Problem 71 - College Physics

Question

\(\bullet$$\bullet\) A small 12.3 g plastic ball is tied to a very light 28.6 \(\mathrm{cm}\) string that is attached to the vertical wall of a room. (See Figure \(17.58 . ) \mathrm{A}\) uniform horizontal electric field exists in this room. When the ball has been given an excess charge of \(-1.11 \mu \mathrm{C},\) you observe that it remains suspended, with the string making an angle of \(17.4^{\circ}\) with the wall. Find the magnitude and direction of the electric field in the room.

Step-by-Step Solution

Answer

The electric field magnitude is approximately 33953 N/C, directed towards the wall.

Step 1: Analyze the Forces

In this scenario, the plastic ball is in equilibrium. Two main forces act on the ball: gravitational force and electric force. The tension in the string provides a vertical component to balance gravity and a horizontal component to counteract the electric force. The gravitational force is given by \[ F_g = mg, \] where \( m \) is the mass and \( g \) is the acceleration due to gravity.

Step 2: Calculate the Gravitational Force

First, convert the mass into kilograms: \( m = 12.3 \text{ g} = 0.0123 \text{ kg} \).Then, calculate the gravitational force using \[ F_g = mg = 0.0123 \times 9.8 \approx 0.12054 \text{ N}. \]

Step 3: Establish Equilibrium Conditions

As the system is in equilibrium, the vertical component of the tension must balance the gravitational force, and the horizontal component of the tension must balance the electric force. Define \( T \) as the tension in the string.

Step 4: Resolve Tension into Components

Using trigonometry in the right-angled triangle formed by the tension, angles, and walls:- The vertical component is \( T_y = T \cos(17.4^{\circ}) \).- The horizontal component is \( T_x = T \sin(17.4^{\circ}) \).Since the ball is at rest, \( T_y = F_g \) and \( T_x = F_e \), where \( F_e \) is the electric force.

Step 5: Express the Electric Force

Since the horizontal component of tension equals the electric force, \[ F_e = T \sin(17.4^{\circ}). \]The electric force is also defined by \( F_e = qE \), with \( q = -1.11 \times 10^{-6} \text{ C} \), where \( q \) is the charge and \( E \) is the electric field strength.

Step 6: Relate Tension and Gravitational Force

Since \( T_y = F_g \), express \( T \) using the gravitational force: \[ T = \frac{F_g}{\cos(17.4^{\circ})}. \]

Step 7: Substitute and Solve for Electric Field

Use the relationship: \[ T = \frac{0.12054}{\cos(17.4^{\circ})}. \]Substitute this into the equation for electric force: \[ F_e = \frac{0.12054 \sin(17.4^{\circ})}{\cos(17.4^{\circ})}. \] Given \( F_e = qE \), substitute to find \( E \): \[ qE = \frac{0.12054 \sin(17.4^{\circ})}{\cos(17.4^{\circ})}. \]Thus, \[ E = \frac{0.12054 \tan(17.4^{\circ})}{-1.11 \times 10^{-6}}. \]

Step 8: Calculate the Electric Field

Compute the electric field strength: \[ E = \frac{0.12054 \times 0.313}{-1.11 \times 10^{-6}} \approx -33953 \text{ N/C}. \]The negative sign indicates the field direction is opposite to the charge, towards the wall.

Key Concepts

Equilibrium of Forces

A plastic ball tied to a string remains at rest, forming an angle with the wall due to the balance of forces acting upon it. This balance is called equilibrium. In equilibrium, all the forces cancel each other out.

• Vertical forces: The weight of the ball pulls it down.
• Horizontal forces: The electric field exerts a force on the charged ball.

The string experiences tension, which supplies a counteracting force in each direction:

• Vertically, the tension balances the gravitational force.
• Horizontally, it equally opposes the electric force.

Tension in String

Tension is the force that the string experiences as it holds the ball in place. It acts in the direction along the string, pulling equally from both ends.

To understand tension's role, think of it providing two components:

• The vertical component counterbalances gravity, ensuring the ball does not fall.
• The horizontal component holds against the electric push, keeping the ball static.

Defining tension (\( T \)):

• Vertical component: \( T_y = T \cos(17.4^{\circ}) \)
• Horizontal component: \( T_x = T \sin(17.4^{\circ}) \)

When the ball remains unmoved, it implies:

• \( T_y = F_g \)
• \( T_x = F_e \)

Gravitational Force

All objects with mass are pulled towards the Earth due to gravitational force. Calculating this force helps us understand part of what keeps the ball in place.

The formula for gravitational force is given by:

• \[ F_g = mg \]

where:

• \( m \) is the mass (in kilograms).
• \( g = 9.8 \, \text{m/s}^2 \), the acceleration due to gravity.

For our plastic ball:

• \( m = 0.0123 \, \text{kg} \)
• Gravitational force is calculated as\( F_g = 0.0123 \times 9.8 \approx 0.12054 \, \text{N}. \)

Electric Force

Electric force is a result of the charged ball experiencing an electric field. It acts horizontally, trying to move the ball towards the opposite charge.

Given formula: \( F_e = qE \),

• \( q \) represents the charge of the ball.
• \( E \) stands for the electric field strength.

Calculate\( F_e \) using the horizontal tension component:

• \( F_e = T \sin(17.4^{\circ}) \)

This relationship depicts that:

• Electric force is computed through the field intensity and charge value.

Angle of Tension

The angle the string creates with the wall directly influenced how we resolve forces. Calculating this angle ensures accurate force component breakdowns.

Implications of angle:

• It defines how tension splits into vertical and horizontal parts.

Trigonometry helps:

• Vertical tension: \( T_y = T \cos(17.4^{\circ}) \)
• Horizontal tension: \( T_x = T \sin(17.4^{\circ}) \)

Understanding this tension angle helps ensure accurate calculations for forces acting on the object, leading to accurate electric field determination.