Problem 82 - University Physics with Modern Physics

Question

CP Two tiny spheres of mass 6.80 \(\mathrm{mg}\) carry charges of equal magnitude, \(72.0 \mathrm{nC},\) but opposite sign. They are tied to the same ceiling hook by light strings of length 0.530 \(\mathrm{m}\) . When a horizontal uniform electric field \(E\) that is directed to the left is turned on, the spheres hang at rest with the angle \(\theta\) between the strings equal to \(50.0^{\circ}(\) Fig. .21 .82\()\) . (a) Which ball (the one on the right or the one on the left) has positive charge? (b) What is the magnitude \(E\) of the field?

Step-by-Step Solution

Answer

(a) The right sphere is positive. (b) The field magnitude is approximately 1.12 × 10⁴ N/C.

Step 1: Understanding the System

We have two spheres with equal and opposite charges, suspended by strings of equal length, forming an angle of 50°. When an electric field is applied, it causes a horizontal force on the charges, leading to a symmetrical configuration.

Step 2: Determining Charge Polarity (a)

With the electric field directed to the left, positive charges experience a force in the same direction. Since the spheres remain at rest, and given the angles formed by the strings, the left sphere repels the positive electric field. Thus, the right sphere must have a positive charge.

Step 3: Break Down Forces

Identify forces acting on one sphere: gravitational force downwards \( F_g = mg \), tension in the string \( T \), and electric force \( F_e = qE \). These need to be in equilibrium both horizontally and vertically.

Step 4: Vertical Force Equilibrium

For the vertical component, the tension provides the force to counteract gravity. The vertical component of tension (\( T \cos\theta \)) must equal the gravitational force \( mg \).

Step 5: Horizontal Force Equilibrium

Horizontally, the electric force \( F_e \) is counteracted by the horizontal component of tension (\( T \sin\theta \)). Thus we have \( qE = T \sin\theta \).

Step 6: Calculating Electric Field (b)

First, express the tension using the vertical force equation: \( T = \frac{mg}{\cos\theta} \). Substitute into the horizontal equation: \( qE = \frac{mg}{\cos\theta} \sin\theta \). Solve for electric field \( E \) to find: \( E = \frac{mg \tan\theta}{q} \).

Step 7: Substitute Values and Solve

Use given values, mass \( m = 6.80 \times 10^{-6} \text{ kg} \), charge \( q = 72.0 \times 10^{-9} \text{ C} \), \( \theta = 25° \) (half of 50° for individual string analysis):\[ E = \frac{6.80 \times 10^{-6} \times 9.8 \times \tan(25°)}{72.0 \times 10^{-9}} \]Calculate to find \( E \approx 1.12 \times 10^4 \text{ N/C} \).

Key Concepts

Charge Polarity

Charge polarity refers to the characteristic of electric charges that determine their interaction with an electric field. In this exercise, each sphere has an equal magnitude of charge but opposite signs. This means one sphere charges is positive, while the other is negative. The electric field applied is directed towards the left, affecting the forces exerted on the spheres.
When a positive charge is exposed to an electric field, it experiences a force in the direction of the field. Conversely, a negative charge will feel a force opposite to the field direction. In our problem, the electric field pushes the positive charged sphere to the left, while the negative one is drawn to the right.
Understanding this principle allows us to identify the charge polarity of the spheres. Since the position of the spheres remains fixed, the one on the right is repelled by the field, indicating it carries the positive charge.

Equilibrium of Forces

In physics, equilibrium of forces occurs when all forces acting on an object balance perfectly, resulting in no net movement. For the spheres in this exercise, equilibrium is key because it keeps them stationary, despite the presence of multiple forces acting upon them, like tension, electric, and gravitational forces.
In the vertical direction, the gravitational force pulling down is balanced by the upward component of the string tension. Horizontally, the electric force exerted by the field is balanced by the horizontal component of tension.

• The vertical equilibrium condition states: the tension's vertical component exactly balances the gravitational force.
• The horizontal equilibrium condition requires the tension's horizontal component to counteract the electric force.

Both conditions must be satisfied to ensure the spheres remain hanging at rest, thereby demonstrating equilibrium.

Electric Force

Electric force is the force experienced by a charged object within an electric field. The magnitude of this force is calculated using the formula: \( F_e = qE \), where \( q \) is the charge and \( E \) is the electric field strength.
In this problem, the electric force pulls the charged spheres in a direction that is dependent on the charge polarity and the external electric field. The positive charge is pulled along with the field direction, while the negative charge is pulled oppositely. This force influences the angle formed by the strings of the spheres since each sphere reacts to these forces differently.
Obtaining the electric force is essential to maintaining equilibrium, as it is balanced precisely by the horizontal component of the tension in the strings.

Gravitational Force

Gravitational force acts on every object with mass, pulling it downward toward the center of the Earth. The calculation of gravitational force is straightforward, given by \( F_g = mg \), where \( m \) is the mass of the object, and \( g \) is the acceleration due to gravity, typically \( 9.8 \, \text{m/s}^2 \) on Earth's surface.
For our spheres, this force competes with the tensions in the strings. The vertical component of tension counters the gravitational pull to maintain vertical equilibrium. Without enough tension, the spheres would drop; thus, it plays an essential role in keeping them hanging at rest.
Gravitational force is omnipresent and acts downwards, which dictates that any counteracting force—such as tension in the string—must be correspondingly structured to prevent rotational movement.