Problem 19 - University Physics with Modern Physics

Question

22.19. How many excess electrons must be added to an isolated spherical conductor 32.0 \(\mathrm{cm}\) in diameter to produce an electric field of 1150 \(\mathrm{N} / \mathrm{C}\) just outside the surface?

Step-by-Step Solution

Answer

About \(3.24 \times 10^{10}\) excess electrons are needed.

Step 1: Understand the Problem

We need to find out how many excess electrons must be added to a spherical conductor in order to produce a specific electric field just outside its surface. The given diameter of the conductor is 32.0 cm, and the electric field strength required is 1150 N/C.

Step 2: Use the Formula for Electric Field

The electric field just outside the surface of a conductor can be calculated using the formula: \( E = \frac{Q}{4\pi\varepsilon_0r^2} \), where \( E \) is the electric field, \( Q \) is the charge, \( \varepsilon_0 \) is the permittivity of free space \( (8.85 \times 10^{-12}\, \text{C}^2/\text{N} \cdot \text{m}^2) \), and \( r \) is the radius of the sphere.

Step 3: Calculate the Radius

Convert the diameter to radius: \( r = \frac{32.0 \text{ cm}}{2} = 16.0 \text{ cm} = 0.16 \text{ m} \).

Step 4: Solve for Charge (Q)

Rearrange the electric field formula to solve for charge: \( Q = E \cdot 4\pi\varepsilon_0r^2 \). Substitute the values: \( Q = 1150 \times 4\pi \times 8.85 \times 10^{-12} \times (0.16)^2 \).

Step 5: Simplify Calculations

Calculate the expression: \( Q = 1150 \times 4\pi \times 8.85 \times 10^{-12} \times 0.0256 \).

Step 6: Compute the Charge

Solve the numerical calculation to find \( Q \). The charge \( Q \approx 5.18 \times 10^{-9} \text{ C} \).

Step 7: Convert Charge to Number of Electrons

The charge of one electron is \( e = 1.60 \times 10^{-19} \text{ C} \). The number of electrons needed is \( n = \frac{Q}{e} = \frac{5.18 \times 10^{-9}}{1.60 \times 10^{-19}} \approx 3.24 \times 10^{10} \).

Key Concepts

Electric Field

The electric field is a fundamental concept in electrostatics. It is a vector quantity, meaning it has both magnitude and direction. The electric field describes the force exerted per unit charge on a test charge placed in the vicinity of another charge. Imagine it as the "aura" or "influence zone" around a charged object.
In the context of our problem, we are concerned with the electric field just outside the surface of a spherical conductor. This electric field is generated by the charge distributed over the sphere's surface. In general, electric fields show us how a charged object will interact with other charged objects nearby.

To calculate the electric field due to a charge, we use the formula:

• \[ E = \frac{F}{q} \]
• where: \( E \) is the electric field, \( F \) is the force, and \( q \) is the test charge.

This formula simplifies to

• \[ E = \frac{Q}{4\pi\varepsilon_0r^2} \]
• where \( Q \) is the charge on the conductor, \( \varepsilon_0 \) is the permittivity of free space, and \( r \) is the radius of the sphere.

By using this relationship, we can determine how strong the electric field will be given a certain distribution of charge.

Charge Calculation

To find out how many excess electrons need to be added to a conductor to create a specific electric field, we must first calculate the necessary charge. This involves rearranging the formula for electric field designed for spherical conductors.
The formula \( E = \frac{Q}{4\pi\varepsilon_0r^2} \) can be rearranged to solve for the charge \( Q \):

• \[ Q = E \cdot 4\pi\varepsilon_0r^2 \]

By substituting the given electric field and known constants into this equation, we compute \( Q \), the charge that must be placed on the conductor.
In our case, substituting the given values yields:

• \[ Q = 1150 \times 4\pi \times 8.85 \times 10^{-12} \times (0.16)^2 \]
• Simplifying, we find \( Q \approx 5.18 \times 10^{-9} \text{ C} \).

Once we have the charge \( Q \), we can easily convert it into the number of electrons, knowing the charge of one electron is \( e = 1.60 \times 10^{-19} \text{ C} \). The number of excess electrons needed is thus:

• \[ n = \frac{Q}{e} = \frac{5.18 \times 10^{-9}}{1.60 \times 10^{-19}} \]
• \[ n \approx 3.24 \times 10^{10} \]

This calculation helps us understand the direct relationship between charge, electric field, and the number of electrons.

Permittivity of Free Space

The permittivity of free space, denoted as \( \varepsilon_0 \), is a key constant in electrostatics equations. It quantifies the ability of the vacuum of space to permit electric field lines. To put it simply, it affects how a charge will "spread out" in space.
The permittivity of free space is used in many standard equations relevant to electrical interactions and forces, like the one for the electric field, Coulomb's Law, and capacitance calculations.
In the electric field formula, it appears as:

• \[ E = \frac{Q}{4\pi\varepsilon_0 r^2} \]
• where \( \varepsilon_0 \) is approximately \( 8.85 \times 10^{-12} \text{ C}^2/\text{N}\cdot\text{m}^2 \).

This constant is crucial in calculating how electric fields behave in a vacuum and helps us predict the force between charged objects.
Understanding \( \varepsilon_0 \) is important because it sets the baseline for how electric fields behave and interact in a world without any other interfering materials. Its role in electrostatics is vast as it establishes the foundational behavior of electric forces in most classical physics models.