The Capacitance of Coaxial Iron Cylinders

Calculating Capacitance and Applied Potential Difference

A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is negatively charged and the outer is positively charged; the magnitude of the charge on each is 11.5 pC. The inner cylinder has a radius of 0.550 mm, the outer one has a radius of 7.20 mm, and the length of each cylinder is 13.0 cm.

(a) What is the capacitance?

Capacitance of coaxial cylinder can be determined using below expression:

C = (2π ε₀ L) / Ln[rb / ra]

Where:

ε₀ = permittivity of free space = 8.85 × 10^-12 Fm^-1

L = length of each cylinder = 13.0 cm = 13 × 10^-2 m

rb = radius of outer cylinder = 7.20 mm = 7.20 × 10^-3 m

ra = radius of inner cylinder = 0.550 mm = 0.550 × 10^-3 m

If we substitute the values we have:

C = (2π ε₀ L) / Ln[rb / ra]

C = (2 × π × 8.85 × 10^-12 × 13 × 10^-2) / Ln[7.20 × 10^-3 / 0.550 × 10^-3]

C = (7.2288 × 10^-12) / 2.5729

C = 2.811 × 10^-12 F

(b) What applied potential difference is necessary to produce these charges on the cylinders?

Vba = Q / C

Where:

Vba = potential difference

Q = charge on each = 11.5 pC = 11.5 × 10^-12 C

C = capacitance = 2.811 × 10^-12 F

If we substitute the values we have:

Vba = (11.5 × 10^-12) / 2.811 × 10^-12

Vba = 4.09V

Do you understand how to calculate the capacitance of coaxial iron cylinders and the applied potential difference?

Yes, I understand the formula and calculations for determining the capacitance of coaxial cylinders and the necessary potential difference based on the given charges.