Problem 23 - Fundamentals of Physics

Question

(a) The drum of a photocopying machine has a length of \(42 \mathrm{~cm}\) and a diameter of \(12 \mathrm{~cm}\). The electric field just above the drum's surface is \(2.3 \times 10^{5} \mathrm{~N} / \mathrm{C}\). What is the total charge on the drum? (b) The manufacturer wishes to produce a desktop version of the machine. This requires reducing the drum length to \(28 \mathrm{~cm}\) and the diameter to \(8.0 \mathrm{~cm}\). The electric field at the drum surface must not change. What must be the charge on this new drum?

Step-by-Step Solution

Answer

(a) The total charge on the original drum is \(3.24 \times 10^{-8}\, \text{C}\). (b) The charge on the desktop drum is \(1.44 \times 10^{-8}\, \text{C}\).

Step 1: Calculate the Surface Area of the Original Drum

The drum is a cylinder, and we want the lateral surface area since the electric field applies to the surface. The lateral surface area \( A \) of a cylinder is given by \( A = \pi d h \), where \( d \) is the diameter and \( h \) is the height (length) of the drum. For the original drum, \( d = 12\, \text{cm} = 0.12\, \text{m} \) and \( h = 42\, \text{cm} = 0.42\, \text{m} \). Thus, \[ A = \pi (0.12) (0.42) = 0.15876\, \text{m}^2. \]

Step 2: Determine the Charge on the Original Drum

The relationship between the electric field \( E \), charge \( Q \), and the area is given by \( E = \frac{\sigma}{\varepsilon_0} \), where \( \sigma = \frac{Q}{A} \) and \( \varepsilon_0 \approx 8.85 \times 10^{-12} \text{ C}^2/\text{N m}^2 \). Solving for \( Q \), we find \[ Q = EA\varepsilon_0 = (2.3 \times 10^5\, \text{N/C})(0.15876\, \text{m}^2)(8.85 \times 10^{-12}\, \text{C}^2/\text{N m}^2) = 3.24 \times 10^{-8}\, \text{C}. \]

Step 3: Calculate the Surface Area of the Desktop Drum

For the desktop version, \( d = 8\, \text{cm} = 0.08\, \text{m} \) and \( h = 28\, \text{cm} = 0.28\, \text{m} \). The surface area is \[ A = \pi (0.08)(0.28) = 0.070528\, \text{m}^2. \]

Step 4: Determine the Charge on the Desktop Drum

Since the electric field remains the same, use the expression from Step 2. For the new surface area \( A \): \[ Q = EA\varepsilon_0 = (2.3 \times 10^5\, \text{N/C})(0.070528\, \text{m}^2)(8.85 \times 10^{-12}\, \text{C}^2/\text{N m}^2) = 1.44 \times 10^{-8}\, \text{C}. \]

Key Concepts

Cylindrical Surface Area

Understanding the cylindrical surface area is crucial for solving problems involving objects like drums and pipes. A cylindrical surface area, especially the lateral one, is used to determine how much space the electric field interacts with. Let's break it down to grasp how we calculate it.
The formula for the lateral surface area of a cylinder is:

• \[ A = \pi d h \]

Here, \(d\) is the diameter and \(h\) is the height or length of the cylinder.
Consider a drum in our exercise. For the original drum, the diameter is 12 cm (converted to meters becomes 0.12 m) and its length is 42 cm (0.42 m).
Plugging these numbers into our formula, we get:

• \[ A = \pi \times 0.12 \times 0.42 = 0.15876 \, \text{m}^2 \]

This surface area is where the electrical charge is distributed, and hence it's vital for our calculations.

Charge Calculation

Charge calculation relies heavily on understanding the relationship between charge, electric field, and surface area. Once you know these, finding the charge becomes straightforward.
In electrostatics, the electric field \(E\) is linked to the surface charge density \(\sigma\) by:

• \[ E = \frac{\sigma}{\varepsilon_0} \]

Here, \(\varepsilon_0\) is the permittivity of free space (approximately \(8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2\)). Surface charge density \(\sigma\) is how much charge is on a certain surface area, defined as \(\sigma = \frac{Q}{A}\), where \(Q\) is the total charge and \(A\) is the surface area.
To find \(Q\), rearrange the equation:

• \[ Q = E A \varepsilon_0 \]

For our first drum setup:

• The electric field is given as \(2.3 \times 10^5 \, \text{N/C}\).
• Substitute our area calculated before \(0.15876 \, \text{m}^2\).

You calculate:

• \[ Q = 2.3 \times 10^5 \times 0.15876 \times 8.85 \times 10^{-12} = 3.24 \times 10^{-8} \, \text{C} \]

This process determines how much charge is necessary for an electric field of this magnitude on the drum's surface.

Electric Field and Charge Relationship

The relationship between the electric field and charge is a fundamental concept in physics. It determines how a charge distributes across a surface and influences the resulting electric field.
The electric field \(E\) results from charges and is dictated by the equation:

• \[ E = \frac{Q}{A\varepsilon_0} \]

In simpler terms, with more charge \(Q\) or a smaller surface area \(A\), the electric field \(E\) becomes stronger.
This explains why the electric field remains unchanged when altering the drum size in our problem—a crucial concept here.
As you reduce the drum's size, ensuring the surface area \(A\) decreases accordingly, the charge must also decrease to maintain the same electric field magnitude.
For the new drum setup:

• We have a reduced area \(0.070528 \, \text{m}^2\).
• The electric field stays at \(2.3 \times 10^5 \, \text{N/C}\).

Maintaining the same electric field:

• \[ Q = E A \varepsilon_0 = 2.3 \times 10^5 \times 0.070528 \times 8.85 \times 10^{-12} = 1.44 \times 10^{-8} \, \text{C} \]

This computation shows the adaptability of charge and field with changing geometry, vital for device miniaturization.