1) Mass: [tex]4.1\cdot 10^{-13}kg[/tex]
2) Electrons: 125
Explanation:
1)
The electric force exerted on the oild drop is given by
[tex]F=qE[/tex]
where
q is the charge on the oil drop
E is the magnitude of the electric field
The electric field between two parallel plates can be written as
[tex]E=\frac{V}{d}[/tex]
where
V is the potential difference
d is the separation between the plates
So the electric force is
[tex]F=\frac{qV}{d}[/tex] (1)
On the other hand, the gravitational force on the oil drop is
[tex]F=mg[/tex] (2)
where
m is the mass of the drop
g is the acceleration due to gravity
The two forces have opposite directions (electric force: upward, gravity: downward), so the oil drop remains in equilibrium if the two forces have same magnitude. So,
[tex]\frac{qV}{d}=mg[/tex]
Here we have
[tex]q=2\cdot 10^{-17}C[/tex] is the charge of the oil drop
[tex]V=10 kV=10000 V[/tex] is the potential difference
[tex]d=5 cm = 0.05 m[/tex] is the separation between the plates
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
Solving for m, we find the mass of the oil drop:
[tex]m=\frac{qV}{dg}=\frac{(2\cdot 10^{-17})(10000)}{(0.05)(9.8)}=4.1\cdot 10^{-13}kg[/tex]
2)
From the text of the problem, we know that the net charge on the oil drop is
[tex]Q=-2\cdot 10^{-17}C[/tex]
Where the charge is negative since it is due to an excess of electrons (which are negatively charged).
The net charge on the oil drop can be written as
[tex]Q=Ne[/tex]
where
N is the number of excess electrons
[tex]e=-1.6\cdot 10^{-19}C[/tex] is the charge on one electron (the fundamental charge)
Therefore, here we can solve the formula for N, to find the number of excess electrons on the oil drop:
[tex]N=\frac{Q}{e}=\frac{-2\cdot 10^{-17}}{-1.6\cdot 10^{-19}}=125[/tex]
So, 125 excess electrons.
