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Chapter 24: Problem 42

Four equal-magnitude \((4.0 \mu \mathrm{C})\) charges in vacuum are placed at the four corners of a square that is \(20 \mathrm{~cm}\) on each side. Find the electric field at the center of the square \((a)\) if the charges are all positive, (b) if the charges alternate in sign around the perimeter of the square, (c) if the charges have the following sequence around the square: plus, plus, minus, minus.

Short Answer

Expert verified
The net electric field is zero in all scenarios at the center of the square.

Step by step solution

01

Understanding the Geometry

The square is formed by four corners with charges placed on each. The center of the square is equidistant from all corners. The distance from any corner to the center of the square, which is the hypotenuse of a right isosceles triangle, is \(\frac{20 \sqrt{2}}{2} = 10\sqrt{2}\) cm or \(0.1\sqrt{2}\) m.
02

Electric Field Due to One Charge

The electric field at a distance \(r\) from a point charge \(q\) is given by \(E = \frac{kq}{r^2}\), where \(k = 8.99 \times 10^9\) N m虏/C虏. Here, \(q = 4 \times 10^{-6}\) C and \(r = 0.1\sqrt{2}\) m. Thus, \(E = \frac{8.99 \times 10^9 \cdot 4 \times 10^{-6}}{(0.1\sqrt{2})^2}\) N/C.
03

Calculating Electric Field Magnitude

Calculating the above expression gives:\[ E = \frac{8.99 \times 10^9 \cdot 4 \times 10^{-6}}{0.01 \times 2} = \frac{35.96 \times 10^3}{0.02} = 1.798 \times 10^6 \text{ N/C} \] This is the magnitude of the electric field due to one charge at the center.
04

Part A: All Charges Positive

If all charges are positive, the electric field vectors from each charge point away from the charges. Since the problem is symmetric, the horizontal and vertical components from opposite charges cancel out, resulting in a net electric field of zero at the center.
05

Part B: Alternate Charges

With alternating charges around the square, each pair of opposite charges cancels out due to symmetry in both direction and field strength, leading to zero net electric field at the center.
06

Part C: Sequential Charges

For the sequence of charges (++, --), opposite charges cancel out each other's effect. Therefore, each pair's contributions to the electric field at the center cancel out, yielding a net electric field of zero.

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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 that describes the electrostatic interaction between two point charges. This law states that the electric force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.

Mathematically, this is expressed as:
  • Force ( \(F\) = \( \frac{k \, |q_1 \cdot q_2|}{r^2} \)
where \(k\) is Coulomb's constant \(8.99 \times 10^9 \, \text{N m虏/C虏}\), \(q_1\) and \(q_2\) are the magnitudes of the charges, and \(r\) is the distance between them.

In the context of the exercise, we are considering the electric field rather than the force, which is a vector quantity indicating the force a charge would experience per unit charge. Here, the electric field \(E\) at a distance \(r\) from a point charge \(q\) is given by:
  • Electric Field ( \(E\) = \( \frac{k \, q}{r^2} \)
This explains how individual charges influence a point in space, which is crucial in determining the electric field at the center of the square.
Superposition Principle
The Superposition Principle is a critical concept in understanding electric fields generated by multiple charges. According to this principle, the total electric field created by a group of charges is the vector sum of the individual fields created by each charge.

In simpler terms:
  • If you have several charges, calculate the electric field each charge produces at the point of interest.
  • Then add these vector fields together to find the total electric field.
For example, in our exercise, the symmetry of the arrangement led to specific results:
- **All charges positive:** The symmetry ensures that opposite fields cancel each other out, resulting in no net field.
- **Alternate charge signs:** Fields again cancel symmetrically.
- **Sequence of charges ++,--:** The fields by opposite charge pairs cancel, leading to zero net electric field.

Understanding this principle is essential to solving complex electric field problems, as it allows breaking down a challenging situation into manageable parts.
Electric Charge
Electric charge is a fundamental property of matter responsible for electric forces. It comes in two types: positive and negative. Like charges repel, and opposite charges attract.
  • In standard units, charge is measured in coulombs (C), where a microcoulomb ( \(\mu\text{C}\)) is \(10^{-6}\,\text{C}\).
The exercise involves manipulating a set of equal charges ( \(4.0\ \mu\text{C}\)) placed at the corners of a square. These charges are crucial since they generate an electric field influencing the net effect at the square's center.

A notable aspect of electric charge is its conservation, meaning the total charge in an isolated system remains constant. The symmetry in the problem allowed for the cancellation of electric fields, highlighting the beauty of charge interactions accross complex arrangements. Understanding the behavior of electric charges is fundamental in fields ranging from basic physics to advanced engineering applications.

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Most popular questions from this chapter

Two charged metal plates in vacuum are \(15 \mathrm{~cm}\) apart as drawn in Fig. 24-7. The electric field between the plates is uniform and has a strength of \(E=3000 \mathrm{~N} / \mathrm{C}\). An electron \(\left(q=-e, m_{e}=9.1 \times 10^{-31}\right.\) \(\mathrm{kg}\) ) is released from rest at point \(P\) just outside the negative plate. (a) How long will it take to reach the other plate? (b) How fast will it be going just before it hits? The electric field lines show the force on a positive charge. (A positive charge would be repelled to the right by the positive plate and attracted to the right by the negative plate.) An electron, being negative, will experience a force in the opposite direction, toward the left, of magnitude $$ F_{E}=|q| E=\left(1.6 \times 10^{-19} \mathrm{C}\right)(3000 \mathrm{~N} / \mathrm{C})=4.8 \times 10^{-16} \mathrm{~N} $$ Because of this force, the electron experiences an acceleration toward the left given by $$ a=\frac{F_{E}}{m}=\frac{4.8 \times 10^{-16} \mathrm{~N}}{9.1 \times 10^{-31} \mathrm{~kg}}=5.3 \times 10^{14} \mathrm{~m} / \mathrm{s}^{2} $$ In the motion problem for the electron released at the negative plate and traveling to the positive plate, $$ v_{i}=0 x=0.15 \mathrm{~m} a=5.3 \times 10^{14} \mathrm{~m} / \mathrm{s}^{2} $$ (a) Fiom \(x=v_{i} t+\frac{a}{a}\) $$ t=\sqrt{\frac{2 x}{a}}=\sqrt{\frac{(2)(0.15 \mathrm{~m})}{5.3 \times 10^{4^{4}} \mathrm{~m} / \mathrm{s}^{2}}}=2.4 \times 10^{-8} \mathrm{~s} $$ (b) \(v=v_{i}+a t=0+\left(5.3 \times 10^{14} \mathrm{~m} / \mathrm{s}^{2}\right)\left(2.4 \times 10^{-8} \mathrm{~s}\right)=1.30 \times 10^{7}\) \(\mathrm{m} / \mathrm{s}\) As you will see in Chapter 41, relativistic effects begin to become important at speeds above this. Therefore, this approach must be modified for very fast particles.

What is the electrostatic force acting on each of two tiny uniformly charged spheres in vacuum if they both carry \(1.00\) C of charge and they are separated, center to center, by \(1.00 \mathrm{~m}\) ?

Two point charges, one \(+400.0 \mathrm{nC}\) and the other \(-400.0 \mathrm{nC}\), located \(20.00 \mathrm{~cm}\) to the right of the first, are in vacuum. Determine the electric field (magnitude and direction) at a point midway between the charges.

A particle of mass \(m\) and charge \(-e\) while in a region of vacuum is projected with horizontal speed \(v\) into an electric field \((E)\) directed downward. Find \((a)\) the horizontal and vertical components of its acceleration, \(a_{x}\) and \(a_{y} ;(b)\) its horizontal and vertical displacements, \(x\) and \(y\), after time \(t\); \((c)\) the equation of its trajectory.

A point charge of \(-3.0 \times 10^{-5} \mathrm{C}\) is placed at the origin of coordinates in vacuum. Find the electric field at the point \(x=5.0\) \(\mathrm{m}\) on the \(x\) -axis.
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