Problem 19 - University Physics with Modern Physics

Question

Two point charges are located on the \(y\) -axis as follows: charge \(q_{1}=-1.50 \mathrm{nC}\) at \(y=-0.600 \mathrm{m},\) and charge \(q_{2}=+3.20 \mathrm{nC}\) at the origin \((y=0) .\) What is the total force (magnitude and direction) exerted by these two charges on a third charge \(q_{3}=+5.00 \mathrm{nC}\) located at \(y=-0.400 \mathrm{m} ?\)

Step-by-Step Solution

Answer

The total force is \(2.5865 \times 10^{-6} \text{ N}\) downward.

Step 1: Identify the Problem and Known Variables

The problem involves calculating the force on a third charge, \( q_3 = +5.00 \text{ nC} \), due to two other point charges, \( q_1 = -1.50 \text{ nC} \) located at \( y = -0.600 \text{ m} \) and \( q_2 = +3.20 \text{ nC} \) at the origin \( y = 0 \). We need to find the net force exerted on \( q_3 \) at location \( y = -0.400 \text{ m} \).

Step 2: Use Coulomb's Law to Find Forces

Coulomb's Law states that the force between two point charges is \( F = k_e \frac{|q_1 q_2|}{r^2} \). Here, \( k_e = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \). Calculate the force between \( q_1 \) and \( q_3 \):\[ F_{13} = k_e \frac{|q_1 q_3|}{r_{13}^2} \]with \( r_{13} = 0.200 \text{ m} \). Similarly, find the force between \( q_2 \) and \( q_3 \):\[ F_{23} = k_e \frac{|q_2 q_3|}{r_{23}^2} \]with \( r_{23} = 0.400 \text{ m} \).

Step 3: Calculate Distance Between Charges

The distance \( r_{13} \) between \( q_1 \) at \( y = -0.600 \text{ m} \) and \( q_3 \) at \( y = -0.400 \text{ m} \) is \( r_{13} = 0.200 \text{ m} \), and the distance \( r_{23} \) between \( q_2 \) at the origin \( y = 0 \) and \( q_3 \) is \( r_{23} = 0.400 \text{ m} \).

Step 4: Calculate Force Magnitudes

First, calculate the force due to \( q_1 \):\[ F_{13} = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \times \frac{|-1.50 \times 10^{-9} \times 5.00 \times 10^{-9}|}{(0.200)^2} = 1.6875 \times 10^{-6} \text{ N} \]. For \( q_2 \):\[ F_{23} = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \times \frac{3.20 \times 10^{-9} \times 5.00 \times 10^{-9}}{(0.400)^2} = 0.899 \times 10^{-6} \text{ N} \].

Step 5: Determine Direction of Forces

Since \( q_1 \) and \( q_3 \) have opposite signs, \( F_{13} \) is attractive, pulling \( q_3 \) down the \( y \)-axis. Since \( q_2 \) and \( q_3 \) have like charges, \( F_{23} \) is repulsive, pushing \( q_3 \) down the \( y \)-axis as well.

Step 6: Compute the Net Force

Both forces act in the same direction (downward). Sum the magnitudes: \[ F_{ ext{net}} = F_{13} + F_{23} = 1.6875 \times 10^{-6} \text{ N} + 0.899 \times 10^{-6} \text{ N} = 2.5865 \times 10^{-6} \text{ N} \]. The resultant force is directed downward along the \( y \)-axis.

Key Concepts

Coulomb's Law

Coulomb's Law is a fundamental principle in the field of electrostatics, describing the force exerted between two point charges. In simple terms, it states that the force (F) between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance (r^2) between them. Mathematically, this is expressed as:
\[F = k_e \frac{|q_1 q_2|}{r^2}\]where \(k_e\)is the electric constant, a value that ensures the units of the equation match up. For most calculations in vacuum or air, \(k_e = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\).
Coulomb's Law helps us understand not just the magnitude of the force but also its attractive or repulsive nature:

• If both charges have the same sign, the force is repulsive, pushing them apart.
• If the charges have opposite signs, the force is attractive, pulling them together.

Understanding Coulomb's Law is crucial to solving problems involving point charges, like the one given in your exercise.

Point Charges

Point charges are idealized charges that are considered to have zero dimension and are located at a single point in space. They are not actual physical entities but rather simplifications used to make calculations easier and more manageable.
In real life, charged bodies often have some volume. Still, when the size of the charge is much smaller than the distances involved in the problem, treating them as point charges is a valid approximation. Using point charges allows us to apply Coulomb's Law without considering the charge's shape or volume, focusing only on their magnitudes and positions.
In the problem presented, each charge (\(q_1\), \(q_2\), and \(q_3\)) is considered a point charge:

• \(q_1 = -1.50 \text{ nC}\) located at \(y = -0.600 \text{ m}\)
• \(q_2 = +3.20 \text{ nC}\) at the origin, \(y = 0\)
• \(q_3 = +5.00 \text{ nC}\) placed at \(y = -0.400 \text{ m}\)

This simplification is essential for calculating the forces between these charges as they are treated as interacting through the very centers of these point charges.

Net Force

The concept of net force is central in analyzing the dynamics of objects under the influence of multiple forces. Simply put, the net force is the vector sum of all individual forces acting upon a charge or object. This affects the object's motion.
For point charges, like in your exercise, the forces calculated using Coulomb's Law for each pair of point charges are summed to find the net force. Each force has both magnitude and direction, which requires careful consideration:

• Forces with opposite directions subtract from each other.
• Forces in the same direction add up.

In the problem you are studying:- \(F_{13}\)due to \(q_1\)is attractive (since one charge is negative and one is positive), so it acts downward.- \(F_{23}\)due to \(q_2\)is repulsive (since both charges are positive), also acting downward.
Both forces thus push \(q_3\)down the \(y\)-axis, and the net effect is obtained by summing their magnitudes:\[F_{\text{net}} = F_{13} + F_{23} = 2.5865 \times 10^{-6} \, \text{N}\]The net force acts downward along the\(y\)-axis, moving \(q_3\) in that direction.