Problem 70 - University Physics with Modern Physics

Question

A charge of \(-\)3.00 nC is placed at the origin of an \(xy\)-coordinate system, and a charge of 2.00 nC is placed on the \(y\)-axis at \(y =\) 4.00 cm. (a) If a third charge, of 5.00 nC, is now placed at the point \(x =\) 3.00 cm, \(y =\) 4.00 cm, find the \(x\)- and \(y\)-components of the total force exerted on this charge by the other two charges. (b) Find the magnitude and direction of this force.

Step-by-Step Solution

Answer

The total force on the 5.00 nC charge is approximately 1.20 µN at an angle of -85.2° from the positive x-axis.

Step 1: Understand the Scenario

There are three charges in play: Charge \( q_1 = -3.00 \text{ nC} \) at the origin \((0,0)\), charge \( q_2 = 2.00 \text{ nC} \) at \((0, 4 \text{ cm})\), and charge \( q_3 = 5.00 \text{ nC} \) at \((3 \text{ cm}, 4 \text{ cm})\). We need to compute the forces exerted on \( q_3 \) by \( q_1 \) and \( q_2 \).

Step 2: Calculate the Force Between Charges

Use Coulomb's law: \( F = k\frac{|q_1 q_2|}{r^2} \). For \( q_1 \) and \( q_3 \), \( r = 3 \text{ cm} = 0.03 \text{ m} \). For \( q_2 \) and \( q_3 \), \( r = 3 \text{ cm} = 0.03 \text{ m} \) along the x-axis, according to the position difference.Electric constant: \( k = 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2 \).

Step 3: Calculate Force from Charge q1 on q3

Convert charges to Coulombs: \( q_1 = -3.00 \text{ nC} = -3.00 \times 10^{-9} \text{ C} \), \( q_3 = 5.00 \text{ nC} = 5.00 \times 10^{-9} \text{ C} \). Using \( r_{13} = 3 \text{ cm} = 0.03 \text{ m} \), the force magnitude is:\[ F_{13} = k \frac{|q_1 q_3|}{r_{13}^2} = 8.99 \times 10^9 \frac{|-3 \times 10^{-9} \cdot 5 \times 10^{-9}|}{(0.03)^2} \approx 1.50 \text{ µN} \]Direction: attractive towards origin \((3 \text{ cm}, 4 \text{ cm})\).

Step 4: Calculate x and y Components of Force from q1 on q3

Force from \( q_1 \) at an angle, needing calculation:\( \theta = \tan^{-1}(4/3) \)\[ F_{13,x} = F_{13} \cos(\theta) \quad \text{and} \quad F_{13,y} = F_{13} \sin(\theta) \]\[ F_{13,x} = -1.50 \times \frac{3}{5} \approx -0.9 \text{ µN}, \quad F_{13,y} = -1.50 \times \frac{4}{5} \approx -1.2 \text{ µN} \]

Step 5: Calculate Force from Charge q2 on q3

Convert charges to Coulombs: \( q_2 = 2.00 \text{ nC} = 2.00 \times 10^{-9} \text{ C} \).Using \( r_{23} = 3 \text{ cm} = 0.03 \text{ m} \), the force magnitude is:\[ F_{23} = k \frac{|q_2 q_3|}{r_{23}^2} = 8.99 \times 10^9 \frac{2 \times 10^{-9} \cdot 5 \times 10^{-9}}{(0.03)^2} \approx 1.00 \text{ µN} \]Direction: repulsive along the x-axis. \[ F_{23,x} = 1.00 \text{ µN}, \quad F_{23,y} = 0 \text{ µN} \]

Step 6: Calculate Total x and y Components of Force on q3

Add the components of the forces on \( q_3 \):\[ F_{x} = F_{13,x} + F_{23,x} = -0.9 \text{ µN} + 1.0 \text{ µN} = 0.1 \text{ µN} \]\[ F_{y} = F_{13,y} + F_{23,y} = -1.2 \text{ µN} + 0 \text{ µN} = -1.2 \text{ µN} \]

Step 7: Calculate Magnitude and Direction of Total Force

Magnitude:\[ F = \sqrt{F_{x}^2 + F_{y}^2} = \sqrt{(0.1 \text{ µN})^2 + (-1.2 \text{ µN})^2} \approx 1.20 \text{ µN} \]Direction (angle \( \phi \)) downwards from positive x-axis:\[ \phi = \tan^{-1}\left(\frac{F_{y}}{F_{x}}\right) = \tan^{-1}\left(\frac{-1.2}{0.1}\right) \approx -85.2^\circ \]

Key Concepts

Electric Forces

Electric forces are the forces that arise from electric charges. They are governed by Coulomb's Law, which states that the magnitude of 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 their centers. This relationship is given by the formula: \[ F = k \frac{|q_1 q_2|}{r^2} \]where:

• \( F \) is the electrostatic force between two charges.
• \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2 \).
• \( q_1 \) and \( q_2 \) are the amounts of the two charges.
• \( r \) is the separation distance between the charges.

Electric forces can be attractive or repulsive. If the charges are of opposite signs, the force is attractive; if they are of the same sign, they repel. This force plays a critical role in the fields of electricity and physics, explaining interactions at both the atomic and macroscopic levels.

Charge Interactions

Charge interactions are the means by which electric forces manifest between charged particles. When dealing with multiple charges, each pair of charges will interact according to Coulomb's Law, resulting in various force vectors acting on each charge. Important points about charge interactions include:

• An attractive force occurs when a positive and a negative charge come together.
• A repulsive force arises when two like charges, either positive-positive or negative-negative, interact.
• The net force on a charge is the vector sum of all the forces acting upon it from other charges.

For example, in the exercise provided, charge \( q_3 \) experiences forces due to both charge \( q_1 \) and charge \( q_2 \). The resulting total force is found by calculating each individual force as a vector and summing them using vector addition. Understanding these interactions is key to solving problems involving multiple charges and finding resultant electric forces.

Vector Components

In physics, vector components are used to simplify and analyze forces, velocities, and other vector quantities that act in multiple directions. Vectors can be divided into their respective components along the axes of the coordinate system, typically the x- and y-axes in a two-dimensional plane.To find the vector components of a force:

• The force vector can be split into an \( x \)-component \( F_x \) and a \( y \)-component \( F_y \) using trigonometry.
• For an angle \( \theta \) with respect to the horizontal, these components are calculated using: \[ F_x = F \cos(\theta) \] \[ F_y = F \sin(\theta) \]
• Each component is a projection of the original vector onto the corresponding axis, allowing for easier mathematical treatment in calculations.

In the provided problem, the x and y components for forces on charge \( q_3 \) from \( q_1 \) and \( q_2 \) are calculated separately, enabling us to determine the net force acting in each respective direction. Summing these components helps in finding the resultant magnitude and direction of the total force.