Problem 72 - University Physics with Modern Physics

Question

A charge \(q_{1}=+5.00 \mathrm{nC}\) is placed at the origin of an \(x y\) - coordinate system, and a charge \(q_{2}=-2.00 \mathrm{nC}\) is placed on the positive \(x\) -axis at \(x=4.00 \mathrm{cm} .\) (a) If a third charge \(q_{3}=+6.00 \mathrm{nC}\) is now placed at the point \(x=4.00 \mathrm{cm}, y=3.00 \mathrm{cm},\) find the \(x\) - and \(y\) -components of the total force exerted on this charge by the other two. (b) Find the magnitude and direction of this force.

Step-by-Step Solution

Answer

The total force on \(q_3\) is \(1.02 \times 10^{-4}\, \text{N}\) at \(-32.0^\circ\) below the x-axis.

Step 1: Identify Forces on Charge q3

Analyze the forces acting on charge \(q_3\) from charges \(q_1\) and \(q_2\). Calculate the distance between \(q_3\) and the other two charges using the Pythagorean theorem.For \(q_3\) to \(q_1\): Distance \(r_{13} = \sqrt{(4.00\, \text{cm})^2 + (3.00\, \text{cm})^2} = \sqrt{16 + 9} = 5.00\, \text{cm}\). For \(q_3\) to \(q_2\): Since \(q_2\) is on the x-axis, the distance is \(r_{23} = 3.00\, \text{cm}\) (straight line on the y-axis).

Step 2: Calculate Force from q1 on q3

Use Coulomb's Law to find the force exerted by \(q_1\) on \(q_3\): \[ F_{13} = k \frac{|q_1 q_3|}{r_{13}^2} \] Where \(k = 8.99 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2\), \(q_1 = 5.00 \times 10^{-9}\, \text{C}\), \(q_3 = 6.00 \times 10^{-9}\, \text{C}\) and \(r_{13} = 0.05 \text{ m}\).\[ F_{13} = 8.99 \times 10^9 \times \frac{5 \times 10^{-9} \cdot 6 \times 10^{-9}}{(0.05)^2} = 1.08 \times 10^{-4} \text{ N} \]

Step 3: Calculate Force from q2 on q3

Use Coulomb's Law for the force exerted by \(q_2\) on \(q_3\): \[ F_{23} = k \frac{|q_2 q_3|}{r_{23}^2} \]Where \(q_2 = -2.00 \times 10^{-9} \text{ C}\) and \(r_{23} = 0.03 \text{ m}\).\[ F_{23} = 8.99 \times 10^9 \times \frac{2 \times 10^{-9} \times 6 \times 10^{-9}}{(0.03)^2} = 1.20 \times 10^{-4} \text{ N} \]

Step 4: Resolve Forces into Components

For the force \(F_{13}\), resolve it into \(x\) and \(y\) components using:- \( F_{13x} = F_{13} \cdot \frac{4}{5} = 8.64 \times 10^{-5}\, \text{N} \)- \( F_{13y} = F_{13} \cdot \frac{3}{5} = 6.48 \times 10^{-5}\, \text{N} \)For the force \(F_{23}\), since it's along a vertical line towards \(q_2\), - \( F_{23x} = 0\) (no x component)- \( F_{23y} = -1.20 \times 10^{-4}\, \text{N} \) (negative because it's attractive and towards q2).

Step 5: Determine Total Force Components

Add the components:- Total Force in \(x\), \(F_{x} = F_{13x} + F_{23x} = 8.64 \times 10^{-5} + 0 = 8.64 \times 10^{-5}\, \text{N}\).- Total Force in \(y\), \(F_{y} = F_{13y} + F_{23y} = 6.48 \times 10^{-5} - 1.20 \times 10^{-4} = -5.52 \times 10^{-5}\, \text{N}\).

Step 6: Calculate Total Force Magnitude and Direction

The magnitude of the total force \(F\) is found using the Pythagorean theorem:\[ F = \sqrt{F_x^2 + F_y^2} = \sqrt{(8.64 \times 10^{-5})^2 + (-5.52 \times 10^{-5})^2} \approx 1.02 \times 10^{-4}\, \text{N} \]The direction \(\theta\) with respect to the x-axis is:\[ \theta = \tan^{-1}\left(\frac{F_y}{F_x}\right) = \tan^{-1}\left(\frac{-5.52 \times 10^{-5}}{8.64 \times 10^{-5}}\right) \approx -32.0^\circ \] (below the x-axis).

Key Concepts

Coulomb's Law

Coulomb's Law is a fundamental principle in physics. It outlines how electric charges interact with one another. This law states that the electric force between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them. Mathematically, it is expressed as:

• \( F = k \frac{|q_1 q_2|}{r^2} \)

Here, \( F \) is the force between the charges, \( k \) is Coulomb's constant \( (8.99 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2) \), \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between them.
This law helps determine the strength and direction of interaction between charged particles.
In our problem, Coulomb's Law is used to calculate the forces between the charges \( q_1 \), \( q_2 \), and \( q_3 \). By employing this fundamental principle, we computed the forces \( F_{13} \) and \( F_{23} \), representing the influences of \( q_1 \) on \( q_3 \) and \( q_2 \) on \( q_3 \), respectively.
Understanding this principle is crucial, as it underlies many phenomena encountered in electromagnetic theory.

Vector Components

When dealing with forces in physics, it's important to consider vector components. A force vector can be resolved into two perpendicular components: the x-component and the y-component. This makes analysis simpler, especially when combining forces from multiple directions.
In our exercise, we calculated the forces exerted by \( q_1 \) and \( q_2 \) on \( q_3 \) and resolved these into components along the x and y axes. For instance, the force from \( q_1 \) on \( q_3 \) was divided into:

• \( F_{13x} = F_{13} \cdot \frac{4}{5} \)
• \( F_{13y} = F_{13} \cdot \frac{3}{5} \)

This breakdown is crucial because it allows us to consider each axis separately when finding the net force.
For \( q_2 \), the force is solely along the y-axis, making the x-component zero:

• \( F_{23x} = 0 \)
• \( F_{23y} = -1.20 \times 10^{-4}\, \text{N} \)

By combining these components, we can easily find the total x and y components of the force on \( q_3 \).
This component-wise approach simplifies calculations and provides clarity in vector addition.

Force Magnitude and Direction

After resolving the forces into vector components, determining the overall force's magnitude and direction is the next step.
To find the magnitude of the resultant force \( F \), use the Pythagorean theorem:

• \( F = \sqrt{F_x^2 + F_y^2} \)

For our charges, this yielded:

• \( F \approx 1.02 \times 10^{-4}\, \text{N} \)

Direction is as vital as magnitude, providing insight into where the force points. The direction angle \( \theta \) can be found using the tangent function:

• \( \theta = \tan^{-1}\left(\frac{F_y}{F_x}\right) \)
• \( \theta \approx -32.0^\circ \)

This angle is measured from the positive x-axis. Here, the negative sign suggests the force is directed below the x-axis, reflecting the attractive interaction between opposite charges.
Understanding how to determine both force magnitude and direction is essential for comprehensively analyzing physical systems involving forces.