/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 (a) At each corner of a square i... [FREE SOLUTION] | 91Ó°ÊÓ

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(a) At each corner of a square is a particle with charge \(q\). Fixed at the center of the square is a point charge of opposite sign, of magnitude \(Q .\) What value must \(Q\) have to make the total force on each of the four particles zero? (b) With \(Q\) taking on the value you just found, show that the potential energy of the system is zero, consistent with the result from Problem 1.6.

Short Answer

Expert verified
The value of \(Q\) that makes the total force on each of the four particles zero is 0. The potential energy of the system is also zero when \(Q\) is 0.

Step by step solution

01

Analyze the forces on one of the particles

In the case of one of the corner particles, the force exerted by the other three corner particles will be balanced out because they are located symmetrically around the particle under consideration. Thus, we need to calculate only the electric force between the charge at the corner, \(q\), and the charge at the center, \(-Q\). According to Coulomb's law, the magnitude of this force is \(F = k \cdot |q \cdot Q| / a^{2}\), where \(a\) is the distance between \(q\) and \(-Q\), and \(k\) is Coulomb's constant.
02

Calculate the force and solve for Q

Since the total force on each corner particle should be zero, we have \(F = 0\). Plugging the force expression obtained in step 1 into this equation gives \(0 = k \cdot |q \cdot Q| / a^{2}\). Solving this equation for \(Q\) yields \(Q = 0\).
03

Calculate the potential energy

The potential energy of the system is the sum of the potential energies due to each pair of charges. Each pair contributes \(U = k \cdot |q1 \cdot q2| / r\), where \(r\) is the distance between the two charges. Since \(Q = 0\), all \(q2\) become 0 and therefore the total potential energy of the system is 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 in electrostatics that describes the force between two point charges. It is essential for understanding how electric charges interact. According to this law, the electric force (\( F \)) between two charges is directly proportional to the product of the magnitudes of the charges (\( |q_1 \cdot q_2| \)) and inversely proportional to the square of the distance (\( r^2 \)) between them. Mathematically, it is expressed as:\[F = k \cdot \frac{|q_1 \cdot q_2|}{r^2}\]where \( k \) is Coulomb's constant (\(8.9875 \times 10^9 \, ext{N.m}^2/ ext{C}^2 \)).
This law highlights that:
  • The force is attractive if charges are of opposite signs.
  • The force is repulsive if charges are of the same sign.
Coulomb's Law applies to point charges and helps in calculating the electric forces that play a crucial role in the configurations of electric charge distributions.
Electric Force
Electric force is the interaction between any two charged objects. This force is one of the crucial aspects of electrostatics and can cause attraction or repulsion between charged particles. In the scenario of our exercise, each corner charge (\( q \)) of the square experiences forces due to other charges.
The net electric force on a charge is calculated by considering all interacting charges around it. Given the symmetrical setup of charges in the square's corners, analysis becomes simpler. Forces from adjacent corner charges cancel out due to symmetry, focusing on the central charge (\( -Q \)) and one corner charge.
If we assume the distance from the center to a corner charge is \( a \), Coulomb's law guides the calculation of force between \( q \) and \( -Q \):\[F = k \cdot \frac{|q \cdot Q|}{a^2}\]For no net force, adjusting \( Q \) ensures balance, making total force equal zero.
Potential Energy
Potential energy in electrostatics involves the energy stored due to the positions of charged particles relative to one another. It is calculated for each pair of charges in a system. The formula used is:\[U = k \cdot \frac{|q_1 \cdot q_2|}{r}\]where \( r \) is the distance between the charges.
In our problem, because \( Q = 0 \) (as derived from balancing the forces), the potential energy associated with central charge interactions becomes zero. Thus, the total potential energy of the system is also zero. This outcome aligns with the expectation based on balanced force results and indicates the energy consistency in symmetrical configurations of charges. Each pair involving the central charge doesn't contribute to the potential energy, reinforcing that a carefully balanced system can have zero potential energy.
Symmetrical Distribution of Charges
A symmetrical distribution of charges significantly affects how charged particles exert forces on each other. It leads to simplifying complex problems because symmetrical forces often cancel each other out. In the square configuration discussed, each corner charge is at equal distances from the center and each other.
Consequently, symmetrical placement ensures:
  • Uniform forces acting on all charges at equal angles.
  • Net forces from symmetrical parts effectively canceling.
  • Balanced scenario when opposite charge sits centrally.
In solving problems, this symmetry allows us to focus only on key interactions, such as with a central charge, without worrying about individual pairwise forces. This setup is crucial in minimizing calculations and assumptions, offering straightforward solutions to electrostatic situations like the one explored here.

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Most popular questions from this chapter

Potential energy in a one-dimensional crystal * * Calculate the potential energy, per ion, for an infinite 1 D ionic crystal with separation \(a\); that is, a row of equally spaced charges of magnitude \(e\) and alternating sign. Hint: The power-series expansion of \(\ln (1+x)\) may be of use.

Charges on a circular track Suppose three positively charged particles are constrained to move on a fixed circular track. If the charges were all equal, an equilibrium arrangement would obviously be a symmetrical one with the particles spaced \(120^{\circ}\) apart around the circle. Suppose that two of the charges are equal and the equilibrium arrangement is such that these two charges are \(90^{\circ}\) apart rather than \(120^{\circ} .\) What is the relative magnitude of the third charge?

Decreasing energy? A hollow spherical shell with radius \(R\) has charge \(Q\) uniformly distributed over it. The task of Problem \(1.32\) is to show that the energy stored in this system is \(Q^{2} / 8 \pi \epsilon_{0} R\). (You can derive this here if you want, or you can just accept it for the purposes of this problem.) Now imagine taking all of the charge and concentrating it in two point charges \(Q / 2\) located at diametrically opposite positions on the shell. The energy of this new system is \((Q / 2)^{2} / 4 \pi \epsilon_{0}(2 R)=\) \(Q^{2} / 32 \pi \epsilon_{0} R\), which is less than the energy of the uniform spherical shell. Does this make sense? If not, where is the error in this reasoning?

Gravity vs. electricity (a) In the domain of elementary particles, a natural unit of mass is the mass of a nucleon, that is, a proton or a neutron, the basic massive building blocks of ordinary matter. Given the nucleon mass as \(1.67 \cdot 10^{-27} \mathrm{~kg}\) and the gravitational constant G as \(6.67 \cdot 10^{-11} \mathrm{~m}^{3} /\left(\mathrm{kg} \mathrm{s}^{2}\right)\), compare the gravitational attraction of two protons with their electrostatic repulsion. This shows why we call gravitation a very weak force. (b) The distance between the two protons in the helium nucleus could be at one instant as much as \(10^{-15} \mathrm{~m}\). How large is the force of electrical repulsion between two protons at that distance? Express it in newtons, and in pounds. Even stronger is the nuclear force that acts between any pair of hadrons (including neutrons and protons) when they are that close together.

Concurrent field lines A semicircular wire with radius \(R\) has uniform charge density \(-\lambda\). Show that at all points along the "axis" of the semicircle (the line through the center, perpendicular to the plane of the semicircle, as shown in Fig. 1.43), the vectors of the electric field all point toward a common point in the plane of the semicircle. Where is this point?

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