Respuesta :
Answer:
(a) [tex]F_{G} = 5.51\times 10^{- 15} N[/tex]
(b) 54 electrons in excess
Solution:
As per the question:
Diameter of the spherical drop, d = [tex]1.5\mu m = 1.5\times 10^{- 6}m[/tex]
Magnitude of the Electric field, E = 640 N/C
Now,
(a) The magnitude of the gravitational force on the spherical drop is given by using Newton's second law:
[tex]F_{G} = mg[/tex] (1)
where
m = mass of the sphere
g = acceleration due to gravity
Also,
[tex]m = \rho V[/tex] (2)
where
[tex]\rho[/tex] = density of the spherical water drop = [tex]1000 kgm^{- 3}[/tex]
V = volume of the sphere = [tex]\frac{4}{3}\pi (\frac{d}{2})^{3}[/tex]
Now, eqn (2) becomes:
[tex]m = \rho \frac{4}{3}\pi (\frac{d}{2})^{3}[/tex]
Thus eqn (1) will be:
[tex]F_{G} = \rho \frac{4}{3}\pi (\frac{d}{2})^{3}g[/tex]
[tex]F_{G} = 1000\times \frac{4}{3}\pi (\frac{1.5\times 10^{- 6}}{2})^{3}\times 9.8[/tex]
[tex]F_{G} = 5.51\times 10^{- 15} N[/tex]
(b) No. of excess electron:
Now, the spherical drop is suspended in the air, both the electrostatic and gravitational forces on the drop must be equal:
Electrostatic force, [tex]F_{E} = QE = neE[/tex]
where
Q = charge = ne
n = no. of electrons
e = electronic charge
Now,
[tex]F_{G} = F_{E}[/tex]
[tex]5.51\times 10^{- 15} = neE[/tex]
[tex]5.51\times 10^{- 15} = n\times 1.6\times 10^{- 19}\times 640[/tex]
[tex]n = {5.51\times 10^{- 15}}{1.6\times 10^{- 19}\times 640}[/tex]
n = 53.8 ≈ 54 electrons
