Problem 34 - College Physics

Question

In a particular Millikan oil-drop apparatus, the plates are 2.25 \(\mathrm{cm}\) apart. The oil used has a density of \(0.820 \mathrm{g} / \mathrm{cm}^{3},\) and the atomizer that sprays the oil drops produces drops of diameter \(1.00 \times 10^{-3} \mathrm{mm}\) . (a) What strength of electric field is needed to hold such a drop stationary against gravity if the drop contains five excess electrons? (b) What should be the potential difference across the plates to produce this electric field? (c) If another drop of the same oil requires a plate potential of 73.8 \(\mathrm{V}\) to hold it stationary, how many excess electrons did it contain?

Step-by-Step Solution

Answer

(a) 5250 N/C, (b) 118.13 V, (c) 8 electrons

Step 1: Calculate the Volume of the Oil Drop

First, find the volume of a spherical oil drop using the formula for the volume of a sphere: \[ V = \frac{4}{3} \pi r^3 \] The diameter is given as \( 1.00 \times 10^{-3} \text{ mm} \), which is \( 1.00 \times 10^{-6} \text{ m} \). The radius \( r \) is half the diameter: \( r = 0.5 \times 10^{-6} \text{ m} \). Plugging into the formula, we get:\[ V = \frac{4}{3} \pi (0.5 \times 10^{-6})^3 \approx 5.24 \times 10^{-19} \text{ m}^3 \]

Step 2: Determine the Mass of the Oil Drop

Next, calculate the mass using the equation \( m = \rho V \), where \( \rho \) is the density of the oil: \( 0.820 \text{ g/cm}^3 = 820 \text{ kg/m}^3 \). Thus, the mass \( m \) of the drop is:\[ m = 820 \times 5.24 \times 10^{-19} \approx 4.29 \times 10^{-16} \text{ kg} \]

Step 3: Calculate the Gravitational Force on the Drop

The gravitational force \( F_g \) can be calculated using \( F_g = mg \), where \( g = 9.8 \text{ m/s}^2 \). Thus:\[ F_g = 4.29 \times 10^{-16} \times 9.8 \approx 4.20 \times 10^{-15} \text{ N} \]

Step 4: Determine the Electric Force Required

For the drop to stay stationary, the electric force \( F_e \) must balance the gravitational force \( F_g \). Therefore, \( F_e = F_g \). The electric force is given by \( F_e = Eq \), where \( q = ne \) is the charge of the drop with \( n = 5 \) excess electrons and \( e = 1.6 \times 10^{-19} \text{ C} \). So, \( q = 5 \times 1.6 \times 10^{-19} \text{ C} = 8 \times 10^{-19} \text{ C} \). Thus, setting \( F_e = 4.20 \times 10^{-15} \text{ N} \), we find:\[ E = \frac{F_e}{q} = \frac{4.20 \times 10^{-15}}{8 \times 10^{-19}} = 5.25 \times 10^3 \text{ N/C} \]

Step 5: Calculate the Potential Difference Across the Plates

The potential difference \( V \) required to create an electric field \( E \) across plates separated by distance \( d \) is given by: \[ V = Ed \] The separation \( d = 2.25 \text{ cm} = 0.0225 \text{ m} \). Thus: \[ V = 5.25 \times 10^3 \times 0.0225 = 118.13 \text{ V} \]

Step 6: Calculate the Number of Excess Electrons for Another Drop

For another drop requiring a potential of \( 73.8 \text{ V} \):Use \( V = Ed \) to find \( E \):\[ E = \frac{V}{d} = \frac{73.8}{0.0225} = 3.28 \times 10^3 \text{ N/C} \] Now, find \( q \) using \( F_e = Eq = F_g = 4.20 \times 10^{-15} \text{ N} \):\[ q = \frac{4.20 \times 10^{-15}}{3.28 \times 10^3} = 1.28 \times 10^{-18} \text{ C} \] The number of excess electrons \( n \) is:\[ n = \frac{q}{e} = \frac{1.28 \times 10^{-18}}{1.6 \times 10^{-19}} \approx 8 \]

Key Concepts

Electric Field

An electric field is a region around a charged object that influences other charged objects within that space. In physics, the strength of the electric field is described by how much force a charge experiences when placed in the field. It is measured in newtons per coulomb (N/C).

In the Millikan experiment scenario, the electric field was responsible for balancing the gravitational force on the oil drop. The electric field exerts a force on any charges within it, calculated by the formula:

• Electric Force (\( F_e \)) = Electric Field (\( E \)) × Charge (\( q \))

The fundamental concept here is that the electric field must be strong enough to exert an upward force equal to the downward gravitational force on the oil droplet, thus keeping it stationary.

Potential Difference

Potential difference, often referred to as voltage, is the work needed to move a unit charge between two points in an electric field. It is measured in volts (V).

In the Millikan oil-drop experiment, the potential difference between two parallel plates creates the electric field that can hold a charged oil droplet stationary. The relationship between potential difference and the electric field across two plates separated by distance (\( d \)) is given by:

• Potential Difference (\( V \)) = Electric Field (\( E \)) × Distance (\( d \))

This equation shows how a greater potential difference is needed for a stronger electric field to maintain the balance of forces on the droplet.

Excess Electrons

Excess electrons refer to the extra electrons that an object, like an oil drop in the Millikan experiment, carries. Electrons are negatively charged particles, and having excess ones gives the droplet a net negative charge.

The number of excess electrons affects the total charge of the droplet, which is calculated by:

• Total Charge (\( q \)) = Number of Excess Electrons (\( n \)) × Elementary Charge (\( e \))

In the Millikan experiment, this charge is pivotal because it determines how strongly the droplet is influenced by the electric field and therefore the strength of the field required to counteract gravity.

Density

Density measures the mass per unit volume of a substance and is expressed in grams per cubic centimeter (g/cm³) or kilograms per cubic meter (kg/m³).

For the oil drop in the Millikan experiment, density is an essential factor since it dictates the mass of the oil drop, calculated as:

• Mass (\( m \)) = Density (\( \rho \)) × Volume (\( V \))

The mass then affects the gravitational force acting on the droplet. Understanding density is crucial to calculating how much force is necessary to counteract gravity and keep the droplet stationary with the electric field.

Gravitational Force

Gravitational force is the force exerted by the Earth on objects towards its center. It is calculated by the formula:

• Gravitational Force (\( F_g \)) = Mass (\( m \)) × Gravitational Acceleration (\( g \))

Here, gravitational acceleration is considered as 9.8 m/s². In the context of the Millikan experiment, the gravitational force pulls the oil drop downward. The goal of the experiment is to set the electric force equal to this gravitational force, allowing the droplet to hover in place without falling or rising. Understanding how gravitational force works is integral to determining the required electric field to achieve equilibrium.