Respuesta :
Answer
Given,
diameter of oil droplet = 0.8 μ m
space between electrodes,d = 11 mm
density of oil = 885 kg/m³
mass of the oil droplet
mass = density x volume
[tex]m = 885\times (\dfrac{4}{3}\pi r^3)[/tex]
[tex]m = 885\times (\dfrac{4}{3}\pi (0.4\times 10^{-6})^3)[/tex]
[tex]m = 2.37\times 10^{-16}\ Kg[/tex]
Charge of the droplet = ?
Electric field expression, [tex]E = \dfrac{V}{d}[/tex]
[tex]E = \dfrac{32}{0.011}[/tex]
[tex]E = 2909.1\ V/m[/tex]
Using the equation of Force
q E = m g
[tex]q = \dfrac{m g}{E}[/tex]
[tex]q = \dfrac{2.37\times 10^{-16}\times 9.81}{2909.1}[/tex]
[tex]q = 7.99 \times 10^{-19}\ C[/tex]
charge of the droplet will be negative.
now,
we know,
q = n e
[tex]n = \dfrac{q}{e}[/tex]
[tex]n = \dfrac{-7.99\times 10^{-19}}{-1.6\times 10^{-19}}[/tex]
[tex]n = 5 [/tex]
The droplet has 5 surplus electron.
