Capacitance of a spherical capacitor with inner radius a and outer radius b is:

Capacitance of a spherical capacitor with inner radius a and outer radius b is:

• A. C = 4π ε0 ab/(b − a)
• B. C = 4π ε0 ab/(a − b)
• C. C = 4π ε0 (a + b)/(b − a)
• D. C = 4π ε0 ab/(b + a)

Answer: (A)

Capacitance is the ratio of charge to the potential difference between conductors. For a spherical capacitor, the inner sphere of radius a carries +Q and the outer shell of radius b carries -Q, so the electric field exists only between them (a < r < b). By Gauss’s law, the field in that region is radial with magnitude E(r) = (1 / (4π ε0)) (Q / r^2). The potential difference between the two conductors is found by integrating the field: V = ∫ from a to b E·dr = ∫_a^b (1 / (4π ε0)) (Q / r^2) dr = (Q / (4π ε0)) (1/a − 1/b) = (Q / (4π ε0)) (b − a) / (ab). The capacitance is C = Q / V, giving C = 4π ε0 ab / (b − a). This is positive because b > a. A useful check: as b → ∞, this reduces to C = 4π ε0 a, the capacitance of an isolated sphere. The other forms would not match the geometry or would yield illogical (negative) results for a < b.

Capacitance is the ratio of charge to the potential difference between conductors. For a spherical capacitor, the inner sphere of radius a carries +Q and the outer shell of radius b carries -Q, so the electric field exists only between them (a < r < b). By Gauss’s law, the field in that region is radial with magnitude E(r) = (1 / (4π ε0)) (Q / r^2).

The potential difference between the two conductors is found by integrating the field: V = ∫ from a to b E·dr = ∫_a^b (1 / (4π ε0)) (Q / r^2) dr = (Q / (4π ε0)) (1/a − 1/b) = (Q / (4π ε0)) (b − a) / (ab).

The capacitance is C = Q / V, giving C = 4π ε0 ab / (b − a). This is positive because b > a. A useful check: as b → ∞, this reduces to C = 4π ε0 a, the capacitance of an isolated sphere. The other forms would not match the geometry or would yield illogical (negative) results for a < b.