You decide to levitate a paperclip (m = 1 g). For that, you generate a vertically oriented (from the ground upward) uniform electric field E = 200 V/m. Then you add an electric charge to the paperclip so that the electric force on the paperclip would cancel gravity.
(a) What is the magnitude of the charge?
(b) What is its sign?
(c) How many elementary charges constitute this charge?
(a) To calculate the magnitude of the charge required to cancel out the force of gravity on the paperclip, we need to set the electric force equal to the force of gravity:
Fe = Fg
qE = mg
Where q is the charge, E is the electric field strength, m is the mass of the paperclip, and g is the acceleration due to gravity.
Plugging in the values:
q * 200 = 0.001 * 9.8
q = 0.001 * 9.8 / 200
q = 4.9 x 10^-6 C
Therefore, the magnitude of the charge required to levitate the paperclip is 4.9 x 10^-6 C.
(b) The sign of the charge depends on the orientation of the electric field and the direction of the gravitational force. Since the electric field is oriented upwards and the gravitational force is oriented downwards, the charge on the paperclip should be negative in order to cancel out the force of gravity. Therefore, the charge is negative.
(c) To find out how many elementary charges constitute this charge, we can use the elementary charge e = 1.6 x 10^-19 C:
Number of elementary charges = |q| / e
Number of elementary charges = 4.9 x 10^-6 / 1.6 x 10^-19
Number of elementary charges ≈ 3.06 x 10^13
Therefore, approximately 3.06 x 10^13 elementary charges constitute the charge on the paperclip.