Problem 31 - Physics

Question

Two charges are placed on the \(x\) axis. One of the charges \(\left(q_{1}=+8.5 \mu \mathrm{C}\right)\) is at \(x_{1}=+3.0 \mathrm{~cm}\) and the other \(\left(q_{2}=-21 \mu \mathrm{C}\right)\) is at \(x_{1}=+9.0 \mathrm{~cm} .\) Find the net electric field (magnitude and direction) at (a) \(x=0 \mathrm{~cm}\) and (b) \(x=+6.0 \mathrm{~cm}\).

Step-by-Step Solution

Answer

At \( x = 0 \, \mathrm{cm} \), the net electric field is \( 10.827 \times 10^6 \mathrm{N/C} \) to the left. At \( x = +6.0 \, \mathrm{cm} \), it is \( 12.503 \times 10^6 \mathrm{N/C} \) to the left.

Step 1: Understanding the Problem

We have two point charges on the x-axis and need to determine the net electric field due to these charges at two specific points on the x-axis. The charges are: \( q_1 = +8.5 \, \mu \mathrm{C} \) at \( x_1 = +3.0 \, \mathrm{cm} \) and \( q_2 = -21 \, \mu \mathrm{C} \) at \( x_2 = +9.0 \, \mathrm{cm} \). We will calculate the electric field at \( x = 0 \, \mathrm{cm} \) and \( x = +6.0 \, \mathrm{cm} \).

Step 2: Formulating the Electric Field Equation

The electric field \( E \) due to a point charge \( q \) at a distance \( r \) is given by the formula: \[ E = \frac{k |q|}{r^2} \] where \( k = 8.99 \times 10^9 \, \mathrm{N \cdot m^2/C^2} \) is Coulomb's constant. The direction of \( E \) is away from the charge if the charge is positive and towards the charge if it is negative.

Step 3: Calculating Electric Field at \( x = 0 \, \mathrm{cm} \)

Calculate the electric field due to each charge at \( x = 0 \, \mathrm{cm} \).1. **Charge \( q_1 \):** - Distance \( r_1 = 3.0 \, \mathrm{cm} = 0.03 \, \mathrm{m} \) - \( E_1 = \frac{k \cdot 8.5 \times 10^{-6}}{0.03^2} = \frac{8.99 \times 10^9 \cdot 8.5 \times 10^{-6}}{0.03^2} = 8.494 \times 10^6 \, \mathrm{N/C} \) directed to the left (negative x-direction).2. **Charge \( q_2 \):** - Distance \( r_2 = 9.0 \, \mathrm{cm} = 0.09 \, \mathrm{m} \) - \( E_2 = \frac{k \cdot 21 \times 10^{-6}}{0.09^2} = \frac{8.99 \times 10^9 \cdot 21 \times 10^{-6}}{0.09^2} = 2.333 \times 10^6 \, \mathrm{N/C} \) directed to the left (negative x-direction).**Net Electric Field:**- \( E = E_1 + E_2 = 8.494 \times 10^6 + 2.333 \times 10^6 = 10.827 \times 10^6 \, \mathrm{N/C} \) to the left.

Step 4: Calculating Electric Field at \( x = +6.0 \, \mathrm{cm} \)

Calculate the electric field due to each charge at \( x = +6.0 \, \mathrm{cm} \).1. **Charge \( q_1 \):** - Distance \( r_1 = 3.0 \, \mathrm{cm} = 0.03 \, \mathrm{m} \) - \( E_1 = \frac{k \cdot 8.5 \times 10^{-6}}{0.03^2} = 8.494 \times 10^6 \, \mathrm{N/C} \) directed to the right (positive x-direction).2. **Charge \( q_2 \):** - Distance \( r_2 = 3.0 \, \mathrm{cm} = 0.03 \, \mathrm{m} \) - \( E_2 = \frac{k \cdot 21 \times 10^{-6}}{0.03^2} = 20.997 \times 10^6 \, \mathrm{N/C} \) directed to the left (negative x-direction).**Net Electric Field:**- \( E = E_2 - E_1 = 20.997 \times 10^6 - 8.494 \times 10^6 = 12.503 \times 10^6 \, \mathrm{N/C} \) to the left.

Key Concepts

Coulomb's Law

Coulomb's Law is fundamental to understanding electric fields and the forces between charged particles. It states that the magnitude of the electric force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. The formula is given by:\[ F = k \frac{|q_1 q_2|}{r^2} \]where:

• \( F \) is the magnitude of the force between the charges.
• \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \mathrm{N \cdot m^2/C^2} \).
• \( q_1 \) and \( q_2 \) are the values of the charges.
• \( r \) is the distance between the centers of the two charges.

It's important to note that the force is attractive if the charges are of opposite signs and repulsive if the charges are of the same sign. This law helps us calculate not only the force between charges but also understand the interaction of charged objects in different setups.

Point Charge

A point charge can be imagined as a charge located at a single point in space. In reality, all charges have dimensions, but when they are small enough compared to the distances involved in a problem, we can model them as point charges for simplicity. The electric field generated by a point charge is considered radially symmetric and always extends in uniform spherical patterns around the charge.The electric field \( E \) due to a point charge \( q \) at a distance \( r \) is defined by the equation:\[ E = \frac{k |q|}{r^2} \]

• The electric field emanates outwards from a positive point charge.
• It converges inward towards a negative point charge.

This field represents how a point charge can exert a force on nearby charges. It's a crucial concept for solving problems involving charge distributions and helps simplify complex interactions among multiple charges.

Electric Field Direction

The direction of an electric field is defined as the direction of the force that would be experienced by a positive test charge placed in the field. This direction depends on the sign of the charges involved:

• For a positive charge, the electric field points away from the charge.
• For a negative charge, the electric field points towards the charge.

Knowing the direction of the electric field is crucial for predicting how other charges will move. In problems like those involving multiple charges, you consider the contributions to the electric field from each charge separately, then use vector addition to find the net electric field. The net direction is determined by the relative magnitudes and directions of individual electric fields contributed by each charge. This concept helps us not only envision how the charge interacts with its surroundings but also assists in solving complex configurations of charges.