Problem 102 - University Physics with Modern Physics

Question

Two very large horizontal sheets are 4.25 \(\mathrm{cm}\) apart and carry equal but opposite uniform surface charge densities of magnitude \(\sigma .\) You want to use these sheets to hold stationary in the region between them an oil droplet of mass 324\(\mu \mathrm{g}\) that carries an excess of five electrons. Assuming that the drop is in vacuum, (a) which way should the electric field between the plates point, and (b) what should \(\sigma\) be?

Step-by-Step Solution

Answer

The electric field should point upwards, and \(\sigma\) should be \(3.51 \times 10^{-6} \mathrm{C/m^2}\).

Step 1: Understanding the Electric Field Direction

The electric field between two charged sheets with opposite charges points from the positively charged sheet to the negatively charged sheet. To hold the negatively charged oil droplet stationary, the electric field should point upwards, opposing the downward gravitational force on the droplet.

Step 2: Calculating the Gravitational Force

The mass of the droplet is given as 324 \(\mu g\) or \(324 \times 10^{-9} \mathrm{kg}\). The gravitational force \(F_g\) acting on the droplet is calculated using \( F_g = mg \), where \( g = 9.8 \mathrm{m/s^2} \). Substitute the values to get \( F_g = 324 \times 10^{-9} \times 9.8 = 3.1752 \times 10^{-6} \mathrm{N} \).

Step 3: Relating Electric Force and Surface Charge Density

The electric force \( F_e \), needed to balance the gravitational force, is given by \( F_e = qe \), where \( q \) is the charge of the droplet. The charge \( q \) is \( 5 \) electrons, or \( 5 \times 1.6 \times 10^{-19} \mathrm{C} = 8 \times 10^{-19} \mathrm{C} \). The electric field \( E \) between two plates is \( \frac{\sigma}{\varepsilon_0} \), so \( F_e = q \cdot E = q \cdot \frac{\sigma}{\varepsilon_0} \).

Step 4: Equating Forces and Solving for \(\sigma\)

Since the gravitational force must be balanced by the electric force, set \( F_e = F_g \). Thus, \( q \cdot \frac{\sigma}{\varepsilon_0} = 3.1752 \times 10^{-6} \). Solving for \( \sigma \) gives \( \sigma = \frac{3.1752 \times 10^{-6} \cdot \varepsilon_0}{q} \). Using \( \varepsilon_0 = 8.85 \times 10^{-12} \mathrm{F/m} \) and substituting the known values, \( \sigma = \frac{3.1752 \times 10^{-6} \cdot 8.85 \times 10^{-12}}{8 \times 10^{-19}} = 3.50985 \times 10^{-6} \mathrm{C/m^2} \).

Key Concepts

Surface Charge Density

Surface charge density is a way of quantifying the amount of electric charge accumulated per unit area on the surface of a conductor or insulator. It is denoted by the Greek letter sigma (\( \sigma \)). In the case of the electric field between two large charged sheets, the surface charge density is crucial in determining the electric field strength.

When the sheets have equal but opposite charges, their surface charge densities are also equal in magnitude but opposite in sign. This configuration creates a uniform electric field between the sheets. The electric field strength \( E \) between two large parallel plates with surface charge density \( \sigma \) is given by the formula

• \( E = \frac{\sigma}{\varepsilon_0} \)

where \( \varepsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \mathrm{F/m} \).

This uniform electric field is instrumental in applications such as capacitors and in the problem we are considering here, where it plays a vital role in keeping an oil droplet stationary by exerting an electric force opposite to gravitational force.

Gravitational Force

Gravitational force is the attractive force that objects with mass exert on each other. This force is calculated using Newton's second law of motion, which is expressed as \( F = mg \), where \( F \) is the gravitational force, \( m \) is the mass of the object, and \( g \) is the acceleration due to gravity, about \( 9.8 \mathrm{m/s^2} \).

In our example, we're considering an oil droplet with a mass of \( 324 \mu\mathrm{g} \), equivalent to \( 324 \times 10^{-9} \mathrm{kg} \). The resulting gravitational force acting on it is

• \( F_g = 324 \times 10^{-9} \times 9.8 = 3.1752 \times 10^{-6} \mathrm{N} \)

This force pulls the droplet downward, and by creating an electric force equal in magnitude but opposite in direction, we can hold the droplet stationary in mid-air. Understanding gravitational force is essential for analyzing and balancing forces in many other real-world applications.

Electric Force

The electric force is a fundamental interaction between charged particles. It is the force that two charges exert on each other. According to Coulomb's law, the electric force \( F_e \) on a charge \( q \) subjected to an electric field \( E \) is given by the equation

• \( F_e = qE \)

In the problem at hand, the electric force is used to counteract the gravitational force on a charged oil droplet.

For our oil droplet carrying an excess of five electrons, the charge \( q \) can be calculated as

• \( q = 5 \times 1.6 \times 10^{-19} \mathrm{C} = 8 \times 10^{-19} \mathrm{C} \)

To balance the gravitational force and keep the droplet stationary, the electric force must be equal in magnitude to the gravitational force. Thus, by setting \( F_g = F_e \), we derive the necessary surface charge density \( \sigma \) to achieve equilibrium and maintain the droplet in position. This principle of balancing forces is prevalent in fields such as physics and engineering.