Question

In a milkans apparatus,an oil drop of weight 2.0×10^-15kg accquires two surplus electrons. when a Potential difference of 620 volt is applied between the pair of horizontal metal plates the drop is in equilibrium. find the distance between plates.​

Answer 1

r = 9.92 mm

Step-by-step explanation:

Given that,

Mass of oil drop,
`m=2* 10^(-15)\ kg`

It acquires 2 surplus electrons, q = +2e
`=3.2* 10^(-19)\ C`

Potential difference, V = 620 V

Thie potential difference is applied between the pair of horizontal metal plates the drop is in equilibrium.

We need to find the distance between the plates.

At equilibrium,

mg = qE

Since, E = V/r (r is distance between plates)

`mg=(qV)/(r)\\\\r=(qV)/(mg)\\\\r=(3.2* 10^(-19)* 620)/(2* 10^(-15)* 10)\\\\=0.00992\ m\\\\=9.92\ mm`

So, the distance between the plates is 9.92 mm.