Calculate the voltage required to balanced an oil drop carrying 10 electrons, when located between plates of a capacitor, which are 5mm apart. Given mass of drop `= 3xx10^(-16) kg`, charge on electron `= 1.6xx10^(-19)C and g = 9.8 m//s^(2)`.

Answer

9.2V

Explanation

Here, `V = ?, n = 10, d = 5 mm = 5xx10^(-3) m`,
`m = 3xx10^(-16) kg, e = 1.6xx10^(-19)C`,
`g = 9.8m//s^(2)`
To balance the oil drop
`F = qE = mg or q((V)/(d)) = mg`
`:. V = (mgd)/(q) = (mgd)/(n e) = (3xx10^(-16)xx9.8xx5xx10^(-3))/(10xx1.6xx10^(-19))`
= 9.2 volt