Problem 98 - University Physics with Modern Physics

Question

A small sphere with mass \(m\) carries a positive charge \(q\) and is attached to one end of a silk fiber of length \(L\) . The other end of the fiber is attached to a large vertical insulating sheet that has a positive surface charge density \(\sigma\) . Show that when the sphere is in equilibrium, the fiber makes an angle equal to arctan \(\left(q \sigma / 2 m g \epsilon_{0}\right)\) with the vertical shee..

Step-by-Step Solution

Answer

The fiber makes an angle \( \theta = \arctan \left( \frac{q\sigma}{2mg\epsilon_0} \right) \) with the vertical sheet.

Step 1: Identify Forces

In equilibrium, the sphere is subjected to several forces: the gravitational force downward, the tension in the silk fiber, and the electric force due to the charged sheet. The gravitational force can be expressed as \( F_g = mg \), and the electric force on the sphere due to the charged sheet can be calculated using the formula for the electric field of an infinite sheet of charge, which is \( E = \frac{\sigma}{2\epsilon_0} \). Therefore, the electric force on the sphere is \( F_e = qE = \frac{q\sigma}{2\epsilon_0} \).

Step 2: Break Down Forces Into Components

The tension in the fiber, \( T \), will have two components: a horizontal component \( T_{x} \) that balances the electric force and a vertical component \( T_{y} \) that balances the gravitational force. Thus, \( T_{y} = mg \) and \( T_{x} = \frac{q\sigma}{2\epsilon_0} \).

Step 3: Express Components in Terms of Angle

Since the components of tension relate to the angle the fiber makes with the vertical, use trigonometry. The angle \( \theta \) with the vertical gives the relations \( T_{x} = T \sin(\theta) \) and \( T_{y} = T \cos(\theta) \). Thus, we have \( \tan(\theta) = \frac{T_{x}}{T_{y}} \).

Step 4: Solve for Angle \( \theta \)

Plug the relations from Step 2 into the equation \( \tan(\theta) = \frac{T_{x}}{T_{y}} \) to get \( \tan(\theta) = \frac{\frac{q\sigma}{2\epsilon_0}}{mg} = \frac{q\sigma}{2mg\epsilon_0} \). Therefore, \( \theta = \arctan \left( \frac{q\sigma}{2mg\epsilon_0} \right) \).

Key Concepts

Electric field

The concept of an electric field is foundational in electrostatics. Imagine an electric field as a region around a charged object where other charges feel a force. For example, our problem involves a large vertical insulating sheet with a positive surface charge density, denoted as \( \sigma \). This sheet creates an electric field around it, impacting the small charged sphere hanging by the fiber.
The electric field \( E \) produced by an infinite sheet of charge is uniform and can be calculated using the formula:

• \( E = \frac{\sigma}{2\varepsilon_0} \)

Here, \( \varepsilon_0 \) denotes the permittivity of free space. This value indicates how easily electric fields can permeate through the vacuum of space. In our scenario, this uniform electric field exerts an electric force \( F_e \) on the sphere, calculated as the product of the charge \( q \) and the electric field \( E \):

• \( F_e = qE = \frac{q\sigma}{2\varepsilon_0} \)

Understanding electric fields helps explain how charges interact at a distance, guiding us in solving equilibrium problems involving charged objects.

Force equilibrium

In physics, equilibrium refers to a state where all forces acting on an object balance each other, resulting in no net motion. This is crucial for analyzing the small sphere connected to the charged sheet in our exercise. The sphere is in equilibrium when the net force is zero, meaning the forces balance perfectly.Three forces act on the sphere:

• Gravitational force, which pulls the sphere downward with a force of \( mg \)
• Electric force from the charged sheet, pushing horizontally with force \( \frac{q\sigma}{2\varepsilon_0} \)
• Tension in the silk fiber, balancing the other two forces

For the sphere to remain stationary, the electric force must be counteracted by the horizontal component of the tension in the fiber, \( T_x \), and the gravitational force by the vertical component, \( T_y \). When these components satisfy \( T_x = F_e \) and \( T_y = mg \), the forces achieve equilibrium, preventing any motion. Understanding this balance illustrates the concept of equilibrium in force dynamics.

Trigonometry in physics

Trigonometry is a mathematical tool that becomes very practical in physics, especially when dissecting forces into components. In this exercise, the angle \( \theta \) the fiber makes with the vertical is crucial for interpreting the forces.The tension in the fiber \( T \) is split into:

• Horizontal component: \( T_x = T \sin(\theta) \)
• Vertical component: \( T_y = T \cos(\theta) \)

By using the trigonometric identity \( \ an(\theta) = \frac{T_x}{T_y} \), we can relate the angle to the ratio of the forces involved:

• \( \tan(\theta) = \frac{\frac{q\sigma}{2\varepsilon_0}}{mg} = \frac{q\sigma}{2mg\varepsilon_0} \)

Thus, solving for \( \theta \) gives us:

• \( \theta = \arctan\left(\frac{q\sigma}{2mg\varepsilon_0}\right) \)

This angle represents how the electric field and gravitational force interact. Trigonometry allows us to express these complex interactions in a simple formula, making it indispensable for analyzing physical problems.