Momentum per unit area per unit time in EM

[athrun200]

Homework Statement

In fact it is problem 8.5 in Griffiths 3rd ed p357

Consider an infinite parallel plate capacitor with the lower plate carrying the charger density - \sigma, and the upper plate carrying the charge density + \sigma.
(a) Determine all nine elements of the stress tensor, in the region between the plates. Display your answer as a 3x3 matrix
(b)Use Eq 8.22,\overrightarrow F = \oint\limits_S {T \cdot d\overrightarrow a } for \overrightarrow S = 0, to determine the force per unit area on the top plate.
(c) What is the momentum per unit area, per unit time, crossing the xy plane?
(d) Find the recoil force per unit area on the top plate.

Homework Equations

Maxwell's Stress Tensor
\overrightarrow F = \oint\limits_S {T \cdot d\overrightarrow a } - {\varepsilon _0}{{\bar \mu }_0}\frac{d}{{dt}}\int {Sd\tau }

Momentum density
{\wp _{em}} = {\varepsilon _0}{\mu _0}\overrightarrow S

\frac{d}{{dt}}({\wp _{em}} + {\wp _{mech}}) = \nabla \cdot T

The Attempt at a Solution

Part (a) and (b) are easy for me.

(a) {T_{xy}} = {T_{xz}} = {T_{yz}} = ... = 0 and by using some equations we can find {T_{xx}},{T_{yy}}{\rm{ and }}{T_{zz}}
So the answer is T = \frac{{{\sigma ^2}}}{{2{\varepsilon _0}}}\left( {\begin{array}{*{20}{c}}<br /> { - 1}&amp;0&amp;0\\<br /> 0&amp;{ - 1}&amp;0\\<br /> 0&amp;0&amp;1<br /> \end{array}} \right)
(b) is also easy, use that equation provided we can find the answer \overrightarrow f = \frac{{{\sigma ^2}}}{{2{\varepsilon _0}}}\widehat z

The problem is (c)
I think the equation \frac{d}{{dt}}({\wp _{em}} + {\wp _{mech}}) = \nabla \cdot T is useful to solve it with {\wp _{mech}}=0
However on the right hand side we have \nabla \cdot T. The div of a tensor would be? I learned only the div over a vector, but not a tensor.

 

[TSny]

The problem is (c)
I think the equation \frac{d}{{dt}}({\wp _{em}} + {\wp _{mech}}) = \nabla \cdot T is useful to solve it with {\wp _{mech}}=0

I think a more relevant equation for part (c) is equation 8.28. Note how Griffiths interprets the last term of this equation (just below equation 8.29). So, how can you interpret $\int_S \overleftrightarrow{T} \cdot d\vec{a}$ over a surface S in terms of momentum?

 

[athrun200]

I think a more relevant equation for part (c) is equation 8.28. Note how Griffiths interprets the last term of this equation (just below equation 8.29). So, how can you interpret $\int_S \overleftrightarrow{T} \cdot d\vec{a}$ over a surface S in terms of momentum?

$\int_S \overleftrightarrow{T} \cdot d\vec{a}$ is the momentum transferred to the surface per unit time

So \overset\leftrightarrow T is the momentum transferred to the surface per unit time, per unit area, thus it is the answer?

The momentum per unit area, per unit time perpendicular to xy is \overset\leftrightarrow {{T_{zz}}} which is \frac{{{\sigma ^2}}}{{2{\varepsilon _0}}}
While \overset\leftrightarrow {{T_{xx}}} = \overset\leftrightarrow {{T_{yy}}} = - \frac{{{\sigma ^2}}}{{2{\varepsilon _0}}} is the momentum per area, per time parallel to the surface?

 

[TSny]

$\int_S \overleftrightarrow{T} \cdot d\vec{a}$ is the momentum transferred to the surface per unit time

So \overset\leftrightarrow T is the momentum transferred to the surface per unit time, per unit area, thus it is the answer?

The momentum per unit area, per unit time perpendicular to xy is \overset\leftrightarrow {{T_{zz}}} which is \frac{{{\sigma ^2}}}{{2{\varepsilon _0}}}

Yes.

While \overset\leftrightarrow {{T_{xx}}} = \overset\leftrightarrow {{T_{yy}}} = - \frac{{{\sigma ^2}}}{{2{\varepsilon _0}}} is the momentum per area, per time parallel to the surface?

Well, $T_{xx}$ is the x-component of momentum per unit area per unit time crossing a surface oriented perpendicular to the x axis. Similarly for $T_{yy}$.

 

[athrun200]

Thanks a lot!