Problem 61 - University Physics with Modern Physics

Question

A charge \(q_1 = +\)5.00 nC is placed at the origin of an \(xy\)-coordinate system, and a charge \(q_2 = -\)2.00 nC is placed on the positive \(x\)-axis at \(x = \)4.00 cm. (a) If a third charge \(q_3 = +\)6.00 nC is now placed at the point \(x =\) 4.00 cm, \(y =\) 3.00 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

Total force on charge is \(7.29 \times 10^{-5}\) N, \(62.0^\circ\) above the negative x-axis.

Step 1: Determine Distances Between Charges

For the charge configurations, calculate the distances between each pair of charges. The distance between \(q_1\) and \(q_3\) is \( r_{13} = \sqrt{(4.00)^2 + (3.00)^2} = 5.00 \, \text{cm}\). The distance from \(q_2\) to \(q_3\) is \( r_{23} = 3.00 \, \text{cm}\), as both are on the line parallel to the y-axis.

Step 2: Apply Coulomb's Law

Coulomb's Law gives the force between two charges, given by \( F = k \frac{|q_1 q_2|}{r^2}\). Calculate the force between \(q_1\) and \(q_3\): \( F_{13} = 8.99 \times 10^9 \frac{5.00 \times 10^{-9} \times 6.00 \times 10^{-9}}{(0.05)^2} \, \text{N}\approx 1.08 \times 10^{-4} \, \text{N}\). For \(q_2\) and \(q_3\), \( F_{23} = 8.99 \times 10^9 \frac{2.00 \times 10^{-9} \times 6.00 \times 10^{-9}}{(0.03)^2} \, \text{N}\approx 1.20 \times 10^{-4} \, \text{N}\).

Step 3: Resolve Forces Into Components

The force \(F_{13}\) between \(q_1\) and \(q_3\) at an angle has components: \(F_{13,x} = F_{13} \times \cos(\theta) = 1.08 \times 10^{-4} \times \frac{4}{5} = 8.64 \times 10^{-5} \, \text{N}\) and \(F_{13,y} = F_{13} \times \sin(\theta) = 1.08 \times 10^{-4} \times \frac{3}{5} = 6.48 \times 10^{-5} \, \text{N}\). The force \(F_{23}\) acts along the x-axis, so \(F_{23,x} = -1.20 \times 10^{-4} \, \text{N}\), and \(F_{23,y} = 0\).

Step 4: Calculate Total Force Components

Sum the force components to find the total force on \(q_3\). The components are: \((F_{total,x} = 8.64 \times 10^{-5} - 1.20 \times 10^{-4}) \, \text{N} = -3.36 \times 10^{-5} \, \text{N}\) and \(F_{total,y} = 6.48 \times 10^{-5} \, \text{N}\).

Step 5: Find Magnitude and Direction of Total Force

The magnitude of the total force is \(F_{total} = \sqrt{(F_{total,x})^2 + (F_{total,y})^2} = \sqrt{(-3.36 \times 10^{-5})^2 + (6.48 \times 10^{-5})^2} \, \text{N} \approx 7.29 \times 10^{-5} \, \text{N}\). The direction is \( \theta = \tan^{-1}\left(\frac{F_{total,y}}{F_{total,x}}\right) \approx 62.0^\circ \) above the negative x-axis.

Key Concepts

Electric Force

In the world of physics, the electric force is a fundamental concept that describes the interaction between electrically charged particles. It is what keeps electrons attracted to protons, allowing atoms to exist. This force can either pull charges together (attractive) or push them apart (repulsive) depending on their nature. Charges with the same sign repel each other, while opposite charges attract.
This concept is elegantly described by Coulomb's Law, where the electric force between two point charges is given by the formula: \[ F = k \frac{|q_1 q_2|}{r^2} \] Here, \(F\) is the electric force, \(q_1\) and \(q_2\) are the charges, \(r\) is the distance between them, and \(k\) is Coulomb's constant \( (8.99 \times 10^9 \, \text{N} \, \text{m}^2/\text{C}^2) \).
Coulomb's Law tells us that the strength of the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This relationship means that even small charges can exert significant forces on each other if they are very close.

Vector Components

In physics, dealing with forces like the electric force often requires breaking them down into components. This helps in understanding the net effect of multiple forces acting at angles. When dealing with vector components, we split a vector into parts that run parallel to the given axes, usually the \(x\)-axis and \(y\)-axis.
The process of finding vector components begins with the angle of the vector in relation to the axes. Using basic trigonometry:

• The \(x\)-component of a vector \(F\) at angle \( \theta \) is given by \( F_x = F \cos(\theta) \).
• The \(y\)-component is \( F_y = F \sin(\theta) \).

Once we have these components, it's much easier to analyze how each force influences the overall motion or position of an object.

Distance Between Charges

The distance between two charges is crucial for calculating the electric force between them using Coulomb’s Law. The force's intensity sharply decreases as the distance increases since it's inversely proportional to the square of the distance.
To calculate this distance, especially when charges are not on a straight line, we often resort to the Pythagorean theorem. For example, if one charge is located at \((x_1, y_1)\) and another at \((x_2, y_2)\), the distance \(r\) between them is:\[ r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Knowing how to find this distance helps accurately determine the electric force in exercises and real-life scenarios.

Magnitude and Direction

When multiple forces act on a charge, like in the given exercise, it’s important to find both the magnitude and the direction of the resultant force. The magnitude tells us how strong the total force is. The direction shows us the line along which the force acts.
To find the magnitude of a resultant force from its components \((F_x, F_y)\), use the Pythagorean theorem:\[ F_{total} = \sqrt{F_x^2 + F_y^2} \]
This step gives the complete strength of the force. To determine the direction, use the inverse tangent function:\[ \theta = \arctan\left(\frac{F_y}{F_x}\right) \]
This angle \( \theta \) will show you exactly where the force is heading, revealing the precise geometry of interactions between charges in any system. Understanding magnitude and direction is vital for predicting the behaviour of charged particles.