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

Answer:

r = 9.92 mm

Explanation:

Given that,

Mass of oil drop, $m=2\times 10^{-15}\ kg$

It acquires 2 surplus electrons, q = +2e $=3.2\times 10^{-19}\ C$

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)

$mg=\dfrac{qV}{r}\\\\r=\dfrac{qV}{mg}\\\\r=\dfrac{3.2\times 10^{-19}\times 620}{2\times 10^{-15}\times 10}\\\\=0.00992\ m\\\\=9.92\ mm$

So, the distance between the plates is 9.92 mm.