Capacitance of a coaxial cable with length L, inner radius a, outer radius b, in vacuum is:

Capacitance of a coaxial cable with length L, inner radius a, outer radius b, in vacuum is:

• A. C = (2π ε0 L)/ln(b/a)
• B. C = (2π ε0 L) ln(b/a)
• C. C = (2π ε0 L)/(b − a)
• D. C = (π ε0 L)/ln(b/a)

Answer: (A)

When a coaxial cable is viewed as a cylindrical capacitor, the electric field between the conductors falls as 1/r. By Gauss’s law, the field at a radius r (between a and b) is E(r) = λ/(2π ε0 r), where λ is the charge per unit length on the inner conductor. The potential difference between the inner and outer conductors is the integral of E from a to b: V = ∫_a^b E(r) dr = (λ/(2π ε0)) ln(b/a). The total charge on a length L is Q = λL, so the capacitance C = Q/V becomes C = (λL) / [ (λ/(2π ε0)) ln(b/a) ] = (2π ε0 L) / ln(b/a). Thus the correct expression is C = (2π ε0 L)/ln(b/a). The other forms don’t fit because they either replace the 1/r field integral with a linear (b−a) term, miss the 2π factor, or place the logarithm in the numerator—none of which matches the cylindrical geometry and the 1/r dependence, which leads to the ln(b/a) in the denominator.

When a coaxial cable is viewed as a cylindrical capacitor, the electric field between the conductors falls as 1/r. By Gauss’s law, the field at a radius r (between a and b) is E(r) = λ/(2π ε0 r), where λ is the charge per unit length on the inner conductor.

The potential difference between the inner and outer conductors is the integral of E from a to b:

V = ∫_a^b E(r) dr = (λ/(2π ε0)) ln(b/a).

The total charge on a length L is Q = λL, so the capacitance C = Q/V becomes

C = (λL) / [ (λ/(2π ε0)) ln(b/a) ] = (2π ε0 L) / ln(b/a).

Thus the correct expression is C = (2π ε0 L)/ln(b/a).

The other forms don’t fit because they either replace the 1/r field integral with a linear (b−a) term, miss the 2π factor, or place the logarithm in the numerator—none of which matches the cylindrical geometry and the 1/r dependence, which leads to the ln(b/a) in the denominator.