Problem 20 - University Physics with Modern Physics

Question

Two point charges are placed on the \(x\)-axis as follows: Charge \(q_1 = +\)4.00 nC is located at \(x =\) 0.200 m, and charge \(q_2 = +\)5.00 nC is at \(x = -\)0.300 m . What are the magnitude and direction of the total force exerted by these two charges on a negative point charge \(q_3 = -\)6.00 nC that is placed at the origin?

Step-by-Step Solution

Answer

The total force on \( q_3 \) is directed to the left with a magnitude of approximately 2.97 µN.

Step 1: Understand Coulomb's Law

Coulomb's Law defines the electrostatic force between two point charges. The force \[ F \] between two charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by:\[F = k \frac{|q_1 q_2|}{r^2}\]where \( k \) is Coulomb's constant, \( k = 8.99 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2 \). The force is attractive if the charges are of opposite signs and repulsive if they are of the same sign.

Step 2: Calculate Force Between q1 and q3

We determine the force exerted by \( q_1 = +4.00 \, \text{nC} \) on \( q_3 = -6.00 \, \text{nC} \), which is located at the origin:\[ r_{13} = |x_1 - x_3| = |0.200 - 0| = 0.200 \, \text{m} \]The magnitude of the force is:\[ F_{13} = k \frac{|q_1 q_3|}{r_{13}^2} = \left(8.99 \times 10^9 \right) \cdot \frac{|4.00 \times 10^{-9} \times (-6.00 \times 10^{-9})|}{(0.200)^2} \] Calculate \( F_{13} \).

Step 3: Calculate Force Between q2 and q3

Now determine the force exerted by \( q_2 = +5.00 \, \text{nC} \) on \( q_3 = -6.00 \, \text{nC} \):\[ r_{23} = |x_2 - x_3| = |-0.300 - 0| = 0.300 \, \text{m} \]The magnitude of the force is:\[ F_{23} = k \frac{|q_2 q_3|}{r_{23}^2} = \left(8.99 \times 10^9 \right) \cdot \frac{|5.00 \times 10^{-9} \times (-6.00 \times 10^{-9})|}{(0.300)^2} \] Calculate \( F_{23} \).

Step 4: Determine Direction of Each Force

Since \( q_3 \) is negative, the force \( F_{13} \) from \( q_1 \) (positive) will attract \( q_3 \) towards \( q_1 \), i.e., to the right. The force \( F_{23} \) from \( q_2 \) (also positive) will attract \( q_3 \) towards \( q_2 \), i.e., to the left.

Step 5: Calculate Total Force on q3

The total force on \( q_3 \) is the vector sum of \( F_{13} \) and \( F_{23} \). Because \( F_{13} \) is acting to the right and \( F_{23} \) to the left, use\[ F_{ ext{total}} = F_{13} - F_{23} \]due to opposite directions, and calculate the resultant force.

Step 6: Evaluate Magnitude and Direction

Calculate the magnitude of \( F_{ ext{total}} \) from the previous step. The direction will be to the right if \( F_{ ext{total}} > 0 \) or to the left if \( F_{ ext{total}} < 0 \). This gives both magnitude and direction of the total electrostatic force on \( q_3 \).

Key Concepts

Electrostatic Force

Electrostatic force is the interaction between two charged particles. This force can be either attractive or repulsive, depending on the signs of the charges involved. According to Coulomb's Law, the magnitude of the 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 mathematically expressed as: \[ F = k \frac{|q_1 q_2|}{r^2} \] where \( F \) is the electrostatic force, \( |q_1 q_2| \) is the absolute product of the charges, \( r \) is the distance between the charges, and \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \). Coulomb's Law helps us understand how charges interact in a vacuum or free space, and is essential for determining the forces acting upon charged bodies.

Point Charges

A point charge is an idealized model of a particle with an electric charge, assumed to be infinitesimally small so it is located at a single point in space. In practice, this concept is useful for simplifying the complex interactions of charged objects, especially when the size of the charged objects is small compared to the distance between them. In our problem:

• We have three point charges: \( q_1 = +4.00 \text{ nC} \), \( q_2 = +5.00 \text{ nC} \), and \( q_3 = -6.00 \text{ nC} \), each placed along a linear path on the x-axis.
• The charges are considered point-like so we can apply Coulomb's Law directly without having to consider the size or distribution of charge.

By treating charges as point charges, calculations become more straightforward, focusing purely on the charges' magnitudes and separation distance.

Vector Sum

When dealing with multiple forces, as in the case of this exercise, it is essential to consider the vector nature of forces. A force is not just a magnitude but also a direction. In this problem, the forces exerted by \( q_1 \) and \( q_2 \) on \( q_3 \) must be combined to determine the net force acting on \( q_3 \). This is done using vector addition:

• The force from \( q_1 \) is directed towards \( q_1 \) (to the right).
• The force from \( q_2 \) is directed towards \( q_2 \) (to the left).
• The vector sum takes into account these opposing directions. Therefore, the net force is calculated by subtracting the two magnitudes due to their opposite directions.

The concept of vector sum helps in determining the precise effect of multiple forces acting on a body, indicating both the resultant direction and magnitude.

Force Direction

Force direction plays a crucial role in understanding how charges will move under the influence of electrostatic forces. In the given scenario:

• The negative charge \( q_3 \), placed at the origin, experiences forces due to the positive charges \( q_1 \) and \( q_2 \).
• As per Coulomb's Law, because \( q_3 \) is negative, it will be attracted towards both positive charges, with \( F_{13} \) acting to the right and \( F_{23} \) acting to the left.
• The resulting motion of the charge is determined by the direction and relative magnitude of these forces.

Understanding force direction can help in predicting the movement and equilibrium position of the charges, which can significantly impact practical applications such as in designing electric circuits and systems.