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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

What is the best explanation for the observation that the electric charge on the stem became positive as the charged bee approached (before it landed)? (a) Because air is a good conductor, the positive charge on the bee's surface flowed through the air from bee to plant. (b) Because the earth is a reservoir of large amounts of charge, positive ions were drawn up the stem from the ground toward the charged bee. (c) The plant became electrically polarized as the charged bee approached. (d) Bees that had visited the plant earlier deposited a positive charge on the stem.

Excess electrons are placed on a small lead sphere with mass 8.00 g so that its net charge is -3.20 x 10\(^{-9}\)C. (a) Find the number of excess electrons on the sphere. (b) How many excess electrons are there per lead atom? The atomic number of lead is 82, and its atomic mass is 207 g/mol.

An average human weighs about 650 N. If each of two average humans could carry 1.0 C of excess charge, one positive and one negative, how far apart would they have to be for the electric attraction between them to equal their 650-N weight?

A \(+\)2.00-nC point charge is at the origin, and a second \(-\)5.00-nC point charge is on the \(x\)-axis at \(x = \)0.800 m. (a) Find the electric field (magnitude and direction) at each of the following points on the x-axis: (i) \(x =\) 0.200 m; (ii) \(x =\) 1.20 m; (iii) \(x = -\)0.200 m. (b) Find the net electric force that the two charges would exert on an electron placed at each point in part (a).

A proton is placed in a uniform electric field of 2.75 \(\times 10^3 \space N/C\). Calculate (a) the magnitude of the electric force felt by the proton; (b) the proton's acceleration; (c) the proton's speed after 1.00 \(\mu\)s in the field, assuming it starts from rest.

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