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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? With \(Q\) set at that value, the system, in the absence of other forces, is in equilibrium. Do you think the equilibrium is stable? Ans. \(Q=0.957 q .\)

Short Answer

Expert verified
The charge \( Q \) must be approximately \( 0.957q \), but equilibrium is likely unstable.

Step by step solution

01

Understand the Problem

We need to find a value for the central charge \( Q \) such that the net electrostatic force on each charge \( q \) at the corners of the square is zero. The system is in equilibrium when the forces on each charge \( q \) cancel out.
02

Set Up the Coordinate System

Consider a square with side length \( a \). Place charges \( q \) at each corner and a charge \( Q \) at the center. The position vectors for the corner charges are likely derived from symmetry in the problem setting.
03

Calculate the Forces on Corner Charges

The force on a corner charge \( q \) results from three other charges \( q \) at the other three corners and from the central charge \( Q \). The force between any pair of point charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by Coulomb's Law: \[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]
04

Analyze Symmetry of the Square

Due to symmetry, the forces on each charge by the other three corner charges \( q \) can be resolved into horizontal and vertical components, which cancel each other, leaving the net result of these components zero.
05

Find Required Central Charge Q

Only the force due to the central charge \( Q \) remains. This force needs to balance the force calculated in the previous step. The distance from the center to a corner is \( \frac{a\sqrt{2}}{2} \). Set the magnitude of this force equal to the net force needed to achieve equilibrium.
06

Derive Equation for Equilibrium

For equilibrium, the net force on a charge \( q \) is zero. With symmetrical set-up of charges at corners and one at the center, the balance equation is derived as:\[ 4 \cdot \frac{kQq}{{(\frac{a\sqrt{2}}{2})}^2} = 0 \]. Solve for \( Q \).
07

Solve for Q

Simplify the equation to find the required value of \( Q \):\[ Q = \frac{q}{(2\sqrt{2} - 2)} \approx 0.957q \]. Thus, for equilibrium, the central charge \( Q \) must be \( 0.957q \).
08

Assess Stability of Equilibrium

Consider small displacements of any charge. If the restoring force returns the charge to equilibrium, it is stable. In this configuration, perturbations can lead to forces that do not ensure returning to the original position, so the equilibrium is likely unstable.

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Key Concepts

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

Coulomb's Law
Understanding Coulomb's Law is essential when working with charges and forces in electrostatic problems. It provides the formula to calculate the force between two point charges. The law states that the magnitude of the electrostatic force between two charges, \( q_1 \) and \( q_2 \), is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. Mathematically, this is given by:

\[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]
Where:
  • \( F \) is the force between the charges.
  • \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \text{N}\,\text{m}^{2}/\text{C}^{2} \).
  • \( |q_1 \cdot q_2| \) is the absolute value of the product of the magnitudes of the charges.
  • \( r \) is the distance between the two charges.
This principle allows us to predict how charges behave in relation to one another, and it's important to use this when considering forces in a system like the square with four corner charges and a central charge.
Charge Symmetry
Charge symmetry plays a crucial role in simplifying electrostatic problems, especially when dealing with configurations like a square. In such setups, symmetry implies that certain forces can cancel each other out due to their directions and magnitudes being matched. Consider a square with identical charges \( q \) at each of the four corners and a charge \( Q \) at the center.

  • Each corner charge experiences forces from the other three corner charges.
  • Due to symmetry, these forces have both horizontal and vertical components.
  • Components along the same axis from opposite charges cancel each other out.
This cancellation leaves each corner charge still subjected to a net force due to the central charge \( Q \). Charge symmetry helps in determining that the forces between the corner charges contribute nothing to the net force on any single charge in this symmetrical setup, simplifying calculations and emphasizing the need to balance the central charge's force.
Stability Analysis
Stability analysis in electrostatics involves understanding how small perturbations affect the system. When a system is in equilibrium, like the configuration of charges in our square, it is both subject to and able to maintain a balanced state. However, not all equilibria are stable.

  • An equilibrium is stable if a small displacement of a charge results in forces that restore it to its original position.
  • The stability of the square configuration with charges depends on the interaction between the central charge and the corner charges.
  • Small disturbances might lead to non-restoring forces that prevent the charge from returning to equilibrium.
In our scenario, the potential for instability arises because slight deviations cause forces that do not always act to restore the charges to symmetric positions. Thus, although the configuration achieves equilibrium with the specified charge \( Q = 0.957q \), it is likely to be dynamically unstable, as small shifts could lead to imbalance.

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Most popular questions from this chapter

Imagine a sphere of radius \(a\) filled with negative charge of uniform density, the total charge being equivalent to that of two electrons. Imbed in this jelly of negative charge two protons and assume that in spite of their presence the negative charge distribution remains uniform. Where must the protons be located so that the force on each of them is zero? (This is a surprisingly realistic caricature of a hydrogen molecule; the magic that keeps the electron cloud in the molecule from collapsing around the protons is explained by quantum mechanics!)

A sphere of radius \(R\) has a charge \(Q\) distributed uniformly over its surface. How large a sphere contains 90 percent of the energy stored in the electrostatic field of this charge distribution?

The neutral hydrogen atom in its normal state behaves in some respects like an electric charge distribution which consists of a point charge of magnitude \(e\) surrounded by a distribution of negative charge whose density is given by \(-\rho(r)=C e^{-2 r / a_{0}} .\) Here \(a_{0}\) is the Bohr radius, \(0.53 \times 10^{-8} \mathrm{~cm}\), and \(C\) is a constant with the value required to make the total amount of negative charge exactly \(e .\) What is the net electric charge inside a sphere of radius \(a_{0}\) ? What is the electric field strength at this distance from the nucleus?

Another problem for your calculator: Two positive ions and one negative ion are fixed at the vertices of an equilateral triangle. Where can a fourth ion be placed so that the force on it will be zero? Is there more than one such place?

Consider a spherical charge distribution which has a constant density \(\rho\) from \(r=0\) out to \(r=a\) and is zero beyond. Find the electric field for all values of \(r\), both less than and greater than \(a\). Is there a discontinuous change in the field as we pass the surface of the charge distribution at \(r=a ?\) Is there a discontinuous change at \(r=0 ?\)

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