Problem 42 - Fundamentals of Physics

Question

Two large metal plates of area \(1.0 \mathrm{~m}^{2}\) face each other, \(5.0\) \(\mathrm{cm}\) apart, with equal charge magnitudes \(|q|\) but opposite signs. The field magnitude \(E\) between them (neglect fringing) is \(55 \mathrm{~N} / \mathrm{C}\). Find \(|q|\).

Step-by-Step Solution

Answer

The magnitude of the charge \(|q|\) is approximately \( 4.9 \times 10^{-10} \mathrm{~C} \).

Step 1: Understand the relation between electric field and charge density

The electric field between two parallel plates with surface charge density \( \sigma \) is given by \( E = \frac{\sigma}{\varepsilon_0} \), where \( \varepsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \mathrm{C}^2/\mathrm{N}\cdot\mathrm{m}^2 \).

Step 2: Express charge using surface charge density equation

Surface charge density \( \sigma \) is related to charge \( q \) and area \( A \) by \( \sigma = \frac{q}{A} \). Here, \( A = 1.0 \mathrm{~m}^2 \), so the formula becomes \( \sigma = q/1.0 \).

Step 3: Relate electric field to charge using formulas

Substitute \( \sigma = q \) into the equation for electric field: \( E = \frac{q}{\varepsilon_0} \). Rearranging gives \( q = E \varepsilon_0 \cdot A \).

Step 4: Substitute given values

Given that \( E = 55 \mathrm{~N}/\mathrm{C} \), \( \varepsilon_0 = 8.85 \times 10^{-12} \mathrm{C}^2/\mathrm{N}\cdot\mathrm{m}^2 \), and \( A = 1.0 \mathrm{~m}^2 \), substitute these into the equation: \( q = 55 \times 8.85 \times 10^{-12} \times 1.0 \).

Step 5: Perform the calculation

Calculate \( q \): \( q = 55 \times 8.85 \times 10^{-12} = 4.8675 \times 10^{-10} \mathrm{~C} \).

Step 6: Round to appropriate significant figures

Since the given data all have two significant figures (e.g., \( E = 55 \mathrm{~N}/\mathrm{C} \)), round \( q \) to two significant figures: \( q = 4.9 \times 10^{-10} \mathrm{~C} \).

Key Concepts

Surface Charge Density

Understanding surface charge density is crucial when dealing with electric fields between two metal plates. Surface charge density, represented by the symbol \( \sigma \), measures how much charge is distributed over a particular surface area. It is defined as the charge \( q \) per unit area \( A \), expressed as \( \sigma = \frac{q}{A} \).

This concept is especially important in physics because it helps explain how electric fields originate between charged surfaces. For example, with two large parallel metal plates, the charges on each plate create an electric field perpendicular to their surfaces. This field is uniform, meaning its strength is constant across the space between the plates. By calculating \( \sigma \), we can predict the field’s intensity, providing insight into how charges interact in various scientific and engineering applications.

In this exercise, since the area is given as \( 1.0 \,\text{m}^2 \), surface charge density simplifies to \( \sigma = q \), helping us relate the charge directly to the electric field strength. This understanding forms the basis for using formulas that link \( \sigma \), \( q \), and electric fields in electrostatics.

Permittivity of Free Space

The term 'permittivity of free space', symbolized as \( \varepsilon_0 \), is fundamental in understanding electric fields between objects. It measures the resistance encountered when forming an electric field in a vacuum. This physical constant helps determine how easily electric forces can propagate through space.

Mathematically, \( \varepsilon_0 \) is approximately \( 8.85 \times 10^{-12} \,\text{C}^2/\text{N} \cdot \text{m}^2 \). This small value reflects the inherent resistance of empty space to electric fields. It serves as a baseline for comparing how different materials affect electric field strength.

In the context of the presented exercise, \( \varepsilon_0 \) plays a pivotal role in calculating the electric field \( E \) between two plates. Using the relationship \( E = \frac{\sigma}{\varepsilon_0} \), we see that field strength depends inversely on \( \varepsilon_0 \). Consequently, understanding \( \varepsilon_0 \) allows us to comprehend how charges influence each other across distances, especially in vacuum environments.

Metal Plates

In electrostatics, metal plates are commonly used to study electric fields and capacitive effects. Their conductive properties make them ideal for generating strong and uniform electric fields when charged. By intuition, any charge on a metal plate evenly spreads across its surface due to the repelling forces between like charges.

The large metal plates in the exercise illustrate a scenario where two plates of area \( 1.0 \, \text{m}^2 \) are positioned facing each other. Such arrangements are often found in parallel plate capacitors, key components in many electronic circuits. With equal and opposite charges on each plate, they create a uniform electric field in the space between them. This simplification allows for precise predictions of the field's behavior, which is particularly valuable in designing devices that rely on consistent electric fields.

By focusing on metal plates' role, one can appreciate how engineers and scientists use their properties to control electric fields and capacitance in practical applications. Understanding this helps us predict how these fields can manipulate charges efficiently.