An oil drop with a charge of 2x10-17C is suspended between 2 parallel plates 5cm apart. The
potential difference between the plates is 10kV. Calculate the mass of the oil drop and the
number of excess electrons it is carrying.

1) Mass:

2) Electrons: 125

Explanation:

1)

The electric force exerted on the oild drop is given by

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

where

V is the potential difference

d is the separation between the plates

So the electric force is

(1)

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

(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,

Here we have

is the charge of the oil drop

is the potential difference

is the separation between the plates

is the acceleration due to gravity

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

2)

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

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

where

N is the number of excess electrons

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:

So, 125 excess electrons.

12000j is work done

c) up is the direction the magnetic force is acting on the charge.