Problem 84 - 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 (\(q\sigma/2mg\epsilon_0\)) with the vertical sheet.

Step-by-Step Solution

Answer

The angle \( \theta \) is \( \arctan\left(\frac{q\sigma}{2mg\epsilon_0}\right) \).

Step 1: Draw a Free-Body Diagram

First, identify all the forces acting on the sphere. The sphere experiences three forces: 1) the gravitational force downwards, \( F_g = mg \), 2) the electric force due to the charged sheet, \( F_e \), and 3) the tension, \( T \), in the fiber, which acts along the fiber.

Step 2: Determine the Electric Force

The electric field \( E \) created by the sheet with charge density \( \sigma \) is given by the formula: \( E = \frac{\sigma}{2\epsilon_0} \). The electric force on the sphere is then \( F_e = qE = \frac{q\sigma}{2\epsilon_0} \). This force acts perpendicular to the surface of the charged sheet.

Step 3: Analyze Forces in Equilibrium

In equilibrium, the net force in both horizontal and vertical directions is zero. Horizontally, the only force component is due to the electric force. Vertically, the tension in the wire has a component that balances the gravitational force.

Step 4: Set up the Force Balance Equations

Write the force balance equations. Horizontally: \( T \sin(\theta) = \frac{q\sigma}{2\epsilon_0} \). Vertically: \( T \cos(\theta) = mg \). These equations ensure there is no net force in either direction.

Step 5: Solve for the Angle

Divide the horizontal force equation by the vertical force equation: \[ \frac{T \sin(\theta)}{T \cos(\theta)} = \frac{\frac{q\sigma}{2\epsilon_0}}{mg} \]This simplifies to \( \tan(\theta) = \frac{q\sigma}{2mg\epsilon_0} \). Take the arctan of both sides to find the angle: \( \theta = \arctan\left(\frac{q\sigma}{2mg\epsilon_0}\right) \).

Key Concepts

Electric field

An electric field is a zone around a charged object where other charges can feel a force. Imagine it like an invisible aura surrounding the object. It is produced by charges, and it influences other charges that enter this zone.

In our problem, the electric field is generated by a large vertical insulating sheet with a positive surface charge density \(\sigma\). The fundamental property of this field is expressed as the force per unit charge. This makes it easy to calculate how much force a charge will feel once inside this field.

For a charged plane, we use the formula:

• \(E = \frac{\sigma}{2\epsilon_0}\)

This equation indicates that the electric field (\(E\)) from the sheet depends on the charge density (\(\sigma\)) and the permittivity of free space (\(\epsilon_0\)).
This field acts perpendicular to the sheet, providing a directional context for the problem.

Coulomb's law

Coulomb's Law helps us understand how charges interact with each other. It explains the relationship between two charged objects. The law states that like charges repel each other, while opposite charges attract. The size of this force depends on the amount of charge and the distance between them.

For our exercise, while Coulomb's Law directly calculates forces between point charges, it establishes the basis for understanding electric forces such as the one between a charged sphere and a charged sheet. Here, the electric force experienced by the sphere can be calculated as:

• \(F_e = qE\)

Substituting the electric field (\(E\)), we get:

• \(F_e = \frac{q\sigma}{2\epsilon_0}\)

This force acts perpendicular to the sheet and pushes the sphere away from it due to the repulsion between like charges.

Equilibrium of forces

In physics, equilibrium occurs when all forces acting on an object are balanced. This means that the net force is zero — the object either remains at rest or moves with constant velocity. In this problem, the charged sphere needs to be in equilibrium.

We organize this by looking at the forces horizontally and vertically:

• Horizontally: The electric force (\(F_e\)) acts as the only force, since it is perpendicular to the charge sheet.
• Vertically: Gravity acts downward, and the tension in the silk fiber counters it. The tension is in the opposite direction vertically with a component that matches the gravitational force.

To maintain this balance, the tension's horizontal component is set to the electric force:\( T \sin(\theta) = \frac{q\sigma}{2\epsilon_0} \), and the vertical component to gravity: \ (T \cos(\theta) = mg)\. These conditions prevent the sphere from moving in any direction.

Gravitational force

Gravitational force is one of the fundamental forces in nature. It attracts any two masses towards each other. In our problem, this force acts on the charged sphere, pulling it downwards towards the Earth. This is expressed simply by the formula:

• \(F_g = mg\)

Here, \(m\) represents the mass of the sphere and \(g\) the acceleration due to gravity (approximately 9.8 m/s² on Earth).

When the sphere is hanging by the silk fiber, this gravitational force is counteracted by the vertical component of the tension in the fiber. This balance allows the sphere to remain in equilibrium, ensuring that it doesn’t fall off or rise up. Gravity thus plays a crucial role in the delicate balance of forces necessary to satisfy equilibrium conditions in our exercise.