If the oil drop is stationary, the net force on it must be Zero or the resultant of all the forces on oil drop is zero.

There are two forces acting on the oil drop, its weight or force due to Earth’s gravity which is pulling it vertically downwards, and Electrostatic force which is acting in vertically upward direction.

Both forces should be equal in magnitude and opposite direction so that they cancel each other.

The arrangement is shown in the figure

Note: Electric Field is in vertically downward direction, because force on an negatively charged body in opposite direction of the field, so force on the drop is in vertically upward direction, which balances weight of the drop acting in vertically downward direction.

There are nine excess electrons which make the drop negatively charged because the electron is negatively charged, the net magnitude charge on a body is given by

q = n × E

n is the excess of electrons or protons on the body,

the charge of an electron is denoted by e

e = 1.60 × 10^–19 C

here since there are twelve excess electrons so

n = 12

i.e. q = 12 × 1.60 × 10^–19 C

= 1.92 × 10^-18 C

Now the magnitude electrostatic force on a charged particle held in an electric field is given by

F = q × E

where, F force is acting on a particle having charge q held in an electric field of magnitude E

here, charge on the oil drop is

q = 1.92 × 10^-18 C

magnitude of Electric field is

E = 2.55 × 10⁴NC^–1

So Electrostatic force on the oil drop is

F = 1.92 × 10^-18 C × 2.55 × 10⁴NC^–1

= 4.896 × 10^-14 N

This force is acting in vertically upward direction, so Weight should have same magnitude of Force and is acting in vertically downward direction.

Let us assume oil drop to be Spherical in shape, so the volume of drop will be

(Volume of Sphere)

Where r is the radius of the Spherical drop

Now we know mass of an object whose volume and density are known is given by

m = V × d

where, m is the mass

V is the volume and d is the density

Here density of the drop is

d = 1.26 gcm^-3

= 1.26 × 10³ Kgm^-3

so mass of the drop is

m = × 1.26 × 10³

Now the downward pull on a body due to earth’s gravitational force is the weight of body given by

W = m × g

Where W is the weight of a body having mass m

The acceleration due to gravity is denoted by g

g = 9.81 ms^-2

so the weight or the downward gravitational force on oil drop is

W = × 1.26 × 10³ Kgm^-3× 9.81ms^-2

Both the forces should be equal in magnitude so equating them

i.e. putting F = W

4.896 × 10^-14 N = (4/3 πr³) × 1.26 × 10³ Kgm^-3 × 9.81ms^-2

we get ,

(1N = 1kgms^-2)

So, r

⇒r = 0.98110^-6 m

= 9.81 × 10^-7 m

= 981 × 10^-4 mm

i.e. radius of the oil drop is 98110^-4 mm