A water droplet of mass ‘m’ and net charge ‘-q’ remains stationary in the air due to Earth’s Electric field.
(a) What must be the direction of the Earth’s electric field?
(b) Find an expression for the Earth’s electric field in terms of the mass and charge of the droplet.

(a) the electric field of the Earth will be directed towards the negatively charged water droplet.

(b) E = (9.8*m)/q

Explanation:

Part (a) the direction of the Earth’s electric field

Electric field is always directed towards negatively charged objects.

The water droplet has a negative charge ‘-q’, therefore the electric field of the Earth will be directed towards the negatively charged water droplet.

Part (b) an expression for the Earth’s electric field in terms of the mass and charge of the droplet

The magnitude of Electric field is given as;

E = F/q

where;

f is force and q is charge

Also from Newton's law, F = mg

where;

m is mass and g is acceleration due to gravity = 9.8 m/s²

E =  F/q = mg/q

E = (9.8*m)/q, Electric field in terms of mass and charge of the droplet.

1 / f = 1 / o + 1 / i

1 / image distance + 1 / object distance = 1  focal length

If the object distance is large the image will be at about the focal length

The value of 1 / f is fixed for any one particular lense

As the object distance decreases the image must increase

The above equation can also be written as

o i / (i + o) = f    or i / (i / o + 1)  = f

If for instance o was very large the image would be at the focal length