Question

Four point charges are placed at the following \(x y\) coordinates: \(Q_{1}=-1 \mathrm{mC},\) at \((-3 \mathrm{~cm}, 0 \mathrm{~cm})\) \(Q_{2}=-1 \mathrm{mC},\) at \((+3 \mathrm{~cm}, 0 \mathrm{~cm})\) \(Q_{3}=+1.024 \mathrm{mC},\) at \((0 \mathrm{~cm}, 0 \mathrm{~cm})\) \(Q_{4}=+2 \mathrm{mC},\) at \((0 \mathrm{~cm},-4 \mathrm{~cm})\) Calculate the net force on charge \(Q_{4}\) due to charges \(Q_{1}, Q_{2}\) and \(Q_{3}\).

Short answer

Based on the given solution, the net force on charge \(Q_{4}\) is \(14.3 N\) at an angle of \(54.5^\circ\).

Step-by-step solution

1. Converting given quantities to SI units:

First, we need to convert all given quantities, charges and coordinates to SI units. -1 mC = -1 × 10⁻³ C 1.024 mC = 1.024 × 10⁻³ C 2 mC = 2 × 10⁻³ C All the coordinates are given in centimeters. We need to convert them to meters. -3 cm = -0.03 m +3 cm = 0.03 m 0 cm = 0 m -4 cm = -0.04 m

2. Calculate the distances between charges:

We need to calculate the distance between \(Q_{4}\) and each of the other charges using the distance formula. $$ d_{ij} = \sqrt{(x_{i}-x_{j})^{2}+(y_{i}-y_{j})^{2}} $$ For \(Q_{1}\) and \(Q_{4}\): $$ d_{14} = \sqrt{(-0.03 - 0)^{2} + (0 - (-0.04))^{2}} = 0.05 \textrm{ m} $$ For \(Q_{2}\) and \(Q_{4}\): $$ d_{24} = \sqrt{(0.03 - 0)^{2} + (0 - (-0.04))^{2}} = 0.05 \textrm{ m} $$ For \(Q_{3}\) and \(Q_{4}\): $$ d_{34} = \sqrt{(0 - 0)^{2} + (0 - (-0.04))^{2}} = 0.04 \textrm{ m} $$

3. Calculate the force for each pair of charges:

Using Coulomb's Law, we'll calculate the force between each pair of charges. $$ F_{ij} = k\frac{Q_{i}Q_{j}}{d_{ij}^{2}} $$ Where \(k = 8.9875517923 × 10^{9} \textrm{ N m² C⁻²}\) For \(Q_{1}\) and \(Q_{4}\): $$ F_{14} = 8.988×10^{9} × \frac{(-1×10^{-3})(2×10^{-3})}{0.05^{2}} = -7.19 \textrm{ N} $$ For \(Q_{2}\) and \(Q_{4}\): $$ F_{24} = 8.988×10^{9} × \frac{(-1×10^{-3})(2×10^{-3})}{0.05^{2}} = -7.19 \textrm{ N} $$ For \(Q_{3}\) and \(Q_{4}\): $$ F_{34} = 8.988×10^{9} × \frac{(1.024×10^{-3})(2×10^{-3})}{0.04^{2}} = 11.6 \textrm{ N} $$

4. Determine the direction of the force:

Now that we have the magnitude of the force for each pair of charges, we need to determine the directions for each force by using the angle between each charge pair. First, we need to find the angle between \(Q_{4}\) and the x-axis for both \(Q_{1}\) and \(Q_{2}\). For \(Q_{1}\) and \(Q_{4}\): $$\tan{\theta_{1}} = \frac{-0.04}{-0.03}$$ $$\theta_{1} = \arctan{\frac{-0.04}{-0.03}} \approx 53.13^{\circ}$$ For \(Q_{2}\) and \(Q_{4}\): $$\tan{\theta_{2}} = \frac{-0.04}{0.03}$$ $$\theta_{2} = \arctan{\frac{-0.04}{0.03}} \approx -53.13^{\circ}$$

5. Calculate force components:

Knowing the magnitude and angle of each force, we can compute the \(x\) and \(y\) components of each force. For \(F_{14}\): $$F_{14x} = F_{14} \cos{53.13^{\circ}} = 4.14 \textrm{ N}$$ $$F_{14y} = F_{14} \sin{53.13^{\circ}} = 5.52 \textrm{ N}$$ For \(F_{24}\): $$F_{24x} = F_{24} \cos{-53.13^{\circ}} = 4.14 \textrm{ N}$$ $$F_{24y} = F_{24} \sin{-53.13^{\circ}} = -5.52 \textrm{ N}$$ For \(F_{34}\): $$F_{34x} = 0 \textrm{ N}$$ $$F_{34y} = 11.6 \textrm{ N}$$

6. Calculate the net force:

Now that we have the components for each force, we can add them together to find the net force on \(Q_{4}\). $$F_{net_x} = F_{14x} + F_{24x} + F_{34x} = 4.14 \textrm{ N} + 4.14 \textrm{ N} + 0 \textrm{ N} = 8.28 \textrm{ N}$$ $$F_{net_y} = F_{14y} + F_{24y} + F_{34y} = 5.52 \textrm{ N} - 5.52 \textrm{ N} + 11.6 \textrm{ N} = 11.6 \textrm{ N}$$ Finally, we have the net force in both components, so we can combine them to find the magnitude and direction of the net force. To find the magnitude: $$F_{net} = \sqrt{F_{net_x}^{2} + F_{net_y}^{2}} = \sqrt{(8.28 \textrm{ N})^{2} + (11.6 \textrm{ N})^{2}} = 14.3 \textrm{ N}$$ To find the direction of the net force, we can compute the angle: $$\tan{\theta_{net}} = \frac{F_{net_y}}{F_{net_x}}$$ $$\theta_{net} = \arctan{\frac{11.6}{8.28}} \approx 54.5^{\circ}$$ So the net force on charge \(Q_{4}\) is \(14.3 N\) at an angle of \(54.5^\circ\).

Key concepts

Electric Force

When you have charges, they exert forces on each other. The strength of these forces is called the electric force. Coulomb's law helps us understand this concept and says that the electric 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 for calculating the electric force is: \[ F = k \frac{Q_1 Q_2}{d^2} \]Here:- \( F \) is the force in Newtons (N).- \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \text{ N m}^2 \text{ C}^{-2} \).- \( Q_1 \) and \( Q_2 \) are the magnitudes of the charges in coulombs (C).- \( d \) is the distance between the charges in meters.Understanding electric force helps in predicting how charged objects will interact with each other. For instance, like charges repel each other, while opposite charges attract.

Point Charges

In physics, a point charge is an idealized model of a particle with an electric charge that has no size or shape—just a position in space. This simplification allows us to calculate electric forces or fields precisely without worrying about the distribution of charge across a volume.Point charges are particularly useful in educational problems. Here, they help demonstrate the principles of Coulomb's law. For example, in our exercise, different point charges were placed in specific positions to calculate the force on one of them, \( Q_4 \). Considering point charges simplifies our calculations, making it easier to understand the interaction of charges separated by specific distances without the complexity of physical dimensions.

Vector Components

Vector components break down a vector into two parts that help us understand direction and magnitude more clearly, typically along the x and y axes. In this exercise, calculating the net electric force on the charge \( Q_4 \) involves such breakdown.Force vectors have both magnitude (how strong the force is) and direction (which way the force is pointing), making them two-dimensional. To analyze them effectively:- Identify the force vector's x-component using \( F_x = F \cos \theta \), where \( \theta \) is the angle with the x-axis.- Identify the force vector's y-component using \( F_y = F \sin \theta \).Adding up these components for all forces influencing \( Q_4 \), you can find the total net force acting on it. This divides complex force directions into manageable parts.

Trigonometry Calculations

Calculations using trigonometry are crucial when dealing with angled forces in physics. Here, they helped determine the exact direction of the net force based on angles.For this exercise:- The tangent function, \( \tan \theta = \frac{opposite}{adjacent} \), was used to find the angles made by the force vectors and the axes.Once you determine these angles:- Use \( \theta = \arctan \frac{F_y}{F_x} \) to find the net angle of the resultant force.These calculations ensure accurate understanding of directions the forces take. By using trigonometry, we solve for unknown angles and then incorporate them into our calculations for both force components, ensuring the solution is fully precise.