Problem 52 - University Physics with Modern Physics

Question

A small sphere with mass \(5.00 \times 10^{-7} \mathrm{kg}\) and charge \(+3.00 \mu C\) is released from rest a distance of 0.400 \(\mathrm{m}\) above a large horizontal insulating sheet of charge that has uniform surface charge density \(\sigma=+8.00 \mathrm{pC} / \mathrm{m}^{2} .\) Using energy methods, calculate the speed of the sphere when it is 0.100 \(\mathrm{m}\) above the sheet of charge?

Step-by-Step Solution

Answer

The speed of the sphere is approximately 7.17 m/s when it is 0.100 m above the sheet.

Step 1: Understand the Problem

We have a small sphere with a given mass and charge that starts from rest and falls toward a charged sheet. We need to calculate its speed at a certain distance above the sheet using energy conservation principles.

Step 2: Identify Relevant Formulas

The energy conservation principle involves equating the change in electric potential energy to the change in kinetic energy. The formula for electric potential energy change \[ \Delta U = q \cdot \Delta V \]where \( q \) is the charge of the sphere and \( \Delta V \) is the change in electric potential. The kinetic energy at a height \( h \):\[ K = \frac{1}{2} m v^2 \].

Step 3: Calculate Electric Field from Surface Charge Density

The electric field \( E \) due to a large sheet with surface charge density \( \sigma \) is given by \[ E = \frac{\sigma}{2\varepsilon_0} \]where \( \varepsilon_0 \) is the permittivity of free space \( 8.85 \times 10^{-12} \, \mathrm{C^2/N \, m^2} \). Substitute \( \sigma = 8.00 \, \mathrm{pC/m^2} \) into the formula to find \( E \).

Step 4: Calculate Change in Electric Potential

To calculate the change in electric potential (\( \Delta V \)), use \[ \Delta V = E \cdot \Delta h \]where \( \Delta h \) is the change in height from \( 0.400 \, \mathrm{m} \) to \( 0.100 \, \mathrm{m} \).

Step 5: Calculate Change in Electric Potential Energy

Use \[ \Delta U = q \cdot \Delta V \],with \( q = 3.00 \times 10^{-6} \, \mathrm{C} \) to find the change in electric potential energy as the sphere moves closer to the sheet.

Step 6: Apply Energy Conservation

The total mechanical energy is conserved, implying that the initial potential energy change is converted into kinetic energy. Therefore, solve \[ \Delta U = \frac{1}{2} m v^2 \]for \( v \), where \( m = 5.00 \times 10^{-7} \, \mathrm{kg} \).

Step 7: Solve for the Speed

Rearrange and solve the equation \[ v = \sqrt{\frac{2 \times \Delta U}{m}} \]to find the speed \( v \) of the sphere when it is \( 0.100 \, \mathrm{m} \) above the charged sheet.

Key Concepts

Kinetic Energy

Kinetic energy is the energy that an object possesses due to its motion. In the context of our small sphere, we are interested in determining its kinetic energy as it approaches the charged sheet from an initial state of rest. The formula for calculating kinetic energy, denoted as \( K \), is given by:

\[ K = \frac{1}{2} m v^2 \]
where:

• \( m \) is the mass of the object (in this case, the sphere).
• \( v \) is the speed of the object.

The sphere starts with zero kinetic energy (since it is at rest) and gains kinetic energy as it falls toward the charged sheet. By the time the sphere is 0.100 meters above the sheet, its kinetic energy is maximized due to the conversion of potential energy into kinetic energy. Understanding kinetic energy helps us calculate the velocity of the sphere by rearranging the formula to solve for \( v \), which is critical for finding the speed of the sphere at the desired height.

Electric Field

An electric field is a region around a charged object where force is exerted on other charged objects. In this problem, the electric field is created by the charged sheet, which has a uniform surface charge density \( \sigma \). The formula to find the magnitude of the electric field \( E \) due to a large sheet is:

\[ E = \frac{\sigma}{2\varepsilon_0} \]
where:

• \( \sigma \) is the surface charge density (given as \( 8.00 \, \mathrm{pC/m^2} \)).
• \( \varepsilon_0 \) is the permittivity of free space (\( 8.85 \times 10^{-12} \, \mathrm{C^2/N \, m^2} \)).

The electric field indicates how strong the influence of the charged sheet is over a certain distance. For this sphere, moving towards the sheet implies experiencing a force due to this field, thereby affecting its motion and energy states. This concept is vital in calculating the electric potential energy changes experienced by the sphere.

Energy Conservation

Energy conservation is a fundamental principle stating that within a closed system, total energy remains constant. It involves converting one form of energy into another without any loss. In the exercise with the sphere, we apply energy conservation by equating the change in electric potential energy to the change in kinetic energy, ensuring the total mechanical energy remains the same throughout the sphere's movement.

The change in electric potential energy \( \Delta U \) is calculated using:
\[ \Delta U = q \cdot \Delta V \]
where:

• \( q \) is the charge of the sphere (\( 3.00 \mu C \)).
• \( \Delta V \) is the change in electric potential.

As the sphere moves closer to the sheet, its potential energy decreases, converting into kinetic energy. By calculating these changes and utilizing the formula:
\[ \frac{1}{2} m v^2 = \Delta U \]
we find the speed of the sphere. This principle helps ensure accurate computations and predictions of the sphere's dynamics as it interacts with the electric field of the charged sheet.