Question

A very small object with mass 8.20 \(\times\) 10\(^{-9}\) kg and positive charge 6.50 \(\times\) 10\(^{-9}\) C is projected directly toward a very large insulating sheet of positive charge that has uniform surface charge density 5.90 \(\times\) 10\(^{-8}\) C/m2. The object is initially 0.400 m from the sheet. What initial speed must the object have in order for its closest distance of approach to the sheet to be 0.100 m?

Short answer

The initial speed must be approximately 109 m/s.

Step-by-step solution

Understanding the Problem

We are asked to find the initial speed of a charged object such that it reaches a closest approach of 0.100 m from a uniformly charged sheet. Given are the mass and charge of the object, and the surface charge density of the sheet.

Using Energy Conservation

We apply the conservation of energy. The initial energy (kinetic + potential) equals the energy at the closest approach (only potential, since it momentarily stops).

Writing the Energy Equation

The initial kinetic energy is \( \frac{1}{2}mv^2 \), where \( m = 8.20 \times 10^{-9} \) kg and \( v \) is the initial speed. The electric potential energy at a distance \( x \) from the sheet is \( U_i = qE x \), where \( q = 6.50 \times 10^{-9} \) C and \( E = \frac{\sigma}{2\varepsilon_0} \) with \( \sigma = 5.90 \times 10^{-8} \) C/m². The permittivity of free space \( \varepsilon_0 \approx 8.85 \times 10^{-12} \) C²/(N·m²).

Electric Field from the Sheet

The electric field \( E \) due to the infinite sheet is \( E = \frac{\sigma}{2\varepsilon_0} = \frac{5.90 \times 10^{-8}}{2 \times 8.85 \times 10^{-12}} \approx 3.33 \times 10^{3} \) N/C.

Calculating Initial Potential Energy

The initial potential energy is due to the distance 0.400 m from the sheet: \( U_i = q \cdot E \cdot 0.400 \approx 6.50 \times 10^{-9} \cdot 3.33 \times 10^{3} \cdot 0.400 = 8.65 \times 10^{-6} \) J.

Calculating Potential Energy at Closest Approach

At the closest approach of 0.100 m, \( U_f = q \cdot E \cdot 0.100 \approx 6.50 \times 10^{-9} \cdot 3.33 \times 10^{3} \cdot 0.100 = 2.16 \times 10^{-6} \) J.

Energy Conservation Equation

Setting initial kinetic energy plus initial potential energy equal to the potential energy at closest approach: \( \frac{1}{2}mv^2 + 8.65 \times 10^{-6} = 2.16 \times 10^{-6} \).

Solve for Initial Speed

Rearrange the equation: \( \frac{1}{2}mv^2 = 2.16 \times 10^{-6} - 8.65 \times 10^{-6} = -6.49 \times 10^{-6} \). Since this wasn't correctly calculated in reality, let us correct the equation: \( \frac{1}{2}mv^2 = 8.65 \times 10^{-6} - 2.16 \times 10^{-6} \). Solve for \( v \): \( v = \sqrt{\frac{2(8.65 \times 10^{-6} - 2.16 \times 10^{-6})}{8.20 \times 10^{-9}}} \approx 1.09 \times 10^2 \) m/s.

Key concepts

Energy Conservation

Energy conservation is a core principle in physics, stating that the total energy in a closed system remains constant over time. In this exercise, we use the principle to analyze the motion of a charged object near a charged sheet.

Initially, the object possesses kinetic energy, which depends on its mass and speed, described by the formula: \[ KE = \frac{1}{2}mv^2 \] where \( m \) is the mass and \( v \) is the speed.

The object also has electric potential energy due to its position relative to the charged sheet. The potential energy \( U \) is given by: \[ U = qEx \] where \( q \) is the charge, \( E \) is the electric field, and \( x \) is the distance to the sheet.

When the charged object approaches the sheet and stops momentarily, it converts its initial kinetic energy into potential energy. This scenario exemplifies energy conservation, where the initial sum of kinetic and potential energy equals the energy at the closest approach point, which is only potential energy at that moment. By balancing these energies, we can solve for the initial speed of the object.

Electric Field

The electric field is an essential concept when analyzing the forces and energy associated with charged objects. It represents the force per charge experienced by a positive test charge in a region of space. For the large charged sheet in our exercise, the electric field \( E \) is constant and directed perpendicular to the sheet. It's calculated using the surface charge density \( \sigma \) with the formula: \[ E = \frac{\sigma}{2\varepsilon_0} \] where \( \varepsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \) C²/(N·m²).

This setup creates a uniform electric field, which simplifies calculations by making \( E \) constant no matter the distance as long as it stays close and parallel to the sheet. In this exercise, \( E \) determines the change in potential energy as the charged object moves towards the sheet. The electric field's strength influences the object's acceleration, illustrating how electric fields govern motion and energy in electrostatic contexts.

Surface Charge Density

Surface charge density, \( \sigma \), refers to the amount of electric charge per unit area on a surface, measured in coulombs per square meter \( \text{C/m}^{2} \). In our problem, it's a given property of the infinitely large insulating sheet.

The value of \( \sigma \) affects the electric field produced by the sheet according to the formula: \[ E = \frac{\sigma}{2\varepsilon_0} \] where \( E \) is the resultant electric field due to the charged sheet, allowing us to compute the electric potential energy when a charged object is nearby.

High surface charge density results in stronger electric fields, leading to larger potential energy changes for charged particles within its influence. Thus, knowing \( \sigma \) is vital when evaluating how charged objects will behave around charged surfaces, making it a critical factor in both calculating forces and the resulting energies involved in such systems. This in turn ties into our energy conservation calculations as we determine energy relations with respect to distances from the charged sheet.