Problem 77 - University Physics with Modern Physics

Question

Three identical point charges \(q\) are placed at each of three corners of a square of side \(L .\) Find the magnitude and direction of the net force on a point charge \(-3 q\) placed (a) at the center of the square and \((b)\) at the vacant corner of the square. In each case, draw a free-body diagram showing the forces exerted on the \(-3 q\) charge by each of the other three charges.

Step-by-Step Solution

Answer

(a) Net force along x-axis, zero y-component. (b) Net force diagonally with equal x and y components.

Step 1: Identify Charge Positions for Part (a)

Given a square of side \(L\), place charges \(q\) at three corners of the square. We are finding the net force on a charge \(-3q\) placed at the center of this square. The charges \(q\) are at positions \((0,0)\), \((L,0)\), and \((0,L)\) with \(-3q\) at the center \((L/2, L/2)\).

Step 2: Calculate Force Due to Each Charge on -3q for Part (a)

Use Coulomb's Law \( F = \frac{{k|q_1 q_2|}}{{r^2}} \). For each charge \(q\) at distance \(r = \frac{L}{\sqrt{2}}\) from the center:1. From \((0,0)\), force is \( F_1 = \frac{{3kq^2}}{{(L/\sqrt{2})^2}} \) along \((L/2, L/2)\).2. From \((L,0)\), force is \( F_2 = \frac{{3kq^2}}{{(L/\sqrt{2})^2}} \) along \((-L/2, L/2)\).3. From \((0,L)\), force is \( F_3 = \frac{{3kq^2}}{{(L/\sqrt{2})^2}} \) along \((L/2, -L/2)\).

Step 3: Determine Net Force Vector for Part (a)

The forces from \((L,0)\) and \((0,L)\) have equal magnitude but opposite directions in the y-component, thus canceling them. The x-components add up:\[ F_{\text{net}} = \frac{{3kq^2}}{{L^2/2}} \hat{x} \].

Step 4: Diagram and Interpretation for Part (a)

Draw a vector diagram at the center showing forces from each point charge. Conclude no net force in the y-direction.

Step 5: Identify Charge Positions for Part (b)

For the square's vacant corner, position charges \(q\) at \((0,0)\), \((L,0)\), \((0,L)\), and place \(-3q\) at \((L,L)\).

Step 6: Calculate Force Due to Each Charge on -3q for Part (b)

Using Coulomb's Law, calculate forces:1. From \((0,0)\), \( F_1 = \frac{{3kq^2}}{{2L^2}} \) along \((1, 1)\).2. From \((L,0)\), \( F_2 = \frac{{3kq^2}}{{L^2}} \) along \((0, 1)\).3. From \((0,L)\), \( F_3 = \frac{{3kq^2}}{{L^2}} \) along \((1, 0)\).

Step 7: Determine Net Force Vector for Part (b)

Forces are added:\[ F_{\text{net}} = \left( \frac{{3kq^2}}{{2L^2}} + \frac{{3kq^2}}{{L^2}} , \frac{{3kq^2}}{{2L^2}} + \frac{{3kq^2}}{{L^2}} \right) = \frac{{9kq^2}}{{2L^2}} \hat{i} + \frac{{9kq^2}}{{2L^2}} \hat{j} \].

Step 8: Diagram and Interpretation for Part (b)

Draw a vector diagram at the vacant corner showing resulting force vectors on \(-3q\). Conclude the net force vector has components in both x and y-directions.

Key Concepts

Electric Forces

In physics, electric forces arise from the interaction between charged particles or objects. This concept is governed by Coulomb's Law, which states that the electric force ( F ) between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Mathematically, this is expressed as: \[ F = \frac{{k |q_1 q_2|}}{{r^2}} \] where:

• \( k \) is the electrostatic constant, approximately equal to \( 8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2 \).
• \( q_1 \) and \( q_2 \) are the magnitudes of the point charges.
• \( r \) is the distance between the charges.

Electric forces can be attractive or repulsive depending on the signs of the charges involved. Like charges repel each other, while opposite charges attract. This fundamental law allows us to calculate the force exerted by multiple charges in complex setups.

Vector Addition

Vector addition is an essential tool in understanding the resultant force when multiple forces act simultaneously on a point charge. Forces, like any other vectors, have both magnitude and direction, and can be combined using vector addition. This involves summing their components.

• In two dimensions, vectors have an x-component and a y-component.
• To add vectors, you add their respective components separately.For example, if vector \( \mathbf{A} = a_x \hat{i} + a_y \hat{j} \) and vector \( \mathbf{B} = b_x \hat{i} + b_y \hat{j} \), then the resultant vector \( \mathbf{R} = (a_x + b_x) \hat{i} + (a_y + b_y) \hat{j} \).

Imagine you have forces acting on a charged particle at different angles or positions, the resultant force or net force is obtained by using this method. This is very handy when determining the net force in exercises involving non-aligned forces like those at the corners of a square.

Net Force

The net force is the vector sum of all individual forces acting on a body. It determines the object's acceleration according to Newton's Second Law: \( F_{ ext{net}} = ma \), where \( m \) is mass and \( a \) is acceleration. In the context of electric forces among multiple charges, evaluating the net force involves several steps:

• Identify each force acting on the charge of interest using Coulomb's Law.
• Break these forces into their respective components.
• Add up all the x-components to find the total x-component of the force and repeat for the y-components.

In the provided example, due to symmetry and the positions of charges, some forces cancel each other out while others enhance, simplifying the calculation of the net force.

Point Charges

Point charges are idealized charges that are assumed to be located at a single point in space. This simplification makes them useful for analyzing electrostatic scenarios without worrying about the physical size of the charge. Point charges are particularly relevant in theoretical and practical physics problems, especially when:

• Dealing with Coulomb's Law, which describes the force between two point charges.
• Calculating electric fields and potential due to these charges.
• Considering setups like the one in the exercise, where charges are placed at geometrically significant points such as the corners of a square.

Understanding point charges facilitates the calculation of forces when charges interact with each other in space, providing clarity and ease in complex calculations like the ones described.