1/2Eo times ( (volume integral)E^2 dtau + (surface integral)E PHI da )
That's the equation, the first integral is over the voume of a surface charged (q) sphere of radius a > R (radius of sphere) so a Gaussian surface beyond the sphere and the second is over the surface, with E vector function and PHI scalr function are the field and potential of q.
I believe this sum would show the total electrostatic energy, since if a => infinity the surface integral goes to zero (as stated in my book) but I'm not 100% sure nor do I know how to show it is really more/less than the total if I'm wrong. If someone can tell me which it is and how to go about proving it I think I can take it from there, just don't know how to start. Thanks!
Yes, you are integrating energy density over volume. That is total energy.
The equation you provided represents the total electrostatic energy in a system. It is given by the sum of two integrals: the volume integral and the surface integral.
The volume integral is given by the expression: ∫ E^2 dτ, where E is the electric field and dτ represents a small volume element. This integral calculates the energy density, which is the energy per unit volume, and is integrated over the entire volume of a charged sphere of radius a.
The surface integral is given by the expression: ∫ E * φ da, where E is the electric field, φ is the electric potential, and da represents a small area element on the surface of the charged sphere. This integral calculates the flux of the electric field through the surface of the sphere and is integrated over the entire surface.
To understand why this equation represents the total electrostatic energy, let's go through the derivation.
We start with the electric field, E, which can be expressed as the gradient of the electric potential, φ: E = -∇φ.
The energy density, u, can be obtained by multiplying the electric field by the electric displacement, D: u = (1/2) * E * D.
Since D = ε0 * E, where ε0 is the permittivity of vacuum, we can write the energy density as: u = (ε0/2) * E^2.
Now, integrate this energy density over the volume, using the volume integral, we have: Uv = ∫ u dτ = (ε0/2) * ∫ E^2 dτ. This represents the energy stored in the volume of the charged sphere.
Next, we consider the energy stored in the electric field outside the charged sphere, represented by the surface integral.
Using Gauss's theorem, the surface integral can be converted into a volume integral: ∫ E * φ da = ∫ ∇ • (E * φ) dτ.
Applying the divergence theorem to the above equation, we get: ∫ E * φ da = ∫ (E • ∇φ) dτ.
Since E = -∇φ, we can simplify further: ∫ E * φ da = -∫ E^2 dτ.
Therefore, the surface integral can be written as -∫ E^2 dτ.
Adding the volume integral and the surface integral together, we obtain the total electrostatic energy: U = Uv + Us = (ε0/2) * ∫ E^2 dτ - ∫ E^2 dτ.
Simplifying this equation, we get: U = (1/2ε0) * ∫ E^2 dτ.
From this derivation, we can conclude that the total electrostatic energy is given by the expression you provided: 1/2ε0 times the volume integral of E^2 over the charged sphere.
To address your question about infinity, if the radius of the charged sphere, a, tends to infinity, the surface integral (∫ E * φ da) will indeed approach zero. This is because as the sphere becomes infinitely large, the electric field and potential become negligible at the boundaries, resulting in a zero flux through the surface. Therefore, in this limit, the total electrostatic energy is solely determined by the volume integral (∫ E^2 dτ).
I hope this explanation helps clarify the concept of electrostatic energy and how it is calculated in the given equation.