A small object with a mass 3.50 · 10−9 kg is suspended by the electric field. (a) Is the excess charge on the mass positive or negative? (b) How many excess electrons or protons reside on the mass?

Question

The space between the plates of a capacitor has an electric field with magnitude 8,480 N/C, directed upwards. A small object with a mass 3.50 · 10⁻⁹ kg is suspended by the electric field. (a) Is the excess charge on the mass positive or negative? (b) How many excess electrons or protons reside on the mass?

Answer 1

For the first part of the question, is the suspended mass positively or negatively charged? The mass must be positively charged. To figure this out, you just need to know that the "direction" of the electrical field lines that you would draw represent the direction that a positively charged test particle would feel a force. So if the electrical field's direction is "up" then a positively charged particle would feel an upward force - basically this is implied in the definition of the "direction" of an electrical field.

For the second part of the question, you now need to know how many coulombs of charge (each electron/proton being worth 1.67*10^-19c) would be required to overcome the force of gravity for this object, in this particular electrical field.

Force of gravity is F = mass * gravity (F=mg), and Electrical force caused by electric field is F = Electric Field Strength * total charge (F= Eq).

Since these two forces must be equal for the object to be suspended in the field, you can say that mg = Eq

Then with mg = Eq you have

(3.5*10^-9)(9.81) = (8480)(q)

You can find q = 4.048938679*10^-12 coulombs from that, pretty easily, where q, the total charge is = (charge of 1 electron/proton)(total number of protons/electrons) (q = e^-*N)

Then, finally just find the total number of protons using q = e^- * N

4.048938679*10^{-12 =} 1.67*10^-19*N

N = 24,245,142 excess protons