Question

Integrated Concepts

In a Millikan oil-drop experiment using a setup like that in Figure 30.9, a 500 - V potential difference is applied to plates separated by 2.50 cm.

(a) What is the mass of an oil drop having two extra electrons that is suspended motionless by the field between the plates?

(b) What is the diameter of the drop, assuming it is a sphere with the density of olive oil?

Short answer

(a) The mass of an oil drop is 6.54 x10^-16 kg.

(b) The diameter of oil drop is 1.108 x 10^-6 m.

Step-by-step solution

Determine the formulas:

The force generated by an electric field is calculated as follows:

$${\mathbf{F}}_{\mathbf{E}}\mathbf{=}\mathbf{q}\mathbf{E}$$

Here, F_E Force due to electric field, q is the charge, and E is the electric Field

Consider the formula for the electric field.

$$\mathbf{E}\mathbf{=}\frac{\mathbf{V}}{\mathbf{d}}$$

Here, d is the separation between the plates.

Consider the formula for the gravitation force.

$${\mathbf{F}}_{\mathbf{G}}\mathbf{=}\mathbf{m}\mathbf{g}$$

Here,F_G is the force due to gravity mis the mass.

Find the mass of an oil drop

(a)

Consider the formula for the forces as:

$$\begin{array}{l}{F}_G=F_E\\ mg=qE\\ mg=q\frac{V}{d}\end{array}$$

Since, there are two extra electrons:

$$mg=2q\frac{V}{d}$$

Mass of oil drop is calculated as:

$$m=\frac{2qV}{gd}$$

Consider the given values:

$$\begin{array}{l}V=500\mathrm{V}\\ q=1.60\times {10}^{-19}\mathrm{C}\\ g=9.8\frac{\mathrm{m}}{{\mathrm{s}}^2}\end{array}$$

$$d=2.5\times {10}^{-2}\mathrm{m}$$

Substitute the values in the above formula,

$$\begin{array}{l}m=\frac{2\left(1.60\times {10}^{-19}\mathrm{C}\right)\left(500\mathrm{V}\right)}{\left(9.8\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)\left(2.5\times {10}^{-2}\mathrm{m}\right)}\\ m=6.54\times {10}^{-16}\mathrm{k}\mathrm{g}\end{array}$$

Therefore, the required mass of an oil drop is 6.54 x10^-16 kg.

Find the diameter of oil drop

(b)

Consider the formulas for the density as:

$$\rho =\frac{m}{V}$$

Consider the volume of the oil drop as:

$$V=\frac{4}{3}\pi r^3$$

Consider the formula for the density as:

$$\begin{array}{l}\rho =\frac{m}{V}\\ \rho =\frac{3m}{4\pi r^3}\\ r^3=\frac{3m}{4\pi \rho }\\ r={\left(\frac{3m}{4\pi \rho }\right)}^{\frac{1}{3}}\end{array}$$

Substitute the values and solve for the radius as:

$$\begin{array}{l}r={\left(\frac{3\left(6.54\times {10}^{-16}\mathrm{k}\mathrm{g}\right)}{4\pi \left(920\frac{\mathrm{k}\mathrm{g}}{{\mathrm{m}}^3}\right)}\right)}^{\frac{1}{3}}\\ r=5.54\times {10}^{-7}\mathrm{m}\end{array}$$

Solve for the diameter as:

$$\begin{array}{l}D=2r\\ D=2\times 5.54\times {10}^{-7}\mathrm{m}\\ D=1.108\times {10}^{-6}\mathrm{m}\end{array}$$

Therefore, the required diameter of oil drop is1.108 x 10^-6 m.