Answer:
option A
Step-by-step explanation:
steel in a conducting ball of diameter
2 r = 1 '' = 2.54 cm = 2.54 × 10⁻² m
Potential due to conducting steel ball



C = 0.1411 × 10⁻¹¹ Farad
C = 1.411 × 10⁻¹² Farad
C = 1.4 pF
Hence, correct answer is option A