Problem 67 - College Physics

Question

\(\bullet$$\bullet\) 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 charges. Find the magnitude and direction of this force.

Step-by-Step Solution

Answer

The total force on \(q_3\) is approximately \(7.26 \times 10^{-5}\) N at \(62.3^\circ\) above the -x axis.

Step 1: Calculate distance between charges

First, calculate the distances between the charge \(q_3\) and the other two charges. The distance between \(q_1\) at the origin and \(q_3\) is given by the Pythagorean theorem: \[ r_1 = \sqrt{(4.00\, \text{cm} - 0)^2 + (3.00\, \text{cm} - 0)^2} = \sqrt{4.00^2 + 3.00^2} = 5.00 \text{ cm}\]The distance between \(q_2\) at \(x = 4.00\, \text{cm}\), \(y = 0\) and \(q_3\) is:\[r_2 = \sqrt{(4.00\, \text{cm} - 4.00\, \text{cm})^2 + (3.00\, \text{cm} - 0)^2} = 3.00 \text{ cm}\]

Step 2: Calculate force from q1 on q3

Using Coulomb's law, calculate the force exerted by \(q_1\) on \(q_3\). The formula: \[F = k \frac{|q_1 q_3|}{r_1^2}\]Substitute values:\[k = 8.99 \times 10^9 \frac{\text{N m}^2}{\text{C}^2}\]\[F_{13} = 8.99 \times 10^9 \frac{5.00 \times 10^{-9} \times 6.00 \times 10^{-9}}{(0.050)^2}\, \text{N} = 10.79 \times 10^{-5} \text{ N}\]The direction of \(F_{13}\) is along the line from \(q_1\) to \(q_3\). Calculating components using trigonometry:\[F_{13x} = F_{13} \cdot \frac{4}{5} \quad \text{and} \quad F_{13y} = F_{13} \cdot \frac{3}{5}\]\[F_{13x} = 8.63 \times 10^{-5} \text{ N}, \quad F_{13y} = 6.47 \times 10^{-5} \text{ N}\]

Step 3: Calculate force from q2 on q3

Calculate the force exerted by \(q_2\) on \(q_3\) using Coulomb's law.The formula:\[F = k \frac{|q_2 q_3|}{r_2^2}\]Substitute values:\[F_{23} = 8.99 \times 10^9 \frac{|-2.00 \times 10^{-9} \times 6.00 \times 10^{-9}|}{(0.030)^2}\, \text{N} = 11.98 \times 10^{-5} \text{ N}\]This force is horizontal to the left since it attracts \(q_3\) towards \(q_2\):\[F_{23x} = -11.98 \times 10^{-5} \text{ N}, \quad F_{23y} = 0 \]

Step 4: Determine total force components on q3

Add the \(x\) and \(y\) components of the forces from \(q_1\) and \(q_2\) to find the total force components on \(q_3\):\[F_{total\, x} = F_{13x} + F_{23x} = 8.63 \times 10^{-5} - 11.98 \times 10^{-5} = -3.35 \times 10^{-5} \text{ N}\]\[F_{total\, y} = F_{13y} + F_{23y} = 6.47 \times 10^{-5} + 0 = 6.47 \times 10^{-5} \text{ N}\]

Step 5: Calculate magnitude and direction of total force

We now calculate the magnitude of the total force using:\[F_{total} = \sqrt{(F_{total\, x})^2 + (F_{total\, y})^2}\]Substitute values:\[F_{total} = \sqrt{(-3.35 \times 10^{-5})^2 + (6.47 \times 10^{-5})^2} \approx 7.26 \times 10^{-5} \text{ N}\]For the direction, calculate the angle \(\theta\) with respect to the negative \(x\) axis:\[\theta = \tan^{-1} \left(\frac{F_{total\, y}}{|F_{total\, x}|}\right) = \tan^{-1} \left(\frac{6.47}{3.35}\right) \approx 62.3^\circ\]The direction is \(62.3^\circ\) above the negative \(x\) axis.

Key Concepts

Electric Charge

Electric charge is a fundamental property of matter that causes it to experience a force when placed in an electric field. It is intrinsic to protons and electrons, the subatomic particles found in atoms. The unit of electric charge is the coulomb (C), but in this exercise, we deal with nano-coulombs (nC), where 1 nC is equal to \(10^{-9}\) C.

There are two types of electric charges: positive and negative. Like charges repel, while opposite charges attract. The charge values for particles and objects are determined by the number of protons and electrons they contain. In our exercise, \(q_1\) and \(q_3\) are positive, while \(q_2\) is negative. Understanding charge interactions helps us predict how particles will move when forces are applied.

Key points to remember:

• Charges come in positive and negative.
• Opposite charges attract, like charges repel.
• Measured in coulombs (C), here in nano-coulombs (nC).
• Important for calculations using Coulomb's Law to determine forces.

Vector Components

When dealing with forces acting in two dimensions, like in this exercise, it's beneficial to break the forces down into their vector components. Vectors have both magnitude and direction, helping us handle forces in different planes.

Vectors can be split into two perpendicular components, typically along the x and y axes. This makes calculations simpler, especially when you're using trigonometry or applying laws like Coulomb's Law. For example, if a force is applied diagonally, using vector components allows you to analyze how much of that force acts horizontally and vertically.

In our example:

• The force from charge \(q_1\) on \(q_3\) is resolved into x and y components using trigonometric functions.
• The horizontal component influences motion along the x-axis, while the vertical component affects the y-axis.
• Arithmetic with vectors often involves adding their x and y components separately to find resultant forces.

This approach simplifies the process of determining net force direction and magnitude.

Electrostatic Force

Electrostatic force is the force between charged objects. It results from interactions of electric charges and is calculated using Coulomb's Law. This law gives us the magnitude of the force between two point charges.

Here's how it works:

• Coulomb's Law states that the magnitude of the force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
• The formula is \(F = k \frac{|q_1 q_2|}{r^2}\), where \(F\) is the force between the charges, \(q_1\) and \(q_2\) are the magnitudes of the charges, \(r\) is the distance between the centers of the two charges, and \(k\) is Coulomb's constant (approximately \(8.99 \times 10^9 \text{ N m}^2/\text{C}^2\)).
• In our scenario, this formula helps us compute the forces exerted by \(q_1\) and \(q_2\) on \(q_3\).

Electrostatic forces can attract or repel, depending on whether the interacting charges are opposite or like, respectively. This understanding allows us to determine direction and interaction at the microscale.

Pythagorean Theorem

The Pythagorean theorem is a cornerstone of trigonometry and geometry. It helps in calculating distances in a coordinate plane, which is crucial in our exercise for determining the separation between charges.

The theorem is expressed as \(a^2 + b^2 = c^2\), where \(c\) is the hypotenuse, and \(a\) and \(b\) are the legs of a right-angled triangle.

In the exercise:

• The distance \(r_1\) between \(q_1\) at the origin and \(q_3\) at \((4.00 \text{ cm}, 3.00 \text{ cm})\) is calculated using this theorem as \(\sqrt{(4.00)^2 + (3.00)^2} = 5.00 \text{ cm}\).
• It provides the basis for determining the separation necessary for calculating forces via Coulomb's Law.

Understanding the Pythagorean theorem is crucial for effective problem-solving in physics, as it regularly appears in calculations involving forces, distances, and vector components.