Two metallic conducors have net charge of `+70 pC and -70 pC`, which result in a potential difference of `20 V` between them. What is the capacitanace of the system ?

Answer

`3.5 pF`

Explanation

Here, `q = 70 pC = 70xx^(-12)C, V = 20 V`,
`C = (q)/(V) = (70xx10^(-12))/(20) = 3.5xx10^(-12) F = 3.5 pF`