Pressure versus stress in uniformly charged sphere

[Rob2024]

Homework Statement

What's the pressure and stress at a distance r within a uniformly charged sphere?

Relevant Equations

$E (r) = \frac{Q r}{4 \pi \epsilon_0 R^3}$
$\sigma(r) = \half \epsilon_0 E^2$

Let the charge density be $\rho$, radius be $R$, total charge be $Q = \rho \frac{4}{3} \pi R^3$. We know
from Gauss's law, $E (r) = \frac{Q r}{4 \pi \epsilon_0 R^3}$.

We also know from Maxwell stress tensor $\sigma(r) = \half \epsilon_0 E^2$.

We can compute the pressure due to the electric field directly, it causes the charge pressing outward due to
repulsion. Take a spherical layer of charge at distance $r$ whose area is $A = 4 \pi r^2$, apply force balance,
$P(r+dr) A - P(r) A = dq E = \rho 4\pi r^2 dr E$
$dP = \rho E dr = \frac{\rho Q r}{4 \pi \epsilon_0 R^3} dr$
$P = \half \frac{k Q \rho r^2}{R^3}$
$P = 3 \half \e\spilon_0 E^2 = 3 \sigma$

Why is the pressure derived directly different from the one from Maxwell's stress tensor? Are they different in physical meaning?

Thanks!

 

[renormalize]

Please rewrite this post in readable LaTeX format. There is a handy LaTeX guide at the left below.

 

[haruspex]

On this forum you need to use a double hash (# #, without the space) instead of $.
And \half is not recognized.
—————————
Homework Statement: What's the pressure and stress at a distance r within a uniformly charged sphere?
Relevant Equations: $E (r) = \frac{Q r}{4 \pi \epsilon_0 R^3}$
$\sigma(r) = \frac 12 \epsilon_0 E^2$

Let the charge density be $\rho$, radius be $R$, total charge be $Q = \rho \frac{4}{3} \pi R^3$. We know
from Gauss's law, $E (r) = \frac{Q r}{4 \pi \epsilon_0 R^3}$

We also know from Maxwell stress tensor $\sigma(r) = \frac 12 \epsilon_0 E^2$

We can compute the pressure due to the electric field directly, it causes the charge pressing outward due to
repulsion. Take a spherical layer of charge at distance $r$ whose area is $A = 4 \pi r^2$, apply force balance,
$P(r+dr) A - P(r) A = dq E = \rho 4\pi r^2 dr E$
$dP = \rho E dr = \frac{\rho Q r}{4 \pi \epsilon_0 R^3} dr$
$P = \frac 12 \frac{k Q \rho r^2}{R^3}$
$P = 3 \frac 12 \epsilon_0 E^2 = 3 \sigma$

Why is the pressure derived directly different from the one from Maxwell's stress tensor? Are they different in physical meaning?

Thanks!

 

[Rob2024]

Hi, I am asking again to see if people have some ideas why the normal stress value is different from pressure? I had thought that they would be the same. Thanks,

 

[haruspex]

$P(r+dr) A - P(r) A = dq E = \rho 4\pi r^2 dr E$

I'm not sure about this. That layer is subject to electrostatic forces from outside as well. Doesn't that reduce the pressure difference required to balance the force from inside?

 

[Rob2024]

Thanks for the correction. With the correction, the pressure is 3/2 of the normal stress. After discussing with a professor, I believe the use of the word stress in this case is not to be confused with stress in elastic material since the Maxwell stress is not 0 in vacuum outside of the sphere. Let me know if any additional enlightenment can be shed on this question. Thanks.

 

[TSny]

We can compute the pressure due to the electric field directly, it causes the charge pressing outward due to
repulsion. Take a spherical layer of charge at distance $r$ whose area is $A = 4 \pi r^2$, apply force balance,
$P(r+dr) A - P(r) A = dq E$

I don’t think this way of deriving $P(r)$ is correct.

The area at $(r+dr)$ is $4 \pi (r+dr)^2 = 4 \pi (r^2 + 2rdr + dr^2)$. The term $dr^2$ in the parentheses can be neglected, but not the term $2rdr$. So, I believe leaving out this term is a mistake.

I’m uncomfortable using an entire spherical shell to deduce $P(r)$. Instead, consider an infinitesimal circular disk of material inside the sphere as shown.

Let $a$ be the (infinitesimal) radius of the disk and $dz$ the thickness of the disk. Let $r$ be the distance from the center of the sphere, $O$, to the center of the bottom surface of the disk. The distance from $O$ to the center of the top of the disk is then $r+dz$. Let $A$ be the area of the bottom (or top) of the disk. So, the volume of the disk is $dV = Adz$

The electric force $F_z^{\rm elec}$ experienced by the disk will be in the $z$ direction: $F_z^{\rm elec} = \rho dV E(r)$.

The net pressure force $F_z^{\rm pr}$ on the disk must be equal and opposite to the electric force: $$F_z^{\rm pr} = P(r)A – P(r+dz)A = - \rho dV E(r).$$ Approximating $P(r+dz) \approx P(r) + P’(r)dz$, show $$P’(r) = \rho E(r) = \frac{\rho^2 r}{3\varepsilon_0}.$$ I think this is what you got. But, as I mentioned above, I don’t believe your derivation is correct.

When you integrate the expression for $P'(r)$, you need to allow for a constant of integration. If there is no externally applied pressure at the surface of the sphere, then $P(R) = 0$. Use this boundary condition to find the constant of integration.

The Maxwell stress tensor components $T_{ij}$ are not directly related to the material pressure $P(r)$. In electrostatics, the Maxwell tensor $\overleftrightarrow{T}$ has the following property. If $S$ is a closed surface, then $\large \oint_S \normalsize \overleftrightarrow{T} \cdot \vec{dA}$ equals the net electric force on the total charge enclosed by $S$. Here, $\vec {dA}$ represents an outward pointing element of area of $S$. For example, if you let $S$ be the surface of the circular disk used in the discussion above, then evaluation of $\large \oint_S \normalsize \overleftrightarrow{T} \cdot \vec{dA}$ gives the result $\rho dV E(r)$ as expected.