Problem 3 - Fundamentals of Physics

Question

A charged belt, \(50 \mathrm{~cm}\) wide, travels at \(30 \mathrm{~m} / \mathrm{s}\) between a source of charge and a sphere. The belt carries charge into the sphere at a rate corresponding to \(100 \mu\) A. Compute the surface charge density on the belt.

Step-by-Step Solution

Answer

The surface charge density on the belt is \( 6.67 \times 10^{-6} \mathrm{~C/m^2} \).

Step 1: Understand the Problem

We are asked to find the surface charge density \( \sigma \) on a conveyor belt that is carrying charge into a sphere. Given parameters include the width of the belt, the speed of the belt, and the current of charge flowing.

Step 2: Analyze Given Information

The problem provides the following: width of the belt \( w = 50 \mathrm{~cm} = 0.5 \mathrm{~m} \), speed of the belt \( v = 30 \mathrm{~m/s} \), and current \( I = 100\mu \mathrm{A} = 100 \times 10^{-6} \mathrm{~A} \).

Step 3: Determine Relationship for Surface Charge Density

The surface charge density \( \sigma \) is defined as the charge per unit area. We can use this with the expression for current \( I = \sigma \cdot A \cdot v \), where \( A \) is the area through which the charge passes per second.

Step 4: Substitute Values into the Formula

Since the belt is carrying charge, the area \( A \) is the width of the belt \( w \) times the distance traveled by the belt per second \( v = 1 \mathrm{~second}\). Thus, \( A = w \cdot 1 \mathrm{~m/s} = 0.5 \mathrm{~m} \cdot 30 \mathrm{~m} = 15 \mathrm{~m^2} \).

Step 5: Use Formula and Solve for Surface Charge Density

Substituting the values into the formula for current: \[I = \sigma \cdot w \cdot v \]\[100 \times 10^{-6} = \sigma \cdot 0.5 \cdot 30\]Solve for \( \sigma \):\[\sigma = \frac{100 \times 10^{-6}}{0.5 \cdot 30} = \frac{100 \times 10^{-6}}{15}\]

Step 6: Simplify to Find Final Solution

Simplifying gives: \[\sigma = \frac{100 \times 10^{-6}}{15} = 6.67 \times 10^{-6} \mathrm{~C/m^2}\]. Thus, the surface charge density on the belt is \( 6.67 \times 10^{-6} \mathrm{~C/m^2} \).

Key Concepts

Electric Current

Electric current is the flow of electric charge. Measured in amperes (A), it represents how much charge passes through a point in a circuit every second. This flow is typically carried by electrons in a wire, but in some cases, like our exercise with the charged belt, the current is transferred through other mediums.
In the relevant problem, electric current is given as 100 microamperes (\(100 \, \mu A\)), which is equivalent to \(100 \times 10^{-6} \, A \). This current results from the movement of charge carried by the belt. According to the exercise, the belt moves at a speed of \(30 \, m/s\).
It's crucial to understand that the formula for electric current in this context is:

• \(I = \sigma \cdot A \cdot v\)

where:

• \(I\) is the electric current,
• \(\sigma\) is the surface charge density,
• \(A\) is the area through which the charge travels,
• \(v\) is the velocity of the moving charge.

Exploring this relationship helps bridge the properties of electric current with the concept of charge density.

Charge per Unit Area

Surface charge density, often represented by the symbol \(\sigma\), measures how much charge exists per unit of area on a surface. It is expressed in coulombs per square meter (C/m²).
Think of it as the concentration of electric charge on a surface. In the context of our problem, it measures how much charge is carried by a section of the belt as it moves.
For the belt problem, the formula to find surface charge density was given by:

• \(I = \sigma \cdot w \cdot v\)

Solving for \(\sigma\) from the given current (\(I\)), width of the belt (\(w\)), and speed (\(v\)):

• \(\sigma = \frac{I}{w \cdot v}\)

By substituting the values into the formula, we obtained the surface charge density as \(6.67 \times 10^{-6} \, \mathrm{C/m^2}\).
The result shows that over each square meter of the belt's surface, this particular amount of charge is distributed.

Physics Problem Solving

Physics problem solving often requires a methodical approach to understand and calculate various parameters. This means breaking down the problem into smaller, manageable parts and identifying what is known and what needs to be found.
For the given transport belt exercise, this approach entailed:

• Recognizing the main quantity to be solved: surface charge density.
• Identifying the given data: belt width, velocity, and electric current.
• Understanding the relationship between these quantities using the correct physical formula.
• Substituting the known values into the formula and simplifying to find the unknown quantity.

This systematic parsing not only clarifies each parameter's role but also drives you to the solution efficiently. When you tackle problems using this method, you're sharpening your ability to translate real-world scenarios into mathematical expressions and solve them accurately. Always remember, context is key; understanding each element's physical significance enhances your comprehension and problem-solving acumen.