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Chapter 18: Problem 26

There are four charges, each with a magnitude of \(2.0 \mu\) C. Two are positive and two are negative. The charges are fixed to the corners of a \(0.30-\mathrm{m}\) square, one to a corner, in such a way that the net force on any charge is directed toward the center of the square. Find the magnitude of the net electrostatic force experienced by any charge.

Short Answer

Expert verified
The net electrostatic force magnitude experienced by each charge is approximately 0.34 N.

Step by step solution

01

Understand the Square Configuration

Visualize the square with sides of length 0.30 m. At each corner, place the charges: two positive and two negative. Assume the arrangement is such that positive and negative charges are diagonal to each other. This will ensure the net force on any charge is directed towards the center due to symmetry.
02

Calculate the Force Between Diagonal Charges

Using Coulomb's Law, calculate the force between a positive charge and a negative charge that are diagonally across the square. Both charges are separated by the diagonal of the square. The diagonal length is computed using the Pythagorean Theorem: \ \[ d = \sqrt{(0.30^2 + 0.30^2)} = 0.30\sqrt{2} \text{ m}. \] Then, apply Coulomb's Law: \ \[ F_{d} = \frac{k \cdot |q_1 q_2|}{d^2} = \frac{(8.99 \times 10^9) \cdot (2.0 \times 10^{-6})^2}{(0.30\sqrt{2})^2}. \]
03

Solve the Force Between Adjacent Charges

Calculate the force between a charge and its next adjacent charge (one positive and one negative) using Coulomb's Law. The distance is 0.30 m. \ The force between adjacent charges is: \ \[ F_{a} = \frac{k \cdot |q_1 q_2|}{(0.30)^2}. \]
04

Determine the Net Force on a Charge

Each charge experiences four forces: two from adjacent charges (one attractive, one repulsive) and two from diagonal charges (both attractive or both repulsive). The net force must be reconstructed using vector addition. The vertical component of the net force from diagonal charges cancel the horizontal components of the adjacent charges. Thus, only the diagonal forces need to be considered for the net inward force: \ \[ F_{net} = 2F_{d} \cos(45^\circ). \]
05

Calculate the Final Force Magnitude

Substitute the known values into the equations and simplify to find: \ \[ F_{net} = 2F_{d} \cdot \frac{\sqrt{2}}{2} = \sqrt{2}F_{d}. \] Calculate using the values found in Step 2 and compute the net force.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coulomb's Law
Coulomb's Law is a fundamental principle in electrostatics, describing the force between two point charges. It states that the magnitude of the force (\( F \)) between two charges is directly proportional to the product of the magnitudes of the charges (\( |q_1 q_2| \)) and inversely proportional to the square of the distance (\( d^2 \)) between them. The formula is:
\[ F = \frac{k \cdot |q_1 q_2|}{d^2} \]
where \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2 \).

This law helps us understand the strength and direction of electrostatic forces in scenarios where charges are fixed in space. It is crucial to note that the force is attractive if the charges have opposite signs and repulsive if they have the same sign. This is the initial step we use to calculate forces between charges fixed at different corners of a square.
Net Electric Force
When multiple forces act on a charge, the net electric force is the vector sum of all these individual forces. To find this, it's important to consider both the magnitude and the direction of each force.

In our exercise, each charge feels forces from the other three charges. The charges are placed such that two positive and two negative are positioned in a symmetrical pattern. For any charge, the diagonal forces play a significant role because they result from pairs of opposite charges facing each other.
  • Diagonally aligned charges exert forces that are either all attractive or all repulsive, simplifying the direction of the net force.
  • The forces between adjacent charges cancel each other's horizontal components, focusing the overall force towards the center of the square.
Given these dynamics, the net force on each charge can be calculated, considering that symmetrical and diagonal arrangements simplify the vector addition process.
Vector Addition
Vector addition allows us to calculate the resultant force when multiple forces act on a point. It is crucial in electrostatics, where forces may pull or push in different directions. The resultant force is found by adding individual force vectors, considering both direction and magnitude.

For the square configuration of charges, each charge experiences forces in different directions:
  • Adjacent charge forces tend to cancel each other in one direction.
  • Diagonal charge forces add constructively due to their symmetrical orientation.
By using vector addition, the net force acting on any charge can be found and often simplified. This simplification is possible because of the square's symmetry, which guides the net force towards the center. Mathematically, this is expressed by combining components of forces using trigonometric functions such as cosine and sine.
Square Configuration of Charges
The square configuration creates a unique symmetry, simplifying the analysis of electrostatic forces. In this exercise, charges are placed at each corner of a square. Two positive and two negative charges are placed diagonally opposite, ensuring symmetry. The importance of this setup includes:
  • Symmetry leads to an inward net force for all charges, pulling them towards the square's center.
  • Forces between diagonal charges are primary contributors to the net force, reducing the need to consider adjacent charge forces.
The square's geometry offers insights into force interactions and vector contributions, making it easier to predict the behavior of charges. This setup demonstrates the powerful role symmetry plays in electrostatics, guiding calculations to simpler solutions involving fewer force components.

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Most popular questions from this chapter

A surface completely surrounds a \(+2.0 \times 10^{-6} \mathrm{C}\) charge. Find the electric flux through this surface when the surface is (a) a sphere with a radius of \(0.50 \mathrm{m},\) (b) a sphere with a radius of \(0.25 \mathrm{m},\) and \((\mathrm{c})\) a cube with edges that are \(0.25 \mathrm{m}\) long.

A spring with an unstrained length of \(0.074 \mathrm{m}\) and a spring constant of \(2.4 \mathrm{N} / \mathrm{m}\) hangs vertically downward from the ceiling. A uniform electric field directed vertically upward fills the region containing the spring. A sphere with a mass of \(5.1 \times 10^{-3} \mathrm{kg}\) and a net charge of \(+6.6 \mu \mathrm{C}\) is attached to the lower end of the spring. The spring is released slowly, until it reaches equilibrium. The equilibrium length of the spring is \(0.059 \mathrm{m} .\) What is the magnitude of the external electric field?

Four point charges have the same magnitude of \(2.4 \times 10^{-12} \mathrm{C}\) and are fixed to the corners of a square that is \(4.0 \mathrm{cm}\) on a side. Three of the charges are positive and one is negative. Determine the magnitude of the net electric field that exists at the center of the square.

Two identical small insulating balls are suspended by separate \(0.25-\mathrm{m}\) threads that are attached to a common point on the ceiling. Each ball has a mass of \(8.0 \times 10^{-4} \mathrm{kg} .\) Initially the balls are uncharged and hang straight down. They are then given identical positive charges and, as a result, spread apart with an angle of \(36^{\circ}\) between the threads. Determine (a) the charge on each ball and (b) the tension in the threads.

A cube is located with one corner situated at the origin of an \(x, y,\) \(z\) coordinate system. One of the cube's faces lies in the \(x, y\) plane, another in the \(y, z\) plane, and another in the \(x, z\) plane. In other words, the cube is in the first octant of the coordinate system. The edges of the cube are \(0.20 \mathrm{m}\) long. A uniform electric field is parallel to the \(x, y\) plane and points in the direction of the \(+y\) axis. The magnitude of the field is \(1500 \mathrm{N} / \mathrm{C}\). (a) Using the outward normal for each face of the cube, find the electric flux through each of the six faces. (b) Add the six values obtained in part (a) to show that the electric flux through the cubical surface is zero, as Gauss' law predicts, since there is no net charge within the cube.
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