Question

An oil drop with a charge of 2x10-17C is suspended between 2 parallel plates 5cm apart. The

Answer 1

1) Mass: $4.1\cdot 10^{-13}kg$

2) Electrons: 125

Explanation:

1)

The electric force exerted on the oild drop is given by

$F=qE$

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

$E=\frac{V}{d}$

where

V is the potential difference

d is the separation between the plates

So the electric force is

$F=\frac{qV}{d}$ (1)

On the other hand, the gravitational force on the oil drop is

$F=mg$ (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,

$\frac{qV}{d}=mg$

Here we have

$q=2\cdot 10^{-17}C$ is the charge of the oil drop

$V=10 kV=10000 V$ is the potential difference

$d=5 cm = 0.05 m$ is the separation between the plates

$g=9.8 m/s^2$ is the acceleration due to gravity

Solving for m, we find the mass of the oil drop:

$m=\frac{qV}{dg}=\frac{(2\cdot 10^{-17})(10000)}{(0.05)(9.8)}=4.1\cdot 10^{-13}kg$

2)

From the text of the problem, we know that the net charge on the oil drop is

$Q=-2\cdot 10^{-17}C$

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

$Q=Ne$

where

N is the number of excess electrons

$e=-1.6\cdot 10^{-19}C$ 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:

$N=\frac{Q}{e}=\frac{-2\cdot 10^{-17}}{-1.6\cdot 10^{-19}}=125$

So, 125 excess electrons.