Electrostatic Energy of a solid sphere with a cavity

Click For Summary
SUMMARY

The discussion focuses on calculating the electrostatic energy of a solid sphere with a cavity using the formula for energy density, specifically ##W = \frac{ε_0}{2} \int E^2 dτ##. The electric field ##E## is derived as ##E = \frac{(R^3 - b^3)\rho}{3ε_0 r^2}## for the region outside the sphere (##r > R##). The user seeks clarification on the applicability of this expression across three distinct regions: inside the cavity (##r < b##), between the cavity and the sphere (##b < r < R##), and outside the sphere (##r > R##). The user also attempts to compute the total energy ##U##, incorporating the electric field expressions for each region.

PREREQUISITES
  • Understanding of electrostatics and electric fields
  • Familiarity with integral calculus in physics
  • Knowledge of charge density and its implications
  • Concept of energy density in electrostatics
NEXT STEPS
  • Study the derivation of electric fields in spherical coordinates
  • Learn about boundary conditions in electrostatics
  • Explore the concept of energy density in electromagnetic fields
  • Investigate the implications of charge density variations in electrostatic systems
USEFUL FOR

Students and professionals in physics, particularly those specializing in electrostatics, electrical engineering, and anyone involved in theoretical physics calculations related to electric fields and energy.

Arman777
Insights Author
Gold Member
Messages
2,163
Reaction score
191
Homework Statement
Determine the electrostatic energy U stored in the sphere
Relevant Equations
##W = ε_0/2 \int E^2d\tau## for all space
1571252938043.png

I tried to use ##W = ε_0/2 \int E^2d\tau## for all space. So I find that ##E = \frac{(R^3 - b^3)\rho}{3ε_0r^2}## where ##\rho## is the charge denisty. So from here when I plug the equation I get something like

$$W = \frac{(R^3 - b^3)^2\rho^2 4 \ pi}{18ε_0} \int_{?}^{\inf}1/r^2dr$$

Is this approach correct ?

At this point I get stuck for the boundary conditions. If I put R I get something meaningless
 
  • Like
Likes   Reactions: Delta2
Physics news on Phys.org
Arman777 said:
I tried to use ##W = ε_0/2 \int E^2d\tau## for all space.
OK
So I find that ##E = \frac{(R^3 - b^3)\rho}{3ε_0r^2}## where ##\rho## is the charge denisty.
For what region of space is this expression for E applicable? You have three distinct regions to consider.
 
TSny said:
OK
For what region of space is this expression for E applicable? You have three distinct regions to consider.
This is for ##r>R## so I have to also find ##E(r<b)##, ##E(b<r<R)## ?

Then The U will be,

$$U = \frac{(R^3 - b^3)^2\rho^2 4 \ pi}{18ε_0} \int_{R}^{\inf}1/r^2dr + \frac{e_0}{2} \int_{b}^{R}\frac{\rho (r - b^3)}{3ε_0}dr$$

Where ##E = \frac{\rho (r - b^3)}{3ε_0}, (b<r<R)##
 
I think the correct expression for the E-field for ##b<r<R## is $$E=\frac{\rho(r^3-b^3)}{3\epsilon_0r^2}$$.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
1
Views
1K
Replies
11
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
7
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K