In Millikan's experiment, an oil drop of radius 1.516 µm and density 0.860 g/cm3 is suspended in chamber C (Figure 22-14) when a downward-pointing electric field of 1.92 105 N/C is applied. Find the charge on the drop, in terms of e. (Include the sign.)

Calulate the mass of the drop as (4/3)*pi r^3*(density), and convert it to kg. Multiply by g = 9.8 m/s^2 to get the gravitational force (weight) in N.

Set the sum of the weight W and the electrostatic force
F = q E,
equal to zero for a non-accelerating (suspended) oil drop. Forces are considered positive downward. E is the electrical field, 1.92*10^5 N/C. Solve for the charge q in Coulombs. It should have a negative sign.

I got it, thanks!

First, let's calculate the mass of the drop:

mass = (4/3) * pi * r^3 * (density)
mass = (4/3) * pi * (1.516 * 10^-6 m)^3 * (0.860 * 10^3 kg/m^3)
mass ≈ 8.928 * 10^-15 kg

Now, let's calculate the gravitational force (weight) in N:

Weight = mass * g
Weight = 8.928 * 10^-15 kg * 9.8 m/s^2
Weight ≈ 8.749 * 10^-14 N

Now, we can set the sum of the weight W and the electrostatic force F equal to zero for a non-accelerating (suspended) oil drop:

Weight + F = 0
8.749 * 10^-14 N - q * E = 0

Solving for the charge q in Coulombs, we get:

q = (8.749 * 10^-14 N) / (1.92 * 10^5 N/C)
q ≈ -4.552 * 10^-19 C

Since the elementary charge e is approximately 1.6 * 10^-19 C, we can find the charge on the drop in terms of e:

q/e ≈ (-4.552 * 10^-19 C) / (1.6 * 10^-19 C/e)
q/e ≈ -2.84

Therefore, the charge on the oil drop is approximately -2.84 times the elementary charge.

You're welcome! If you have any more questions, feel free to ask.

You're welcome! If you have any more questions, feel free to ask.