Problem 8 - Fundamentals of Physics

Question

splashing of the water on the bare tub can fill the room's air with negatively charged ions and produce an electric field in the air as great as \(1000 \mathrm{~N} / \mathrm{C}\). Consider a bathroom with dimensions \(2.5 \mathrm{~m} \times\) \(3.0 \mathrm{~m} \times 2.0 \mathrm{~m}\). Along the ceiling, floor, and four walls, approximate the electric field in the air as being directed perpendicular to the surface and as having a uniform magnitude of \(600 \mathrm{~N} / \mathrm{C}\). Also, treat those surfaces as forming a closed Gaussian surface around the room's air. What are (a) the volume charge density \(\rho\) and (b) the number of excess elementary charges \(e\) per cubic meter in the room's air?

Step-by-Step Solution

Answer

(a) \( \rho \approx 1.31 \times 10^{-8} \text{ C/m}^3 \); (b) about \( 8.19 \times 10^{10} \text{ charges/m}^3 \).

Step 1: Understand the problem and identify given values

We are given a bathroom with dimensions 2.5 m x 3.0 m x 2.0 m, and an electric field of magnitude 600 N/C. We need to find two things: (a) the volume charge density \( \rho \) and (b) the number of excess elementary charges per cubic meter.

Step 2: Recall Gauss's Law

Gauss's Law states \( \Phi_E = \frac{Q_{ ext{enclosed}}}{\varepsilon_0} \), where \( \Phi_E \) is the electric flux, \( Q_{\text{enclosed}} \) is the enclosed charge, and \( \varepsilon_0 \) is the permittivity of free space (\(8.85 \times 10^{-12} \text{C}^2/\text{N}\cdot\text{m}^2\)).

Step 3: Calculate Electric Flux

The electric flux \( \Phi_E \) through a closed surface is given by \( \Phi_E = EA\), where \( E = 600 \text{ N/C} \) and the surface area \( A \) consists of all 6 surfaces of the room.The area \( A \) is calculated as: \[A = 2(lw + lh + wh) = 2(2.5 \times 3.0 + 2.5 \times 2.0 + 3.0 \times 2.0) = 2(7.5 + 5.0 + 6.0) = 2 \times 18.5 = 37.0 \text{ m}^2\] Thus, the electric flux is: \[ \Phi_E = 600 \times 37.0 = 22200 \text{ Nm}^2/\text{C} \]

Step 4: Find the enclosed charge \(Q_{\text{enclosed}}\)

Using Gauss's Law \( \Phi_E = \frac{Q_{\text{enclosed}}}{\varepsilon_0} \), we can find: \[ Q_{\text{enclosed}} = \Phi_E \cdot \varepsilon_0 = 22200 \cdot 8.85 \times 10^{-12} \approx 1.9657 \times 10^{-7} \text{ C} \]

Step 5: Calculate volume charge density \(\rho\)

The charge density is given by \( \rho = \frac{Q_{\text{enclosed}}}{V} \), where \( V \) is the volume of the room. The volume \( V \) is \[ V = 2.5 \times 3.0 \times 2.0 = 15.0 \text{ m}^3 \]Thus,\[ \rho = \frac{1.9657 \times 10^{-7}}{15.0} \approx 1.31 \times 10^{-8} \text{ C/m}^3 \]

Step 6: Calculate number of excess elementary charges

The number of excess elementary charges \( n \) per cubic meter is given by \( n = \frac{\rho}{e} \), where \( e \approx 1.6 \times 10^{-19} \text{ C} \).Thus,\[ n = \frac{1.31 \times 10^{-8}}{1.6 \times 10^{-19}} \approx 8.19 \times 10^{10} \text{ charges/m}^3 \]

Key Concepts

Electric Flux

Electric flux is a crucial concept in the study of electric fields, particularly when examining how they interact with surfaces. It is a measure of how much electric field "flows" through a certain area. Here, our interest is the flux through the surfaces of a room, forming a closed surface around the air within.

• Electric flux, \( \Phi_E \), is calculated as \( \Phi_E = E \times A \), where \( E \) is the electric field magnitude and \( A \) is the surface area through which the field lines pass.
• The surface area of the room is the combined area of the ceiling, floor, and walls, totaling 37.0 m² in this example.
• In our specific problem, a uniform electric field of 600 N/C results in an electric flux \( \Phi_E = 22200 \text{ Nm}^2/\text{C} \).

The remarkable part about electric flux is that it links directly with Gauss's Law, a powerful tool for calculating enclosed charges based on the flux passing through a closed surface.

Volume Charge Density

Volume charge density, represented as \( \rho \), describes the amount of electric charge per unit volume within a region. It's a vital quantity when assessing how charges are distributed in a given space like the air in our room example.

• The volume charge density is computed using the formula \( \rho = \frac{Q_{\text{enclosed}}}{V} \), where \( Q_{\text{enclosed}} \) is the total charge within the volume, and \( V \) is the volume itself.
• For our room problem, the enclosed charge \( Q_{\text{enclosed}} \) was found using Gauss’s Law and was approximately \( 1.9657 \times 10^{-7} \text{ C} \).
• The room volume is calculated as \( V = 2.5 \times 3.0 \times 2.0 = 15.0 \text{ m}^3 \).
• Therefore, the volume charge density \( \rho \) is \( 1.31 \times 10^{-8} \text{ C/m}^3 \).

Volume charge density offers a way to understand how sparsely or densely charge carriers are distributed in the space, providing insight into the electrical characteristics of that area.

Elementary Charge

The concept of an elementary charge is an essential building block in understanding how electric charge is quantified at the atomic level. The elementary charge, symbolized by \( e \), is the charge of a single proton, which is roughly \( 1.6 \times 10^{-19} \text{ C} \).

Linking to the Number of Charges

In practical situations like our example of a room, the number of excess elementary charges per cubic meter (\( n \)) can be found from the volume charge density.

• The formula for determining \( n \) is \( n = \frac{\rho}{e} \), linking the charge density directly back to individual elementary charges.
• With our room's charge density of \( 1.31 \times 10^{-8} \text{ C/m}^3 \), the number of elementary charges is computed as approximately \( 8.19 \times 10^{10} \text{ charges/m}^3 \).

This tells us that for each cubic meter of air within the room, there are roughly 81 billion excess ions, providing a tangible measure of ionization in the environment.