- #1
Rob2024
- 11
- 1
- 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!
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!