In Millikan's experiment, an oil drop of radius 1.75 ƒÊm and density 0.882 g/cm3 is suspended in chamber C (Fig. 22-14) when a downward electric field of 1.35 �~ 105 N/C is applied. Find the charge on the drop, in terms of e.
Calculate the number of coulombs of positive charge in 200 cm3 of (neutral) water. (Hint: A hydrogen atom contains one proton; an oxygen atom contains eight protons.)
To find the charge on the oil drop in terms of e, we need to use Millikan's oil drop experiment formula. The formula is given by:
q = (6πηvr)/E
where:
q is the charge on the oil drop,
π is the mathematical constant (approximately equal to 3.14159),
η is the viscosity of air (equal to 18.6 x 10^-6 N s/m^2),
v is the velocity of the oil drop,
r is the radius of the oil drop, and
E is the electric field strength.
In this case, the radius of the oil drop (r) is given as 1.75 μm, the density of the oil drop is not needed to calculate the charge, and the electric field strength (E) is given as 1.35 x 10^5 N/C.
Let's plug in the values and solve for the charge on the oil drop:
q = (6πηvr)/E
q = (6 * 3.14159 * (18.6 x 10^-6 N s/m^2) * v * (1.75 x 10^-6 m)) / (1.35 x 10^5 N/C)
Now, to make the answer in terms of e, we need to convert it to elementary charge units.
The elementary charge (e) is equal to 1.6 × 10^⁻19 C. So, we divide the charge (q) by the elementary charge (e) to get the charge in terms of e.
q = [(6 * 3.14159 * (18.6 x 10^-6 N s/m^2) * v * (1.75 x 10^-6 m)) / (1.35 x 10^5 N/C)] / (1.6 × 10^⁻19 C)
Now, all you need to do is substitute the known values and perform the calculations to find the charge on the oil drop in terms of e.