What Potential Should Be Used for Energy Stored in a Charged Sphere?

Click For Summary
SUMMARY

The energy stored in a charged sphere can be calculated using the equation U = (1/2) ∫ φ(r) ρ(r) dV, where φ(r) represents the electrostatic potential. It is essential to use the potential inside the sphere, as outside the sphere ρ(r) equals zero. The alternative expression U = (ε/2) ∫ E² dV is also valid, and both methods yield the same result when applying Gauss's theorem and the relationship E = -∇φ. The constant in the potential can be fixed by setting U = 0 for a homogeneously charged sphere.

PREREQUISITES
  • Understanding of electrostatics and electric fields
  • Familiarity with the Legendre equation and its solutions
  • Knowledge of Gauss's theorem in three dimensions
  • Basic calculus for evaluating integrals
NEXT STEPS
  • Study the derivation of the Legendre equation in spherical coordinates
  • Learn about the application of Gauss's theorem in electrostatics
  • Explore the relationship between electric fields and potentials in electrostatics
  • Investigate methods for calculating energy stored in different charge distributions
USEFUL FOR

Students and professionals in physics, particularly those focusing on electrostatics, electrical engineering, and anyone involved in theoretical physics or applied mathematics related to electric fields and potentials.

Kosta1234
Messages
45
Reaction score
1
Homework Statement
Energy stored in a charged sphere
Relevant Equations
$$ U = \frac {1}{2} \cdot \int \phi (r) \cdot \rho(r) dV $$
Hi.
When I am asked to figure out the Energy stored in a charged sphere and I use this equation: ## U = \frac {1}{2} \cdot \int \phi (r) \cdot \rho(r) dV ##
what is the potential ## \phi ( r) ## stands for? I tried to use the potential inside the sphere, because out side of the sphere ## \rho (r) = 0 ##, and I tried to sum those to up.
I'm not getting the same answer as in this equation:
$$ U = \frac {\varepsilon }{2} \cdot \int_{all space} E^2 dV $$so what ## \phi (r) ## I've to use and why?Edit: I got it right, I think.
was I right when I said to use the potential inside the sphere, because out side of the sphere ## \rho (r) = 0 ##
and to use the ## U = \frac {\varepsilon }{2} \cdot \int_{all space} E^2 dV ## to all space?
 
Last edited:
Physics news on Phys.org
It's the electrostatic potential you should use, i.e., the solution of the Legendre equation,
$$\Delta \phi(\vec{x})=-\frac{1}{\epsilon} \rho(\vec{x}).$$
This potential is defined up to a constant, and so is ##U##.

It's simple to show that (up to a constant) your two expressions deliver the same result. To see this, just use
$$\vec{E}=-\vec{\nabla} \phi$$
in one of the factors in the integral
$$\tilde{U}=\frac{\epsilon}{2} \int_{\mathbb{R}^3} \mathrm{d}^3 x \vec{E}^2(\vec{x}),$$
and then Gauss's theorem for 3D partial integration! It's a good exercise!

BTW: You can fix the constant by demanding that ##U=0## for ##\rho=Q/V## (where ##Q## is the total charge, and ##V## the volume of the sphere; I guess you mean a homogeneously charged sphere).

You can also find the exact solution of this problem by solving the Legendre equation in shperical coordinates.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
720
  • · Replies 2 ·
Replies
2
Views
4K
Replies
11
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
11K
Replies
13
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K