Problem 39 - Physics

Question

A rectangle has a length of \(2 d\) and a height of \(d\). Each of the following three charges is located at a corner of the rectangle: \(+q_{1}\) (upper left corner), \(+q_{2}\) (lower right corner), and \(-q\) (lower left corner). The net electric field at the (empty) upper right corner is zero. Find the magnitudes of \(q_{1}\) and \(q_{2}\). Express your an swers in terms of \(q\).

Step-by-Step Solution

Answer

The magnitudes are \( q_1 = \frac{q}{5} \) and \( q_2 = q \).

Step 1: Identify the Electric Field Contributions

Firstly, identify the electric field contributions at the upper right corner due to the three charges. We need to consider the electric field due to each charge at the empty corner.

Step 2: Electric Field Due to Charge +q_1

The electric field due to the charge \(+q_1\) at the upper right corner can be computed using Coulomb's Law:\[ E_{q1} = \frac{k \, q_1}{(2d)^2} \]This field is directed to the right because the charge is positive and located diagonally opposite at the upper left corner.

Step 3: Electric Field Due to Charge +q_2

The electric field due to the charge \(+q_2\) at the upper right corner is:\[ E_{q2} = \frac{k \, q_2}{d^2} \]This field is directed downward because it is at the opposite side of the rectangle on the lower right corner.

Step 4: Electric Field Due to Charge -q

The electric field due to the charge \(-q\) at the upper right corner is:\[ E_{q} = \frac{k \, q}{(\sqrt{5}d)^2} \]This field is directed toward the upper left corner, noting the synthesis of vectors in the negative x and positive y directions.

Step 5: Find the Condition for Zero Net Electric Field

For there to be no net electric field at the upper right corner, the sum of the vector fields must equal zero. Therefore:- The x-components must cancel out: \[ q_1 = \frac{q}{5} \] - The y-components must cancel out: \[ q_2 = q \]

Key Concepts

Electric Field

An electric field can be thought of as a force field around a charged object. It dictates how other charges will experience forces in its presence. The strength and direction of these forces are characterized by the electric field vector. The electric field, denoted as \( E \), due to a point charge \( q \) is calculated using:
\[E = \frac{k \, q}{r^2} \]Here:

• \( k \) is the Coulomb's constant \( (8.99 \times 10^9 \, \text{N m}^2/\text{C}^2) \).
• \( q \) is the charge.
• \( r \) is the distance from the charge to the point where the field is being calculated.

The direction of the electric field depends on the sign of the charge:- For a positive charge, the field radiates outward.- For a negative charge, it points inward towards the charge.Understanding an electric field is crucial because it is the medium through which charges exert forces on each other.

Coulomb's Law

Coulomb's Law provides the foundational equation for understanding the force between two point charges in electrostatics. The law states that like charges repel and opposite charges attract each other. The magnitude of the force between two charges is given by:
\[F = \frac{k \, |q_1 q_2|}{r^2} \]

• \( F \) is the force between the charges.
• \( q_1 \) and \( q_2 \) are the values of the charges.
• \( r \) is the separation distance between the charges.

This inverse square law indicates that the force decreases rapidly as the distance between the charges increases. Coulomb's Law is integral to electrostatics because it establishes how charges act at a distance and is used to derive the expressions for electric fields, such as the ones calculated in the original exercise.

Vector Addition

When dealing with multiple charges, as in the exercise, we use vector addition to find the resultant electric field at a point. Each charge exerts an electric field, which has both magnitude and direction.

• Vectors are added by combining their respective components. For electric fields, this includes both the x and y components.
• The x-component of a vector is found using \( E_x = E \cdot \cos(\theta) \).
• The y-component is found using \( E_y = E \cdot \sin(\theta) \).

When the fields due to all charges are known, their vector sum gives the net electric field at a point. This is crucial for determining conditions such as when the net electric field is zero, as in the exercise where the x and y components must cancel each other to produce no resultant field.

Charge Distribution

Understanding how charges are distributed in space, or charge distribution, is essential in predicting the behavior of electric fields. In the exercise, charges are placed at the corners of a rectangle, creating different field strengths at certain points based on their locations and magnitudes.
- The arrangement of "+q_1", "+q_2", and "-q" in the rectangle affects the direction and magnitude of the electric fields at the empty corner. - Depending on whether a distribution is symmetrical or asymmetrical, the resultant electric field may vary significantly. - Charge distribution helps in visualizing and calculating the net electric field by showing how each charge contributes independently to the field at any given point. Understanding charge distribution is vital to correctly applying vector addition, utilizing Coulomb's Law, and calculating the resultant electric field.