Question

One method to measure the electric charge on a small oil drop is to place it in an electric field that counterbalances gravity. If an oil drop has a mass of 0.12 milligrams and a net charge of +10 electrons, find the electric field (magnitude and direction) required to make the oil drop float in mid-air.

Answer 1

Given : Charge on the drop is +10 e Coulomb ( e represents charge on an electron, e 1.6 *10^-19 Coulombs)

g represents acceleration due to gravity. ( g=9.8m/s² )

Mass of the drop is 0.12mg or 0.12 * 10^-3 g

We know that weight of drop is given by 'mg'

And the Electrostatic force on a charged particle placed in an Electric Field is 'qE' (where q is the charge on the particle and E is the magnitude of electric field.)

Now, to make drop float the Electrostatic force need to balance the weight of the drop.

i.e. mg = qE

Now, plugging in the given values, we get

We know that direction of electrostatic force on a positive charge kept in an electric field is in the direction of Electric field.

Therefore, if the direction of gravity is in -y direction (or, ) , then Electric field should point in +y direction (or, ). As shown in the figure. (in red)