frac{\partial }{\partial \mathrm{x}}+{\mathrm{p}}_{\mathrm{y}}\frac{\partial }{\partial \mathrm{y}}+{\mathrm{p}}_{\mathrm{z}}\frac{\partial }{\partial \mathrm{z}}\right)\frac{\mathrm{x}\widehat{\mathrm{x}}+\mathrm{y}\widehat{\mathrm{y}}+\mathrm{z}\widehat{\mathrm{z}}}{{\mathrm{r}}^3}\\ ={\mathrm{p}}_{\mathrm{x}}\frac{{\mathrm{r}}^3\mathrm{x}-3{\mathrm{r}}^2-\mathrm{r}}{{\mathrm{r}}^6}+...\\ ={\mathrm{p}}_{\mathrm{x}}\frac{\mathrm{r}\widehat{\mathrm{x}}-3\mathrm{x}\widehat{\mathrm{r}}}{{\mathrm{r}}^4}+...\end{array}$

Summing the terms:

$\frac{1}{r^4}|p_x(r\widehat{x}-3x\widehat{r})+p_y(r\widehat{y}-3y\widehat{r})+p_z(r\widehat{z}-3z\widehat{r})|=\frac{1}{r^3}\left|\overrightarrow{p}(t_0-3(\overrightarrow{p}\left(t_0\right)\cdot \widehat{r})\widehat{r})\right|$

Similarly:

$\begin{array}{c}(\dot{\overrightarrow{\mathrm{p}}}\left({\mathrm{t}}_0\right)\cdot \nabla )\frac{\widehat{\mathrm{r}}}{\mathrm{rc}}=\left({\mathrm{p}}_{\mathrm{x}}\frac{\partial }{\partial \mathrm{x}}+{\mathrm{p}}_{\mathrm{y}}\frac{\partial }{\partial \mathrm{y}}+{\mathrm{p}}_{\mathrm{z}}\frac{\partial }{\partial \mathrm{z}}\right)\frac{\mathrm{x}\widehat{\mathrm{x}}+\mathrm{y}\widehat{\mathrm{y}}+\mathrm{z}\widehat{\mathrm{z}}}{{\mathrm{r}}^2\mathrm{c}}\\ ={\dot{\mathrm{p}}}_{\mathrm{x}}\frac{1}{\mathrm{c}}\frac{\widehat{\mathrm{x}}{\mathrm{r}}^2-2\mathrm{r}\frac{\mathrm{x}}{\mathrm{r}}\overrightarrow{\mathrm{r}}}{{\mathrm{r}}^4}+...\\ ={\dot{\mathrm{p}}}_{\mathrm{x}}\frac{1}{\mathrm{c}}\frac{\mathrm{r}\widehat{\mathrm{x}}-2\mathrm{x}\widehat{\mathrm{r}}}{{\mathrm{r}}^3}+...\end{array}$

Summing the terms:

$\frac{1}{r^{3}c}[{\dot{p}}_x(\widehat{x}r-2x\widehat{r})+{\dot{p}}_y(\widehat{y}r-2y\widehat{r})+{\dot{p}}_z\left(z\widehat{r}\right)]=\frac{1}{r^{2}c}[\dot{\overrightarrow{p}}\left(t_0\right)-2(\dot{\overrightarrow{p}}\left(t_0\right)\cdot \widehat{r})\widehat{r}]$

Now, using the rule of BAC-CAB and (1) and (2) we have:

$\begin{array}{l}-\frac{\widehat{r}}{r^2}\times \left(\frac{\widehat{r}}{c}\times \dot{\overrightarrow{p}}\left(t_0\right)\right)=-\frac{\widehat{r}}{c}\frac{\widehat{r}\cdot \dot{\overrightarrow{p}}\left(t_0\right)}{r^2}+\dot{\overrightarrow{p}}\left(t_0\right)\frac{1}{r^{2}c}\\ -\frac{\widehat{r}}{rc}\times \left(\frac{\widehat{r}}{c}\times \dot{\overrightarrow{p}}\left(t_0\right)\right)=-\frac{\widehat{r}}{c}\frac{\widehat{r}\cdot \dot{\overrightarrow{p}}\left(t_0\right)}{r^2}+\dot{\overrightarrow{p}}\left(t_0\right)\frac{1}{r{c}^2}\end{array}$

Using this with (3), (4), (5) and (6) the gradient of potential is:

$\nabla V=\frac{1}{4\pi {\epsilon }_0}\left\{-\frac{\widehat{r}(\widehat{r}\cdot \ddot{\overrightarrow{p}}\left(t_0\right))}{c^{2}r}+\frac{1}{r^3}\left[(\overrightarrow{p}+\frac{r}{c}\dot{\overrightarrow{p}}\left(t_0\right))-3\widehat{r}(\widehat{r}\cdot (\overrightarrow{p}-\frac{r}{c}\dot{\overrightarrow{p}}\left(t_0\right)))\right]\right\}$

The electric field is finally, when the vector potential term is included and$c^2=1/{\epsilon }_0{\mu }_0$ :

role="math" $\overrightarrow{E}=-\frac{{\mu }_0}{4\pi }\left\{\frac{\overrightarrow{p}\left(t_0\right)-\widehat{r}(\widehat{r}\cdot \ddot{\overrightarrow{p}}\left(t_0\right))}{r}+\frac{c^2}{r^3}\left[(\overrightarrow{p}+\frac{r}{c}\dot{\overrightarrow{p}}\left(t_0\right))-3\widehat{r}(\overrightarrow{r}\cdot (\overrightarrow{p}-\frac{r}{c}\dot{\overrightarrow{p}}\left(t_0\right)))\right]\right\}$

Therefore, the value of electric field of a time-dependent ideal electric dipole$p\left(t\right)$ at the origin is $\overrightarrow{E}=-\frac{{\mu }_0}{4\pi }\left\{\frac{\overrightarrow{p}\left(t_0\right)-\widehat{r}(\widehat{r}\cdot \ddot{\overrightarrow{p}}\left(t_0\right))}{r}+\frac{c^2}{r^3}\left[\left(\overrightarrow{p}+\frac{r}{c}\dot{\overrightarrow{p}}\left(t_0\right)\right)-3\widehat{r}\left(\overrightarrow{r}\cdot \left(\overrightarrow{p}-\frac{r}{c}\dot{\overrightarrow{p}}\left(t_0\right)\right)\right)\right]\right\}$