ρ₀ = 20ε₀ for |z| ≤ 1m and ρ₀ = 0 elsewhere
ρ₀ = 40ε₀ in the entire region
ρ₀ = 10ε₀ for |z| ≤ 1m and ρ₀ = 0 elsewhere
ρ₀ = 20ε₀ in the entire region
Since, V(z) = 30z² for |z| ≤ 1m and V(z) = 35|z| for |z| ≥ 1m
That means electric field: E(z) = -dV/dz = -60z; |z| ≤ 1m and E(z) = -35; |z| ≥ 1m (considering z>0 for simplicity)
For |z| ≤ 1m, E(z) is proportional to z ⇒ continuous charge distribution for |z| ≤ 1m
|z| ≥ 1m, E(z) is constant. (Electric field due to an infinite sheet outside it)
Hence, consider a cylindrical Gaussian surface:
Applying Gauss's law for |z| ≤ 1m
∫E⋅dA = Q/ε₀
E × 2A = ρ × 2z × A / ε₀ ⇒ 60z × 2A = ρ × 2z × A / ε₀ ⇒ ρ = 60ε₀
However, this is incorrect because we have to consider only the charge inside the Gaussian surface which is not given directly by ρ x 2z x A. This solution is wrong.
Let's apply Gauss' law for |z| ≤ 1m
E(z) = -dV/dz = -60z for |z| ≤ 1m
Consider a cylindrical Gaussian surface of radius r and height 2z centered at z=0 with |z| ≤ 1m.
The flux through the surface is given by:
Φ = ∫ E⋅dA = E(z) × 2πr²
The charge enclosed is given by:
Q_enc = ρ₀ × 2πr²z
By Gauss's law:
Φ = Q_enc / ε₀
-60z × 2πr² = (ρ₀ × 2πr²z) / ε₀
ρ₀ = -60ε₀
The negative sign indicates the charge distribution is opposite the direction of the electric field which is only possible if the electric field is in the opposite direction.
For |z| ≥ 1m, E(z) is constant. This represents the field due to an infinite sheet of charge.
Applying Gauss's law for |z| ≥ 1m:
E × 2A = σ/ε₀, where σ is the surface charge density.
In this case the E field is discontinuous at |z|=1m, and Gauss' law shows ρ = 0 for |z| > 1m. The charge density is non-zero only within |z| ≤ 1m. Let's use the divergence theorem instead.
∇⋅E = ρ/ε₀
For |z| ≤ 1m, ∇⋅E = -60
For |z| > 1m, ∇⋅E = 0
Therefore, ρ₀ = -60ε₀ for |z| ≤ 1m and ρ₀ = 0 elsewhere. The options do not contain the correct answer (-60ε₀). There might be a mistake in the question or options provided.