EEI

Tutorial 2: Worked solutions

Electrical Engineering I · Module 92315 · Tutorial 2

These solutions build total charge by integrating the volume charge density over the region where it exists. Track the coordinate-dependent volume element and the density’s support.

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Task 4: Cylinder coordinates

The space charge density is constant, ϱ_V = ϱ₀, in the hollow cylindrical region a ≤ ρ ≤ b, 0 ≤ φ ≤ 2π, 0 ≤ z ≤ c, and zero outside. Assume 0 ≤ a < b and c > 0.

Connected topic: Charge densities (§1.6): the total charge is the volume integral of the density. The cylindrical volume element is in Appendix A.5.

Only the wall is charged. The density ϱ₀ fills the region between the radii a and b (shaded on top) over the height c; the bore inside radius a is empty, so the radial integral starts at a.

4: Total charge in the hollow cylinder

Try first: write the cylindrical volume element, then integrate across the inner and outer radii, the full angle, and the height.

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4 · Cylindrical volume integral

Total charge is the volume integral of volume charge density over the charged region. In cylindrical coordinates, a small volume is a wedge-shaped piece with volume ρ dρ dφ dz:
Q = ∭_V ϱ_V dV, dV = ρ dρ dφ dz
Substitute the constant density and the stated limits. The inner radius a excludes the empty central cylinder:
Q = ϱ₀ ∫₀^c ∫₀^2π ∫_a^b ρ dρ dφ dz
Q = ϱ₀ ∫₀^c ∫₀^2π [ρ²2]_a^b dφ dz
Evaluate the radial term, then the full revolution and cylinder height:
Q = ϱ₀ b² − a²2 (2π) c
Q = ϱ₀ π(b² − a²)c
The sign follows the sign of ϱ₀. Check: the factor π(b² − a²)c is exactly the hollow cylinder’s volume, and ϱ₀ times volume has units C.

Task 5: Spherical coordinates

The density varies radially as ϱ_V(r) = ϱ₀r²/a² inside a sphere of radius a and is zero for r > a. Assume a > 0.

Connected topic: Charge densities (§1.6): each thin spherical shell contributes its density times its volume. The spherical volume element is in Appendix A.5.

Add up thin shells. The shell at radius r and thickness dr (shaded) has volume 4πr² dr and density ϱ₀r²/a²; the outer shells are both larger and denser.

5: Total charge in the sphere

Try first: multiply the radial density by the spherical shell volume 4πr²dr and integrate from the centre to a.

Attempted it? Open the worked steps

5 · Spherical volume integral

For a spherically symmetric density, integrate over concentric shells. A shell at radius r has volume 4πr²dr, and only radii 0 through a contain charge:
Q = ∭_V ϱ_V dV = 4π ∫₀^a ϱ_V(r) r² dr
Insert the given density. One factor r² comes from the shell area and another from ϱ_V:
Q = 4π ∫₀^a ϱ₀ r²a² r² dr
Q = 4πϱ₀a² ∫₀^a r⁴ dr
Integrate r⁴ and apply both limits:
Q = 4πϱ₀a² [r⁵5]₀^a
Q = 4πϱ₀ a³5
The sign follows ϱ₀. Check: the average density is 3ϱ₀/5, giving (3ϱ₀/5)(4πa³/3), the same result.

Personal worked-solution guide based on the supplied EE1 Tutorial Task sheet, written in my own words; no tutorial text or figures are reproduced. Not an official FAU publication; check course details against current material. Made by Thanate Amornjiraporn (Focus).