Question

It is desired to levitate a 1 cubic centimeter object with a mass of 10 grams and -10 microcoulombs of charge ten millimeters over a metal plate. How many electrons must be applied to the metal plate?

Answer 1

Solution :

Given :

mass of the object : m = 10 g = 10 x 10^-3 kg

Charge on the object : q = - 10 μC = - 10 x 10^-6 C

And, r = 10 mm = 10 x 10^-3 m

.

Let the number of electrons applied to the metal plate be n.

Then charge on the plate will be : Q = n e

.

Now, To levitate the cube. The electrostatic force on the cube must be equal and opposite to the force of gravity on the cube.

Thus : F_e = m g