Question

The classic Millikan oil drop experiment was the first to obtain an accurate measurement of the charge on an electron. In it, oil drops were suspended against the gravitational force by a vertical electric field. (See Figure 18.58.) Given the oil drop to be\(1.00{\rm{ }}\mu {\rm{m}}\)in radius and have a density of\(920{\rm{ kg}}/{{\rm{m}}^3}\): (a) Find the weight of the drop. (b) If the drop has a single excess electron, find the electric field strength needed to balance its weight.

Figure 18.58 In the Millikan oil drop experiment, small drops can be suspended in an electric field by the force exerted on a single excess electron. Classically, this experiment was used to determine the electron charge\({q_e}\)by measuring the electric field and mass of the drop.

Short answer

(a) The weight of the drop is\(3.78 \times {10^{ - 16}}{\rm{ N}}\).

(b) The strength of the electric field is \(2.36 \times {10^5}{\rm{ N}}/{\rm{C}}\).

Step-by-step solution

Mass

Mass is a fundamental quantity of the object which remains constant throughout the universe. The mass is related to the density of the material as,

\(m = \rho V\)

Here, \(\rho \) is the density of the material and \(V\) is the volume of the object.

(a) Weight of the drop

The volume of the sphere is,

\(V = \frac{4}{3}\pi {r^3}\)

Here,\(r\)is the radius of the sphere.

From equation (1.1) and (1.2), the mass of the object is,

\(m = \frac{4}{3}\pi \rho {r^3}\)

The weight of the object is,

\({F_g} = mg\)

Using equation (1.3),

\({F_{_g}} = \frac{4}{3}\pi \rho {r^3}g\)

Substitute\(920{\rm{ kg}}/{{\rm{m}}^3}\)for\(\rho \),\(1.00{\rm{ }}\mu {\rm{m}}\)for\(r\), and\(9.81{\rm{ m}}/{{\rm{s}}^2}\)for g.

\(\begin{array}{c}{F_g} = \frac{4}{3}\pi \times \left( {920{\rm{ kg}}/{{\rm{m}}^3}} \right) \times {\left( {1.00{\rm{ }}\mu {\rm{m}}} \right)^3} \times \left( {9.81{\rm{ m}}/{{\rm{s}}^2}} \right)\\ = \frac{4}{3}\pi \times \left( {920{\rm{ kg}}/{{\rm{m}}^3}} \right) \times {\left[ {\left( {1.00{\rm{ }}\mu {\rm{m}}} \right) \times \left( {\frac{{{{10}^{ - 6}}{\rm{ m}}}}{{1{\rm{ }}\mu {\rm{m}}}}} \right)} \right]^3} \times \left( {9.81{\rm{ m}}/{{\rm{s}}^2}} \right)\\ = 3.78 \times {10^{ - 16}}{\rm{ N}}\end{array}\)

Hence, the weight of the drop is \(3.78 \times {10^{ - 16}}{\rm{ N}}\).

(b) The strength of the electric field

The weight of the oil drop is supported by the electrostatic force due to the electric field. Therefore,

\({F_g} = qE\)

Here,\(q\)is the magnitude of the charge on electron, and\(E\)is the electric field.

The expression for the electric field is,

\(E = \frac{{{F_g}}}{q}\)

Substitute\(3.78 \times {10^{ - 14}}{\rm{ N}}\) for\({F_g}\)and\(1.6 \times {10^{ - 19}}{\rm{ C}}\)for\(q\),

\(\begin{array}{c}E = \frac{{\left( {3.78 \times {{10}^{ - 14}}{\rm{ N}}} \right)}}{{\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)}}\\ = 2.36 \times {10^5}{\rm{ N}}/{\rm{C}}\end{array}\)

Hence, the strength of the electric field is \(2.36 \times {10^5}{\rm{ N}}/{\rm{C}}\).