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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 \(-3q\) 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 \(-3q\) charge by each of the other three charges.

Short Answer

Expert verified
(a) Net force is 0. (b) Force is \(\frac{3\sqrt{2}kq^2}{L^2}\) toward the square's center.

Step by step solution

01

Understand the Problem Setup for Part (a)

We have four corners of a square with three identical point charges labeled as \(q\) placed at corners A, B, and C, and an empty corner D. A charge \(-3q\) is placed at the center. We need to find the net force on the \(-3q\) charge.
02

Determine Forces Exerted on \(-3q\) in Part (a)

The force exerted by each \(q\) on \(-3q\) is due to Coulomb's Law: \[ F = k \frac{|q \cdot (-3q)|}{r^2} = \frac{3kq^2}{r^2} \]where \(r\) is the distance from each corner to the center of the square (\(r = \frac{L}{\sqrt{2}}\)), so\[ F = \frac{6kq^2}{L^2} \]. Each force is directed towards each \(q\) charge.
03

Calculate Net Force on \(-3q\) in Part (a)

The forces by the three charges on the center charge cancel each other out due to symmetry. Each force is balanced by the force exerted by the charge on the opposite side of the center, resulting in a net force of 0. The diagram would show three equal and opposite forces, demonstrating no net force.
04

Understand Problem Setup for Part (b)

Now place the charge \(-3q\) at the vacant corner D, with charges \(q\) at corners A, B, C. We need to determine the net force on \(-3q\).
05

Assess Forces on \(-3q\) in Part (b)

Using Coulomb's Law again, the forces exerted by each charge \(q\) on \(-3q\) are each \(F = \frac{3kq^2}{L^2} \) due to the distance \(L\) from each corner to the vacant corner D.
06

Calculate Net Force on \(-3q\) in Part (b)

The forces between adjacent charges (e.g., A to D and B to D) can be added vectorially. For horizontal and vertical components, add: the resultant force in these directions points diagonally toward the square's center. The magnitude of this resultant diagonal force can be found using vector addition of each pair of adjacent forces:\[ \sqrt{(\frac{3kq^2}{L^2})^2 + (\frac{3kq^2}{L^2})^2} = \frac{3\sqrt{2}kq^2}{L^2} \]

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Key Concepts

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

Net Force
In physics, the concept of "net force" is vital to understanding how objects move. It represents the sum total of all forces acting on an object. Net force takes into account both the magnitude and direction of these forces. When multiple forces act on a point charge, such as in the provided exercise, you must add them vectorially.
This means considering both the size and direction of each force. For example, if three forces are applied to a point charge at the center of a square, and they are equal in magnitude and opposite in direction, their net force will cancel out to zero. This is a common result when forces are symmetrically arranged around a point.
Net force determines the acceleration and ultimate motion of the object. If the net force is zero, the object will remain at rest or move with constant velocity, according to Newton's first law. However, if there's a net force, the object will accelerate in the direction of this resultant force. Understanding net force helps us predict how charges and particles will behave under multiple interactions.
Point Charge
A "point charge" is a theoretical concept where the charge is assumed to be concentrated at a single point in space. This simplification is handy for analyzing electrostatic problems and calculating forces and potentials. For instance, in Coulomb's Law, point charges provide a straightforward way to determine the electric force between charges.
In our exercise, point charges are used at each corner of a square and at the center (or vacant corner). The charge values are represented by symbols such as \(q\) and \(-3q\). Point charges make it easier to apply mathematical equations to solve for forces between them. They allow for easy usage of the inverse square law in Coulomb's description, where the force magnitude depends on the distance between charges:
  • The force is directly proportional to the product of the two charges.
  • The force is inversely proportional to the square of the distance between them.
Point charges are crucial in simplifying complex electric field problems and focusing on the core interactions without being distracted by real-world size or distribution specifics.
Vector Addition
Vector addition is indispensable in physics, especially when determining forces. Vectors have both magnitude and direction. They represent quantities such as forces, which need to be combined accurately to find out the net effect. Each force acting on a point charge can be thought of as a vector.
When calculating forces such as in the provided square problem, each force from a point charge must be added vectorially. Essentially, you assess both horizontal and vertical components. If done correctly, the result is a single vector representing the net force.
For example, the exercise demonstrates how forces at 90-degree angles can be added using vector addition, revealing their combined effect on a point charge. The Pythagorean theorem often helps to calculate the magnitude of the resultant vector:
  • Add the square of each perpendicular component.
  • Take the square root of the sum to find the resultant vector's magnitude.
Understanding vector addition allows you to solve for the overall effect of forces in complex setups, a skill highly applicable in numerous practical scenarios.

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Most popular questions from this chapter

Two small aluminum spheres, each having mass 0.0250 kg, are separated by 80.0 cm. (a) How many electrons does each sphere contain? (The atomic mass of aluminum is 26.982 g\(/\)mol, and its atomic number is 13.) (b) How many electrons would have to be removed from one sphere and added to the other to cause an attractive force between the spheres of magnitude 1.00 \(\times\) 10\(^4\) N (roughly 1 ton)? Assume that the spheres may be treated as point charges. (c) What fraction of all the electrons in each sphere does this represent?

(a) What must the charge (sign and magnitude) of a 1.45-g particle be for it to remain stationary when placed in a downward-directed electric field of magnitude 650 N\(/\)C? (b) What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight?

Suppose you had two small boxes, each containing 1.0 g of protons. (a) If one were placed on the moon by an astronaut and the other were left on the earth, and if they were connected by a very light (and very long!) string, what would be the tension in the string? Express your answer in newtons and in pounds. Do you need to take into account the gravitational forces of the earth and moon on the protons? Why? (b) What gravitational force would each box of protons exert on the other box?

Two small plastic spheres are given positive electric charges. When they are 15.0 cm apart, the repulsive force between them has magnitude 0.220 N. What is the charge on each sphere (a) if the two charges are equal and (b) if one sphere has four times the charge of the other?

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