Perturbation of Maxwell stress from voltage distribution

[chowdhury]

TL;DR Summary

I want to find the perturbation of Maxwell stress for ac perturbation of voltage on top of DC.

I have a voltage distribution $V(x,y) = V_{dc}(x,y)+ V_{ac}(x,y) \cos(\omega t)$, I have derived the Matrix e. But I do not know how to extract it from the voltage, meaning I do not know how to find $E_{x0} , E_{y0}, \delta E_{x}, \delta E_{y}$ in terms of $V_{dc}(x,y), V_{ac}(x,y)$.


\begin{equation*}
\mathbf{E} = -\nabla \negthinspace V = \begin{bmatrix}
E_{x}\\
E_{y} \end{bmatrix} = \begin{bmatrix}
E_{x0} + \delta E_{x}\\
E_{y0} + \delta E_{y}
\end{bmatrix}
\end{equation*}

\begin{equation*}
\mathbf{T}^{\mathrm{Maxwell}} = \epsilon \mathbf{E} \otimes \mathbf{E} - \frac{1}{2} \epsilon \mathbf{E}^2 \mathcal{I}
\end{equation*}

Here $\mathbf{E}^2 = E_{x}^2 + E_{y}^2$.

\begin{align*}
\mathbf{T}^{\mathrm{Maxwell}} &= \epsilon \begin{bmatrix}
\frac{1}{2} \left( E_x^2 - E_y^2 \right) & E_{x} E_{y} \\
E_{x} E_{y} & -\frac{1}{2} \left( E_x^2 - E_y^2 \right)
\end{bmatrix} \\
&= \mathbf{T}^{\mathrm{Maxwell}}_{0} + \delta\mathbf{T}^{\mathrm{Maxwell}}
\end{align*}

\begin{align*}
\delta\mathbf{T}^{\mathrm{Maxwell}} &= \epsilon \begin{bmatrix}
E_{x0} \delta E_{x} - E_{y0} \delta E_{y} & E_{x0} \delta E_{y} + E_{y0} \delta E_{x}\\
E_{x0} \delta E_{y} + E_{y0} \delta E_{x} & -\left( E_{x0} \delta E_{x} - E_{y0} \delta E_{y} \right )
\end{bmatrix} \\
&= \begin{bmatrix}
\delta T_{1}^{\mathrm{Maxwell}} & \delta T_{6}^{\mathrm{Maxwell}}\\
\delta T_{6}^{\mathrm{Maxwell}} & \delta T_{2}^{\mathrm{Maxwell}}
\end{bmatrix}
\end{align*}

The e-coefficient is
\begin{align*}
e_{iI} = \begin{bmatrix}
e_{x1} & e_{x2} & e_{x6} \\
e_{y1} & e_{y2} & e_{y6}
\end{bmatrix}
\end{align*}

In component form $e_{iI} = e_{Ii}$, we can write

\begin{align*}
e_{x1} &= -\delta T_{1}^{\mathrm{Maxwell}} / \delta E_{x}, \qquad e_{y1} = - \delta T_{1}^{\mathrm{Maxwell}} / \delta E_{y} \\
e_{x2} &= -\delta T_{2}^{\mathrm{Maxwell}} / \delta E_{x}, \qquad e_{y2} = - \delta T_{2}^{\mathrm{Maxwell}} / \delta E_{y} \\
e_{x6} &= -\delta T_{6}^{\mathrm{Maxwell}} / \delta E_{x}, \qquad e_{y6} = - \delta T_{6}^{\mathrm{Maxwell}} / \delta E_{y} \\
\end{align*}

\begin{align*}
e_{x1} &= -\epsilon \left( E_{x0} - E_{y0} \frac{\delta E_{y}}{\delta E_{x}} \right) \\
e_{x2} &= \epsilon \left( E_{x0} - E_{y0} \frac{\delta E_{y}}{\delta E_{x}} \right) \\
e_{x6} &= -\epsilon \left( E_{x0} \frac{\delta E_{y}}{\delta E_{x}} + E_{y0} \right) \\
e_{y1} &= -\epsilon \left( E_{x0} \frac{\delta E_{x}}{\delta E_{y}} - E_{y0} \right) \\
e_{y2} &= \epsilon \left( E_{x0} \frac{\delta E_{x}}{\delta E_{y}} - E_{y0} \right) \\
e_{y6} &= -\epsilon \left( E_{x0} + E_{y0} \frac{\delta E_{x}}{\delta E_{y}} \right)
\end{align*}

Now I have the problem, given $V(x,y) = V_{dc}(x,y)+ V_{ac}(x,y) \cos(\omega t)$, how to find,

• $E_{x0}$
• $E_{y0}$
• $\delta E_{x}$
• $\delta E_{y}$

in terms of $V_{dc}(x,y), V_{ac}(x,y)$For just 1D, we have
$E_{x} = E_{0} + \delta E = \frac{V_{dc} + V_{ac}}{d}$, then $$\mathbf{T}^{\mathrm{Maxwell}} = {T}_{0} + \delta T$$ and we obtain $e = -\epsilon E_{dc}$, which I can clearly find out, but in 2D, how to find the above four things?

 

[Valerie Park]

To find $E_{x0}, E_{y0}, \delta E_{x}$, and $\delta E_{y}$ in terms of $V_{dc}(x,y), V_{ac}(x,y)$, you need to solve the following two equations:\begin{align*}E_{x} &= \frac{V_{dc}(x,y) + V_{ac}(x,y) \cos(\omega t)}{d} \\E_{y} &= \frac{V_{dc}(x,y) + V_{ac}(x,y) \sin(\omega t)}{d}\end{align*}The solution will be:\begin{align*}E_{x0} &= \frac{V_{dc}(x,y)}{d} \\E_{y0} &= \frac{V_{dc}(x,y)}{d} \\\delta E_{x} &= \frac{V_{ac}(x,y) \cos(\omega t)}{d} \\\delta E_{y} &= \frac{V_{ac}(x,y) \sin(\omega t)}{d}\end{align*}