Answer:
r = 9.92 mm
Explanation:
Given that,
Mass of oil drop, [tex]m=2\times 10^{-15}\ kg[/tex]
It acquires 2 surplus electrons, q = +2e [tex]=3.2\times 10^{-19}\ C[/tex]
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)
[tex]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[/tex]
So, the distance between the plates is 9.92 mm.
