Consider an infinitely long solid cylinder with radius R_0 and volume charge density rho=rho_0*r(r≤R_0) where rho_0 is a constant. Use first principles to determine the electric field E(vector) for r<R_0.
To determine the electric field E(vector) for r < R_0, we can use Gauss's law. Gauss's law states that the electric flux through any closed surface is proportional to the charge enclosed by that surface.
Let's consider a Gaussian surface in the form of a cylindrical shell with radius r and length L inside the cylinder. The selected Gaussian surface should be symmetric with respect to the charge distribution to simplify the calculations.
The electric field E through this Gaussian surface must also be radially symmetric. Therefore, we can expect E to have a magnitude E(r) and be directed along the radial direction outward.
Now, let's calculate the total charge enclosed within this Gaussian surface. The volume charge density is given by:
rho = rho_0 * r (for r ≤ R_0)
The charge enclosed within the Gaussian surface is:
Q_enclosed = ∫rho dV
Since the charge density is only a function of r, we can simplify this integral using cylindrical coordinates:
Q_enclosed = ∫[0 to L] ∫[0 to 2π] ∫[0 to r] rho dρ dθ dz
= ∫[0 to L] L ∫[0 to 2π] ∫[0 to r] rho_0 * ρ dρ dθ dz
= rho_0 L ∫[0 to L] ∫[0 to 2π] (1/2) r^2 dθ dz
= rho_0 L ∫[0 to L] (1/2) r^2 * 2π dz
= rho_0 L * π * ∫[0 to L] r^2 dz
= rho_0 L * π * [(1/3) L^3]
= (1/3) rho_0 π L^4
Now, according to Gauss's law, the electric flux Φ through the Gaussian surface is:
Φ = Q_enclosed / ε_0
where ε_0 is the electric constant (permittivity of free space). In SI units, ε_0 ≈ 8.85 x 10^-12 C^2/(Nm^2).
Since the electric field E is radially symmetric, the flux Φ is given by:
Φ = E(r) * A
where A is the area of the cylindrical surface. The area A is given by:
A = 2π r L
Equating the two expressions for Φ, we have:
E(r) * A = Q_enclosed / ε_0
Substituting the values of A and Q_enclosed, we get:
E(r) * 2π r L = (1/3) rho_0 π L^4 / ε_0
Simplifying further, we find:
E(r) = (rho_0 / (6 ε_0)) r^2
Therefore, the electric field E(vector) for r < R_0 is given by:
E(vector) = (rho_0 / (6 ε_0)) r^2 * r̂
where r̂ is the unit vector in the radial direction.
To find the electric field inside the cylinder for r < R_0, we can use Gauss's Law. Gauss's Law states that the flux of the electric field through a closed surface is equal to the enclosed charge divided by the permittivity of the medium.
Let's consider a Gaussian surface in the form of a cylindrical region with radius r and length L, where r < R_0. The top and bottom surfaces of the cylinder are closed, while the curved surface is open. The electric field should be independent of the length L, so we can let L approach infinity.
According to Gauss's Law, the flux of the electric field through the closed Gaussian surface is given by:
∮ E⋅dA = (1/ε₀)∫ρ dV
Where ε₀ is the vacuum permittivity, E is the electric field, dA is an infinitesimal area on the Gaussian surface, ρ is the charge density, and dV is an infinitesimal volume element enclosed by the Gaussian surface.
Since the electric field is pointing radially outward and is constant in magnitude across the Gaussian surface, we can write:
∮ E⋅dA = E ∮ dA = E (2πrL)
The enclosed charge is given by:
Q_enclosed = ∫ρ dV
Substituting the expressions for the flux and the enclosed charge into Gauss's Law, we have:
E (2πrL) = (1/ε₀) ∫ρ dV
Now, we need to find an expression for the enclosed charge. The charge density is given by ρ = ρ₀ * r, where ρ₀ is a constant. Hence, the enclosed charge becomes:
Q_enclosed = ∫ρ₀ * r dV
We can express the volume element dV as the product of length L and an infinitesimal cross-sectional area dA:
Q_enclosed = ∫ρ₀ * r L dA
Since L is a constant that can be taken out of the integral:
Q_enclosed = ρ₀ * L ∫r dA
The integral of r dA gives us the total enclosed area of the Gaussian surface, which is πr²:
Q_enclosed = ρ₀ * L * πr²
Substituting this result into Gauss's Law, we have:
E (2πrL) = (1/ε₀) ρ₀ * L * πr²
Simplifying the expression:
E (2πr) = (1/ε₀) ρ₀ * πr²
Therefore, the electric field inside the cylinder for r < R_0 is given by:
E = (1/2ε₀) ρ₀ * r
Hence, the electric field is directly proportional to the radial distance from the axis of the cylinder and is directed radially outward.