Problem 50 - University Physics with Modern Physics

Question

A small sphere with mass 5.00 \(\times 10^{-7}\) kg and charge \(+\)7.00 \(\mu\)C is released from rest a distance of 0.400 m above a large horizontal insulating sheet of charge that has uniform surface charge density \(\sigma = +\)8.00 pC\(/\)m\(^2\). Using energy methods, calculate the speed of the sphere when it is 0.100 m above the sheet.

Step-by-Step Solution

Answer

The speed of the sphere when it's 0.100 m above the sheet is approximately 1.95 m/s.

Step 1: Determine Change in Electric Potential Energy

The change in electric potential energy is calculated using the formula:\[ \Delta U = q \Delta V \] where \( \Delta V \) is the change in electric potential and \( q \) is the charge of the sphere. The change in electric potential \( \Delta V \) can be determined from the electric field \( E \) generated by the charged sheet. The electric field due to an infinite sheet of charge is given by: \[ E = \frac{\sigma}{2\varepsilon_0} \] where \( \varepsilon_0 \) is the permittivity of free space, \( 8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2 \). The potential difference \( \Delta V \) between two points at distances \( d_1 \) and \( d_2 \) from the sheet is: \[ \Delta V = E \times (d_1 - d_2) \] with \( d_1 = 0.400 \text{ m} \) and \( d_2 = 0.100 \text{ m} \). Substituting, we find \( \Delta V \).

Step 2: Calculate the Electric Field

First, compute the electric field \( E \) using \( \sigma = 8.00 \times 10^{-12} \text{ C/m}^2 \):\[ E = \frac{8.00 \times 10^{-12}}{2 \times 8.85 \times 10^{-12}} \approx 0.452 \text{ N/C} \].

Step 3: Compute Change in Electric Potential

Now determine the change in potential \( \Delta V \) over the distance change \( 0.400 \text{ m} \) to \( 0.100 \text{ m} \):\[ \Delta V = E \times (0.400 \text{ m} - 0.100 \text{ m}) = 0.452 \times 0.300 \approx 0.136 \text{ V} \].

Step 4: Determine Change in Potential Energy

Substitute the value of \( q \) and \( \Delta V \) into \( \Delta U = q \Delta V \):\[ \Delta U = (7.00 \times 10^{-6} \text{ C}) \times 0.136 \text{ V} \approx 9.52 \times 10^{-7} \text{ J} \]. Ensure the change is negative as the potential energy decreases.

Step 5: Apply Energy Conservation

The work done by the electric field results in an increase in the kinetic energy of the sphere. Therefore:\[ \Delta K = -\Delta U \] where \( \Delta K = \frac{1}{2} mv^2 \) and \( m = 5.00 \times 10^{-7} \text{ kg} \). Solve for the speed \( v \):\[ \frac{1}{2} \times 5.00 \times 10^{-7} \times v^2 = 9.52 \times 10^{-7} \].

Step 6: Solve for Speed

Rearrange to solve for \( v \):\[ v^2 = \frac{2 \times 9.52 \times 10^{-7}}{5.00 \times 10^{-7}} \approx 3.808 \] Then, calculate \( v = \sqrt{3.808} \approx 1.95 \text{ m/s} \).

Key Concepts

Electric Field

The electric field is a fundamental concept in physics, representing the force per unit charge exerted on a charged object in the space surrounding it. In this exercise, we deal with an electric field generated by a large, uniformly charged insulating sheet.

To find the strength of the electric field (\( E \)) generated by such a sheet, we use the formula:

• \[ E = \frac{\sigma}{2\varepsilon_0} \]

Here, \( \sigma \) is the surface charge density, and \( \varepsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2 \).

This formula tells us that the electric field strength depends on how much charge is spread across a given area of the sheet. The key point is that the field created by an infinite sheet is constant and not dependent on the distance from the sheet.

This is why the current scenario uses the electric field calculation to determine potential differences and energy changes, crucial for finding the sphere's speed.

Surface Charge Density

Surface charge density (\( \sigma \)) is a measure of how much electric charge is accumulated per unit area on a surface. For an insulating sheet, it describes the distribution of electric charge across its surface.

In this problem, the surface charge density is given as \( +8.00 \text{ pC/m}^2 \). This positive value indicates that the sheet is positively charged, resulting in an outward electric field.

Calculating \( \sigma \) is crucial because it directly influences the electric field produced by the sheet. Recall, a higher surface charge density produces a stronger electric field:

• If the charge density were higher, the electric field strength would increase.
• Conversely, a lower charge density would decrease the field strength.

The uniform nature of \( \sigma \) ensures that the field strength remains constant across different distances parallel to the sheet. This characteristic simplifies calculations concerning electric potential and energy conservation.

Energy Conservation

Energy conservation is a principle stating that the total energy in a closed system remains constant. When dealing with electric fields and potentials, energy conservation helps us understand how electric potential energy converts into kinetic energy or vice versa.

In this exercise, the small sphere's potential energy decreases as it moves closer to the charged sheet. Here's how it works:

• Initially, the sphere has a potential energy due to its position in the electric field.
• As it moves towards the charged sheet, potential energy \( \Delta U \) decreases.

This change in potential energy becomes kinetic energy, increasing the sphere's speed. Using the calculated change in potential energy (\( 9.52 \times 10^{-7} \text{ J} \)), we apply energy conservation to find the sphere's speed at a lower height. This is symbolic of converting stored energy into motion, a key practicality of energy conservation.

Overall, this transformation characterizes how energy conservation governs the dynamics of charged objects in electric fields.