Problem 66 - 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 force on \(q_3\) is 10.27 N at \(-32.5^\circ\) from the positive \(x\)-axis.

Step 1: Calculate the distance between charges

First, calculate the distance between charge \(q_1\) at the origin and charge \(q_3\) at the point \((4.00 \, \text{cm}, 3.00 \, \text{cm})\). Use the distance formula which gives \(r_{13} = \sqrt{(4.00)^2 + (3.00)^2} = 5.00 \, \text{cm}\).

Step 2: Determine direction vectors

Determine the unit vector components from charge \(q_1\) to charge \(q_3\), which has components: \(\hat{i}_{13} = \frac{4}{5}\) and \(\hat{j}_{13} = \frac{3}{5}\). Similarly, for charge \(q_2\) to \(q_3\), the unit vector is purely in the \(y\)-direction, \(\hat{j}_{23} = 1\), since \(q_2\) and \(q_3\) are vertically aligned.

Step 3: Calculate forces using Coulomb's Law

Apply Coulomb's law \(F = \frac{k \cdot |q_1q_3|}{r_{13}^2}\) for both pairs of charges \(q_1, q_3\) and \(q_2, q_3\).For \(q_1, q_3\): \(F_{13} = \frac{(8.99 \times 10^9) \cdot (5.00 \times 10^{-9})(6.00 \times 10^{-9})}{(0.05)^2} = 10.79 \, \text{N}\).For \(q_2, q_3\): \(F_{23} = \frac{(8.99 \times 10^9) \cdot (2.00 \times 10^{-9})(6.00 \times 10^{-9})}{(0.03)^2} = 11.99 \, \text{N}\).

Step 4: Resolve forces into components

Resolve each force into its components. For \(F_{13}\), using the unit vector components: \(F_{13x} = 10.79 \times \frac{4}{5} = 8.63 \, \text{N}\) and \(F_{13y} = 10.79 \times \frac{3}{5} = 6.47 \, \text{N}\).For \(F_{23}\), since it only acts in the \(y\)-direction: \(F_{23x} = 0\) and \(F_{23y} = 11.99 \, \text{N}\). Since \(q_2\) is negative, the direction is reversed.

Step 5: Calculate total force components

Add the components of the forces to find the total force on \(q_3\):\(F_{3x} = F_{13x} + F_{23x} = 8.63 + 0 = 8.63 \, \text{N}\).\(F_{3y} = F_{13y} - F_{23y} = 6.47 - 11.99 = -5.52 \, \text{N}\).

Step 6: Calculate magnitude and direction of total force

Find the magnitude of the total force using Pythagorean theorem:\[ F = \sqrt{(8.63)^2 + (-5.52)^2} = 10.27 \, \text{N} \]Find the direction by calculating the angle \(\theta\) from the positive \(x\)-axis with \(\tan \theta = \frac{-5.52}{8.63}\), giving an angle of \(\theta = \arctan\left(\frac{-5.52}{8.63}\right) = -32.5^\circ\).

Key Concepts

Electric Charge

Electric charge is a fundamental property of matter that causes it to experience a force when placed in an electromagnetic field. Charges are typically described as positive or negative, like in this exercise where we have three charges: \( q_1 = +5.00 \, ext{nC} \), \( q_2 = -2.00 \, ext{nC} \), and \( q_3 = +6.00 \, ext{nC} \). Charged objects interact with each other through the electrostatic force, which is governed by Coulomb's Law.

The interaction between charges depends on their magnitude and sign. Like charges repel each other, while opposite charges attract. For example, \( q_1 \) and \( q_3 \) have like charges (+), leading to repulsion, whereas \( q_2 \) and \( q_3 \) have opposite charges, causing attraction. Understanding the nature of charges is crucial for analyzing the forces at play.

Vector Components

In physics, vectors are quantities that have both magnitude and direction. For the problem at hand, we must resolve the forces between charges into their vector components to understand how they affect \( q_3 \). Each force can be broken into parts along the \( x \)- and \( y \)-axis.

Determining these components involves using unit vectors and trigonometric relationships. For instance, the force \( F_{13} \) between \( q_1 \) and \( q_3 \) needs to be split into \( F_{13x} \) and \( F_{13y} \). We use unit vector properties, where \( \hat{i}_{13} = \frac{4}{5} \) and \( \hat{j}_{13} = \frac{3}{5} \), to calculate these. This breakdown is essential for adding forces adequately. The sum of the \( x \)-components gives the total force in the \( x \)-direction, and similarly for the \( y \)-components.

Force Magnitude and Direction

After resolving the components of each force, the next task is to determine the overall effect on \( q_3 \) by calculating the force's magnitude and direction. The total force components we found are \( F_{3x} = 8.63 \, \text{N} \) and \( F_{3y} = -5.52 \, \text{N} \).

The magnitude of the force is calculated using the Pythagorean theorem: \[ F = \sqrt{(F_{3x})^2 + (F_{3y})^2} = \sqrt{(8.63)^2 + (-5.52)^2} = 10.27 \, \text{N} \] The direction can be found using trigonometry with the tangent function: \[ \tan \theta = \frac{F_{3y}}{F_{3x}} = \frac{-5.52}{8.63} \] Solving for \( \theta \), we find \( \theta = \arctan\left(\frac{-5.52}{8.63}\right) \approx -32.5^\circ \), meaning the force is directed below the \( x \)-axis. This visualization helps predict the resultant impact on the charge.