Answer:
option A
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
[tex]V = \dfrac{1}{4\pi \epsilon_0}\dfrac{q}{r}[/tex]
[tex]\dfrac{q}{V} = {4\pi \epsilon_0}{r}[/tex]
[tex]C = \dfrac{1}{9\times 10^9}\times\dfrac{2.54\times 10^{-2}}{2}[/tex]
C = 0.1411 × 10⁻¹¹ Farad
C = 1.411 × 10⁻¹² Farad
C = 1.4 pF
Hence, correct answer is option A
