# A Capacitor Made from Two Hollow, Coaxial Iron Cylinders

## Calculating the Capacitance of the Capacitor

**Part A: What is the capacitance?**Using the formula C = Q / ΔV and the given values: - Inner radius (a): 0.350 mm = 0.00035 m - Outer radius (b): 7.20 mm = 0.0072 m - Charge (Q): 18.5 pC = 18.5 x 10^-12 C - Length (L): 17.0 cm = 0.17 m - Permittivity of free space (ε₀): 8.854 x 10^-12 F/m First, we need to calculate the potential difference (ΔV) between the cylinders: The electric field in the region r_inner < r < r_outer is given by E = Q / (2πε₀r^2). Then, using ∆V = 2k(Q/L) ln(b/a), where k is the Coulomb constant, we can find the potential difference. Finally, the capacitance C is calculated as C = L / (2k ln(b/a)). Plugging in the values, we get: C = 0.17 / (2 * 9 * 10^9 * ln(0.0072 / 0.00035)) C ≈ 3.12 * 10^-12 F or 3.12 pF Therefore, the capacitance of the capacitor made from two hollow, coaxial, iron cylinders is approximately 3.12 pF.

What are the key components involved in calculating the capacitance of the capacitor made from two hollow, coaxial, iron cylinders?

The key components involved in calculating the capacitance of the capacitor are the charge on each cylinder, the radii of the inner and outer cylinders, the length of the cylinders, and the permittivity of free space. Additionally, the formula for capacitance (C = Q / ΔV) and the formula for potential difference between the cylinders (ΔV = 2k(Q/L) ln(b/a)) are crucial in determining the capacitance value.