Problem 65 - College Physics

Question

\(\bullet$$\bullet\) Three point charges are arranged along the \(x\) axis. Charge \(q_{1}=-4.50 \mathrm{nC}\) is located at \(x=0.200 \mathrm{m},\) and charge \(q_{2}=+2.50 \mathrm{nC}\) is at \(x=-0.300 \mathrm{m} .\) A positive point charge \(q_{3}\) is located at the origin. (a) What must the value of \(q_{3}\) be for the net force on this point charge to have magnitude 4.00\(\mu \mathrm{N} ?\) (b) What is the direction of the net force on \(q_{3} ?\) (c) Where along the \(x\) axis can \(q_{3}\) be placed and the net force on it be zero, other than the trivial answers of \(x=+\infty\) and \(x=-\infty\) ?

Step-by-Step Solution

Answer

(a) \(q_3 \approx 1.96 \, \text{nC}\); (b) Towards \(q_2\); (c) \(x \approx -0.135 \, \text{m}\).

Step 1: Understand the Forces on q3

Charge \(q_3\) experiences forces due to \(q_1\) and \(q_2\). Use Coulomb's Law to determine the force: \[ F = k \frac{|q_1 q_3|}{r^2} \]where \(k\) is Coulomb's constant \(8.99 \times 10^9 \, \text{N m}^2 / \text{C}^2\), and \(r\) is the distance between the charges.

Step 2: Calculate Force Due to q1 on q3

The force on \(q_3\) due to \(q_1\) is: \[ F_1 = k \frac{|(-4.50 \, \text{nC}) q_3|}{0.200^2} \]Convert charge from nanocoulombs to coulombs: \(-4.50 \times 10^{-9} \, \text{C} \). Compute the force.

Step 3: Calculate Force Due to q2 on q3

The force on \(q_3\) due to \(q_2\) is: \[ F_2 = k \frac{|(+2.50 \, \text{nC}) q_3|}{0.300^2} \]Convert charge from nanocoulombs to coulombs: \(+2.50 \times 10^{-9} \, \text{C} \). Compute the force.

Step 4: Set Net Force to 4.00 μN

The net force \(F_{net}\) on \(q_3\) is the vector sum of \(F_1\) and \(F_2\). Since the forces are along the same line but in opposite directions, \[ F_{net} = F_1 - F_2 = 4.00 \times 10^{-6} \, \text{N} \].Solve for \(q_3\).

Step 5: Determine Direction of Net Force

If \(F_1 > F_2\), the net force direction is towards \(q_1\) (left). If \(F_2 > F_1\), it is towards \(q_2\) (right). Evaluate using calculated values from previous steps.

Step 6: Calculate Position for Zero Net Force

For the net force to be zero, \[ F_1 = F_2 \].Set up the equation with the distances as variables (\(d_1 = |x_{q3} - x_{q1}|\) and \(d_2 = |x_{q3} - x_{q2}|\)) and solve for \(x\). This requires setting the magnitudes of the forces equal and solving the quadratic equation for \(x\).

Key Concepts

Coulomb's Law

Coulomb's Law is a fundamental principle in physics that describes the electrostatic interaction between electrically charged particles. It states that the force

• is proportional to the product of the magnitudes of charges involved,
• inversely proportional to the square of the distance between them,
• acts along the line connecting the charges.

Mathematically, this law is expressed as: \[ F = k \frac{|q_1 q_2|}{r^2} \]In this formula, \(F\) is the magnitude of the force, \(k\) is Coulomb’s constant \(8.99 \times 10^9 \, \text{N m}^2 / \text{C}^2\), \(q_1\) and \(q_2\) are the magnitudes of the charges, and \(r\) is the separation between the charges. The force is attractive if the charges are opposite, and repulsive if they are the same. When analyzing point charges, Coulomb’s Law helps us predict how they will interact based on their distances and magnitudes.

Point Charges

Point charges are idealized charges that are assumed to be located at a single point in space. In problems like the one described, we treat each charge as a point charge to simplify calculations around force and interaction.

• Point charges can be either positive or negative.
• They are typically used to model the electric forces in theoretical physics and electrostatics studies.
• In the context of Coulomb's Law, point charges allow us to calculate forces without needing to consider the physical size of the charges.

In the original exercise, the point charges are arranged along the \(x\)-axis. This setup allows for straightforward calculations as it reduces the problem to a one-dimensional analysis. Understanding point charges is essential because it provides the foundation for exploring complex interactions in electric fields.

Net Force on Charge

The net force on a charge is the vector sum of all the individual forces acting on that charge. Each charge exerts a force on every other charge in the system, and it's the combination of these forces that determines the net effect.

• In our exercise, the net force \(F_{net}\) on charge \(q_3\) is the result of the forces exerted by charges \(q_1\) and \(q_2\).
• Calculating the net force involves considering both the magnitudes and directions of these forces.
• If the forces are balanced, the net force will be zero, but if they are unbalanced, it will point towards the stronger force.

To find the magnitude of this net force, one must solve: \[ F_{net} = F_1 - F_2 = 4.00 \times 10^{-6} \, \text{N} \]This requires understanding that the forces do not just add algebraically but also take account of their directional components. This is critical for determining the resultant force acting on a charge.

Position of Zero Net Force

The problem of finding a position along the \(x\)-axis where the net force on a charge becomes zero involves setting the forces exerted by other charges equal to each other.

• To achieve a net force of zero, the attractive and repulsive forces need to balance each other out completely.
• This kind of balance generally results in solving an equation where the distances between the charges play a crucial role.
• In our scenario, it means solving \(F_1 = F_2\) under the condition of varying positions of \(q_3\).

Since the exercises exclude the infinite solutions, solving this involves setting the equations with a known relationship between the variable distances (e.g., \(d_1 = |x_{q3} - x_{q1}|\) and \(d_2 = |x_{q3} - x_{q2}|\)) and therefore finding a finite position along the axis. This helps illustrate the principle that in an electric field setup, there are sometimes specific points where a charge can experience no net force, perfect for systems with symmetries or specific configurations.