0
$\begingroup$

Given a capacitor with large plates (area $A$) separated by a dielectric with relative permittivity $\varepsilon_r$ and thickness $g$, I believe the Maxwell stress tensor states that the force on each capacitor plate is: $$F_{zz} = A\sigma_{zz} = A\left(\frac{\varepsilon_0}{2} (E_{zz}^2 - E_{xx}^2 - E_{yy}^2)\right) = \frac{1}{2}\left(\varepsilon_0 A\right)\left(\frac{V}{g}\right)^2$$ , where $E_{xx} = E_{yy} = 0$ and $E_{zz} = \frac{V}{d}$.

However, the force is commonly derived using an energy method: $$F_{zz} = \frac{1}{2} \left(-\frac{\partial C}{\partial g}\right)V^2 -= \frac{1}{2} \left(\varepsilon_0\varepsilon_rA\right)\left(\frac{V}{g}\right)^2$$ , where $C = \frac{\varepsilon_0\varepsilon_rA}{g}$ is the capacitance. I've tried using an FEA software (Altair Flux), and found that the second equation seems to be quite accurate with what the software predicts given voltage boundary conditions on both sides of the dielectric.

My question: Why does my derivation using the Maxwell stress tensor omit the $\varepsilon_r$? I couldn't find any reference that says the Maxwell stress tensor is different inside a dielectric (and, trying to follow the original derivation, I don't immediately see where a $\varepsilon_r$ term would come from), but I would love any pointers/explanations on how to derive the first equation using the Maxwell stress tensor.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
0
$\begingroup$

$E_{zz}= \varepsilon_r$V/d so that's where the relative permittivity is hiding.

The Feynman lectures may be helpful.

Improve this answer
$\endgroup$
1
  • $\begingroup$ I don't believe that's correct? My understanding of the Feynman lectures is that, assuming a constant voltage, the charge on each plate increases by $\varepsilon_r$, but that the electric field actually decreases by $\varepsilon_r$ relative to the charge on each plate (Eq. 10.27). From the equation above 10.10 (and from my FEA sims), $V = Ed$ still even with a dielectric. Every derivation I've seen for the force uses the energy method (Eq. 10.28/10.30), which indeed gives the factor $\varepsilon_r^2/\varepsilon_r = \varepsilon_r$, but the Maxwell stress tensor method remains a mystery to me. $\endgroup$
    – asyndeton256
    Commented Jun 30 at 5:18
0
$\begingroup$

I figured out the answer to my question! Although I couldn't find any reference to the Maxwell stress tensor in a dielectric in my normal Griffiths textbook, I found that "Electrodynamics of Continuous Media", 2nd edition (1984), by Landau and Lifshitz states on Page 66 (Eq. 16.7) that the Maxwell stress tensor is indeed scaled by the material's permittivity $\varepsilon$, not the vacuum permittivity $\varepsilon_0$ (the textbook gives the Maxwell stress tensor in cgs units, but I believe the takeaway is still the same): Screenshoot from Landau and Lifshitz, showing the modified Maxwell stress tensor inside a dielectric

Improve this answer
New contributor
asyndeton256 is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.