Problem 27 - Physics

Question

A tiny ball (mass \(=0.012 \mathrm{~kg}\) ) carries a charge of \(-18 \mu \mathrm{C}\). What electric field (magnitude and direction) is needed to cause the ball to float above the ground?

Step-by-Step Solution

Answer

The electric field needed is \( 6540 \, \text{N/C} \) directed downward.

Step 1: Understand the Problem

The problem involves finding the electric field required for a charged ball to float. This means the electric force must balance the gravitational force acting on the ball.

Step 2: Calculate Gravitational Force

First, calculate the gravitational force acting on the ball using the equation \( F_g = m \cdot g \), where \( m = 0.012 \, \text{kg} \) and \( g = 9.81 \, \text{m/s}^2 \). So, \( F_g = 0.012 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 0.11772 \, \text{N} \).

Step 3: Set Electric Force Equal to Gravitational Force

The electric force \( F_e \) that causes the ball to float must be equal in magnitude and opposite in direction to the gravitational force (to provide an upward force). So, set \( F_e = F_g = 0.11772 \, \text{N} \).

Step 4: Use Electric Force Formula

Use the formula for electric force: \( F_e = q \times E \), where \( q = -18 \, \mu \text{C} = -18 \times 10^{-6} \, \text{C} \) and \( E \) is the electric field. Rearrange to find \( E \): \( E = \frac{F_e}{q} \).

Step 5: Substitute Values to Find Electric Field

Substitute the known values into the equation: \( E = \frac{0.11772 \, \text{N}}{-18 \times 10^{-6} \, \text{C}} \). This gives \( E = -6540 \, \text{N/C} \).

Step 6: Determine Direction of Electric Field

The negative sign in the charge indicates that for the force to be upward (opposing gravity), the electric field must be directed downward. Therefore, the electric field magnitude is \( 6540 \, \text{N/C} \), directed downward.

Key Concepts

Gravitational Force

Gravitational force, commonly referred to as weight, is the force exerted by the Earth's gravity on an object with mass. Every object with mass experiences this force pulling it towards the center of the Earth. The gravitational force is calculated using the equation:

• \( F_g = m \cdot g \)

where \( F_g \) is the gravitational force, \( m \) is the mass of the object, and \( g \) is the acceleration due to gravity, which is approximately \(9.81 \, \text{m/s}^2\).
In our exercise, when a tiny charged ball with mass \(0.012 \, \text{kg}\) is subjected to Earth's gravity, the force it experiences is \(0.11772 \, \text{N}\).
This force works downward, towards the Earth's surface.

Electric Force

The electric force acts upon charged particles due to an electric field. This force can be attractive or repulsive, depending on the nature of the charges involved. When a charged object is placed in an electric field, it experiences a force given by the formula:

• \( F_e = q \times E \)

where \( F_e \) is the electric force, \( q \) is the electric charge of the object, and \( E \) is the electric field strength.
In the exercise, the tiny ball carries a charge of \(-18 \, \mu\text{C}\), which is \(-18 \times 10^{-6} \, \text{C}\).
This negative charge suggests that the direction of the electric field needed to balance the gravitational force and cause the ball to float would be downwards, counteracting the upward electric force.

Equilibrium of Forces

Equilibrium of forces occurs when all the forces acting on an object are balanced, resulting in no net force and therefore no acceleration. For the ball to float, the electric force must exactly counterbalance the gravitational force. This means:

• The magnitude of the electric force \( F_e \) should equal the gravitational force \( F_g \), or \( F_e = F_g \).
• The direction of the forces should be opposite—one upward, one downward.

In this scenario, the electric force will counteract the downward pull of gravity.
To achieve equilibrium for the floating ball, the trajectory required entails creating an electric field strong enough to produce an electric force of \(0.11772 \, \text{N}\) upwards, effectively cancelling out the gravitational pull.
This takes advantage of the negative charge to determine the electric field's downward direction, resulting in a specified electric field strength of \(6540 \, \text{N/C}\).