A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of 8380 N/C. The mass of the water drop is 3.40 10-9 kg.
How many excess electrons or protons reside on the drop?
Enough (n) to make n*e*E equal to the drop's weight, M g
Solve for n.
e is the electron charge, E is the electric field strength, and you know what g is.
To determine the number of excess electrons or protons residing on the water drop, we need to use the concept of electric force.
The electric force acting on the water drop can be calculated using the equation:
F = q * E
Where:
F is the electric force,
q is the charge on the water drop,
E is the magnitude of the electric field.
We can rearrange this equation to solve for the charge:
q = F / E
Given:
F = m * g (where m is the mass of the water drop and g is the acceleration due to gravity)
m = 3.40 * 10^(-9) kg
g = 9.8 m/s^2
E = 8380 N/C
Substituting the values into the equation:
q = (m * g) / E
q = (3.40 * 10^(-9) kg * 9.8 m/s^2) / 8380 N/C
Calculating the value:
q ≈ 4.01 * 10^(-16) C
Since electrons have a charge of -1.6 * 10^(-19) C and protons have a charge of +1.6 * 10^(-19) C, we can calculate the number of excess electrons or protons:
Number of excess electrons or protons = q / (charge of an electron or proton)
Number of excess electrons or protons ≈ (4.01 * 10^(-16) C) / (1.6 * 10^(-19) C)
Number of excess electrons or protons ≈ 2.51 * 10^3
Therefore, there are approximately 2.51 * 10^3 excess electrons or protons residing on the water drop.