Question

Point charge \(q_1 = -6.00 \times \space 10^{-6}\) C is on the \(x\)-axis at \(x = -0.200\space \mathrm{m}\). Point charge \(q_2\) is on the \(x\)-axis at \(x = +0.400 \space \mathrm{m}\). Point charge \(q_3 = +3.00 \times \space 10^{-6}\) C is at the origin. What is \(q_2\) (magnitude and sign) (a) if the net force on \(q_3\) is \(6.00 \mathrm{N}\) in the \(+x-\mathrm{direction}\); (b) if the net force on \(q_3\) is \(6.00 \mathrm{N}\) in the \(-x-\mathrm{direction}\)?

Short answer

(a) \( q_2 = +6.00 \times 10^{-6} \) C; (b) \( q_2 = -1.166 \times 10^{-6} \) C.

Step-by-step solution

Identify Force Directions

We know there are two forces acting on charge \( q_3 \) due to charges \( q_1 \) and \( q_2 \). Let's denote these forces as \( F_{13} \) (force due to \( q_1 \)) and \( F_{23} \) (force due to \( q_2 \)). Since \( q_1 = -6.00 \times 10^{-6} \) C is negative and it's located at \(-0.200\space \mathrm{m}\), it will attract \( q_3 \), so \( F_{13} \) points toward \( q_1 \) along the negative \( x \)-direction.\( q_2 \) is at \( +0.400 \space \mathrm{m}\), so the direction of \( F_{23} \) will depend on the sign of \( q_2 \). If \( q_2 \) is positive, it repels and points toward the positive \( x \)-direction. If \( q_2 \) is negative, it attracts and points towards negative \( x \)-direction.

Apply Coulomb's Law

Coulomb's Law states that the magnitude of the force between two point charges is \( F = k \frac{|q_1 q_2|}{r^2} \), where \( k = 8.99 \times 10^9 \, \mathrm{N}\,\mathrm{m}^2/\mathrm{C}^2 \) and \( r \) is the distance between the charges. - The distance from \( q_1 \) to \( q_3 \) is \( 0.200 \, \mathrm{m} \).- The distance from \( q_2 \) to \( q_3 \) is \( 0.400 \, \mathrm{m} \).

Calculate Force from q_1 (F_13)

The force due to \( q_1 \) on \( q_3 \) is:\[ F_{13} = \frac{k \cdot |q_1 \cdot q_3|}{(0.200)^2} \]\[ F_{13} = \frac{8.99 \times 10^9 \cdot 6.00 \times 10^{-6} \cdot 3.00 \times 10^{-6}}{0.0400} \]\[ F_{13} = 4.05 \, \mathrm{N} \]This force is in the negative \( x \)-direction.

Set Up Force Equation for Part (a)

Since the net force on \( q_3 \) is \( +6.00 \mathrm{N} \), the equation for the forces is:\[ F_{23} - F_{13} = 6.00 \]Substitute \( F_{13} = 4.05 \) N:\[ F_{23} - 4.05 = 6.00 \]So:\[ F_{23} = 10.05 \]

Solve for q_2 in Part (a)

Calculate \( q_2 \) using the expression for \( F_{23} \):\[ F_{23} = \frac{k \cdot |q_2 \cdot q_3|}{(0.400)^2} \]Substitute \( F_{23} = 10.05 \) N and \( q_3 = 3.00 \times 10^{-6} \) C:\[ 10.05 = \frac{8.99 \times 10^9 \cdot |q_2| \cdot 3.00 \times 10^{-6}}{0.160} \]\[ |q_2| = \frac{10.05 \times 0.160}{8.99 \times 10^9 \times 3.00 \times 10^{-6}} \]\[ |q_2| = 6.00 \times 10^{-6} \] C.Since \( F_{23} \) is positive, \( q_2 \) is positive \((+6.00 \times 10^{-6} \text{ C})\).

Set Up Force Equation for Part (b)

Now, the net force on \( q_3 \) is \( -6.00 \mathrm{N} \), so the equation is:\[ F_{23} - F_{13} = -6.00 \]Substitute \( F_{13} = 4.05 \) N:\[ F_{23} - 4.05 = -6.00 \]So:\[ F_{23} = -1.95 \]

Solve for q_2 in Part (b)

Use the expression for \( F_{23} \):\[ F_{23} = \frac{k \cdot |q_2 \cdot q_3|}{(0.400)^2} \]Substitute \( F_{23} = -1.95 \) N (as magnitude):\[ 1.95 = \frac{8.99 \times 10^9 \cdot |q_2| \cdot 3.00 \times 10^{-6}}{0.160} \]\[ |q_2| = \frac{1.95 \times 0.160}{8.99 \times 10^9 \times 3.00 \times 10^{-6}} \]\[ |q_2| = 1.166 \times 10^{-6} \] C.Since \( F_{23} \) is negative, \( q_2 \) is negative \((-1.166 \times 10^{-6} \text{ C})\).

Key concepts

Coulomb's Law

Coulomb's Law is fundamental to understanding the forces between electric charges. It describes how the magnitude of the electric force between two charges is proportional to the product of their absolute values and inversely proportional to the square of the distance between them. The formula is represented as:
\[ F = k \frac{|q_1 q_2|}{r^2} \]
where:

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

This law helps us calculate the force's size, but to understand its direction, we must also consider the nature of the charges, whether they are positive or negative. Like charges repel each other, while opposite charges attract. This principle is consistent with the exercise's analysis of forces between point charges at different positions on a one-dimensional axis.

Point Charges

Point charges are idealized models of charges that are assumed to occupy a single point in space and have no spatial extent. In physics, they are very useful because they simplify the analysis of electric forces which can otherwise become quite complex.
The concept of point charges was used to place charges \( q_1 \), \( q_2 \), and \( q_3 \) on the x-axis at specific positions in the exercise. These charges exert forces on each other that are easy to compute using the formula derived from Coulomb's Law.
In practical terms, a point charge is an approximation to treat real-world objects when the distance between them is significantly larger than their size, making the effect of the objects' dimensions negligible. Thus, the concept of point charges allows students and scientists to predict and calculate interactions without unnecessary complications stemming from the object's physical dimensions.

Vector Analysis

Vector analysis is crucial when dealing with forces because forces are vector quantities, meaning they have both magnitude and direction. Understanding how forces interact involves breaking them down into their vector components and adding them accordingly.
In the exercise, vector analysis was essential to determine the net force on \( q_3 \). As we've seen, it involves considering forces like \( F_{13} \) and \( F_{23} \), where directions need to be aligned or opposed based on the charges' properties and positions.

• Direction matters: Attraction and repulsion depend on signs of the charges.
• Superposition principle applies: The net force is the vector sum of all individual forces.
• Sign conventions must be understood to predict correctly how forces interact.

By mastering vector analysis, students can solve complex force systems, predicting the movement and interaction of charged particles, making it indispensable in physics problem-solving.