Calculate the voltage needed to balance on all drop carrying 10 electrons when located between the plates of a capacitor which are 5 mm apart. Mass of oil drop is `3xx10^(-16) kg `(take `g = 10m//s^(2))`

Explanation

Here, `q =n e = 10xx1.6xx10^(-19)`
`= 1.6xx10^(-18)C`
`m = 3xx10^(-16) kg r = 5 mm = 5xx10^(-3)m`
`E = (V)/(r ) = (V)/(5xx10^(-3)) Vm^(-1)`
For drop to remain stationary, mg = qE
`3xx10^(-16)xx10 = 1.6xx10^(-18) xx (V)/(5xx10^(-3))`
`V = (3xx10^(-16)xx10xx5xx10^(-3))/(1.6xx10^(-18)) = 9.47V`