Electrostatics - Proof of Gauss's Law
1 Answer
Gauss's Law is a fundamental principle in electrostatics that relates the electric flux through a closed surface to the total electric charge enclosed by that surface. It is a mathematical statement that helps us understand the behavior of electric fields and charges in various situations.
The formal proof of Gauss's Law involves using the concept of the divergence theorem, which relates the flux of a vector field through a closed surface to the volume integral of its divergence over the volume enclosed by that surface. Here's a sketch of the proof:
Step 1: Definition of Electric Flux
Consider a closed surface S enclosing a volume V. The electric flux ΦE through S due to an electric field E is defined as the surface integral of the dot product of E and the differential area vector dA over the closed surface:
ΦE = ∬S E ⋅ dA
Step 2: Divergence Theorem
The divergence theorem relates the flux of a vector field through a closed surface to the volume integral of its divergence over the enclosed volume. Mathematically, it can be expressed as:
∬S F ⋅ dA = ∭V (div F) dV
where F is a vector field, div F is its divergence, and the integrals are over the closed surface S and the enclosed volume V, respectively.
Step 3: Applying the Divergence Theorem to Electric Flux
Consider a point charge q located at the origin within the enclosed volume V. The electric field due to this point charge, at a distance r from the origin, is given by Coulomb's law:
E = k * q / r^2 * r̂
where k is Coulomb's constant and r̂ is the unit vector pointing from the origin to the observation point on the closed surface.
The divergence of the electric field is given by:
div E = ∇ ⋅ E = (1/r^2) * d(r^2E)/dr = 4πk * δ(r)
where δ(r) is the Dirac delta function.
Step 4: Applying the Divergence Theorem
Now, applying the divergence theorem to the electric field E, we have:
∬S E ⋅ dA = ∭V (div E) dV = ∭V (4πk * δ(r)) dV
Since the Dirac delta function only contributes when the point charge is within the volume V, the volume integral becomes:
∭V (4πk * δ(r)) dV = 4πk * q
Step 5: Final Step
Substitute the result back into the electric flux formula:
ΦE = ∬S E ⋅ dA = 4πk * q
This is the final expression for the electric flux through the closed surface S due to the point charge q at the origin.
Finally, since Gauss's Law states that the electric flux through a closed surface is proportional to the total enclosed charge, we can write:
ΦE = 4πk * q_enclosed
Comparing this with the expression for electric flux, we obtain Gauss's Law:
∮E ⋅ dA = 4πk * q_enclosed
where the integral is taken over a closed surface enclosing a volume with total charge q_enclosed.
Keep in mind that this is a simplified sketch of the proof, and the actual mathematical derivations might involve more rigorous mathematical steps. The key idea is to relate the electric flux through a closed surface to the divergence of the electric field and then apply the divergence theorem to establish the connection between the flux and the enclosed charge.
The formal proof of Gauss's Law involves using the concept of the divergence theorem, which relates the flux of a vector field through a closed surface to the volume integral of its divergence over the volume enclosed by that surface. Here's a sketch of the proof:
Step 1: Definition of Electric Flux
Consider a closed surface S enclosing a volume V. The electric flux ΦE through S due to an electric field E is defined as the surface integral of the dot product of E and the differential area vector dA over the closed surface:
ΦE = ∬S E ⋅ dA
Step 2: Divergence Theorem
The divergence theorem relates the flux of a vector field through a closed surface to the volume integral of its divergence over the enclosed volume. Mathematically, it can be expressed as:
∬S F ⋅ dA = ∭V (div F) dV
where F is a vector field, div F is its divergence, and the integrals are over the closed surface S and the enclosed volume V, respectively.
Step 3: Applying the Divergence Theorem to Electric Flux
Consider a point charge q located at the origin within the enclosed volume V. The electric field due to this point charge, at a distance r from the origin, is given by Coulomb's law:
E = k * q / r^2 * r̂
where k is Coulomb's constant and r̂ is the unit vector pointing from the origin to the observation point on the closed surface.
The divergence of the electric field is given by:
div E = ∇ ⋅ E = (1/r^2) * d(r^2E)/dr = 4πk * δ(r)
where δ(r) is the Dirac delta function.
Step 4: Applying the Divergence Theorem
Now, applying the divergence theorem to the electric field E, we have:
∬S E ⋅ dA = ∭V (div E) dV = ∭V (4πk * δ(r)) dV
Since the Dirac delta function only contributes when the point charge is within the volume V, the volume integral becomes:
∭V (4πk * δ(r)) dV = 4πk * q
Step 5: Final Step
Substitute the result back into the electric flux formula:
ΦE = ∬S E ⋅ dA = 4πk * q
This is the final expression for the electric flux through the closed surface S due to the point charge q at the origin.
Finally, since Gauss's Law states that the electric flux through a closed surface is proportional to the total enclosed charge, we can write:
ΦE = 4πk * q_enclosed
Comparing this with the expression for electric flux, we obtain Gauss's Law:
∮E ⋅ dA = 4πk * q_enclosed
where the integral is taken over a closed surface enclosing a volume with total charge q_enclosed.
Keep in mind that this is a simplified sketch of the proof, and the actual mathematical derivations might involve more rigorous mathematical steps. The key idea is to relate the electric flux through a closed surface to the divergence of the electric field and then apply the divergence theorem to establish the connection between the flux and the enclosed charge.