An oil drop is stationary within an electric field of 490 V/m that is set up between two parallel plates. If the volume of the oil drop is 4 X 10 (-19) m3, how many excess electrons does it carry?
(I'll give ya'll just one hint: the fundamental unit of charge e = 1.6 X 10 (-19) C. )
downward force= mg
upward force= Eq
so what is the mass? density*volume
so you need the density of the oil
q=density*volume*9.8/490 coulombs
number electrons= q/e
To determine how many excess electrons the oil drop carries, we can start by finding the charge on the oil drop. The charge can be calculated using the equation:
Charge = Electric field * Volume
Given that the electric field is 490 V/m and the volume of the oil drop is 4 x 10^(-19) m^3, we can calculate the charge:
Charge = 490 V/m * 4 x 10^(-19) m^3
Next, we need to convert the charge to the number of excess electrons. We know that the fundamental unit of charge, represented by e, is equal to 1.6 x 10^(-19) C. Therefore, the number of excess electrons can be calculated by dividing the charge by the fundamental unit of charge:
Number of excess electrons = Charge / e
Substituting the values:
Number of excess electrons = (490 V/m * 4 x 10^(-19) m^3) / (1.6 x 10^(-19) C)
By simplifying the expression, we get:
Number of excess electrons = (490 V/m * 4) / 1.6
Calculating:
Number of excess electrons = 1220 / 1.6
Number of excess electrons ≈ 762.5
Rounding it to the nearest whole number, the oil drop carries approximately 763 excess electrons.