A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of 8000 N/C. The mass of the water drop is 3.54 10-9 kg
How many excess electrons or protons reside on the drop?
mg=Eq=E*e*n
n is the number of charges of e charge.
e is the fundamental charge on an electron.
solve for n
n= mg/Ee
((3.54 10-9 kg)(9.8))/ 8000 N/C
and it gave me the wrong answer so im not sure what im doing wrong.
What you have written does no thave the electron charge "e" in it.
To determine the number of excess electrons or protons residing on the drop, we need to calculate the net charge of the drop.
The net force acting on the drop is the electric force, given by F = qE, where F is the force, q is the charge, and E is the electric field.
Since the drop is suspended motionless, the net force is zero. Therefore, we can write the equation as:
F = qE = 0
Rearranging the equation, we have:
q = -F/E
Substituting the given values, we have:
q = -(3.54 * 10^(-9) kg * 9.8 m/s^2) / 8000 N/C
Simplifying the equation, we find:
q = -4.3475 * 10^(-14) C
The negative sign indicates that the excess charge on the drop is negative, meaning there is an excess of electrons.
To find the number of excess electrons or protons, we need to know the elementary charge, which is the charge of a single electron or proton.
The elementary charge is given by e = 1.6 * 10^(-19) C.
Dividing the net charge by the elementary charge, we get:
number of excess electrons or protons = q / e
Substituting the values, we have:
number of excess electrons or protons = (-4.3475 * 10^(-14) C) / (1.6 * 10^(-19) C)
Calculating the result, we find:
number of excess electrons or protons ≈ -2.7171875 * 10^5
Since the number of excess electrons or protons cannot be negative, we take the absolute value:
number of excess electrons or protons ≈ 2.7171875 * 10^5
Therefore, there are approximately 2.7171875 * 10^5 excess electrons or protons residing on the drop.