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Chapter 23: Problem 58

A point charge \(Q\) is placed at the centre of a circular wire of radius \(R\) having charge \(q\). The force of (a) \(\frac{q Q}{4 \pi \varepsilon_{0} R^{2}}\) point charge and the wire is : electrostatic interaction between (b) zero (c) \(\frac{q^{2}}{4 \pi \varepsilon_{0} R}\) (d) none of these

Short Answer

Expert verified
The correct answer is (b): zero.

Step by step solution

01

Understand the Problem

We have a point charge \( Q \) at the center of a circular wire of radius \( R \) with a charge \( q \). We need to determine the electrostatic interaction force between the point charge and the wire.
02

Apply Gauss's Law

By symmetry, the net electric field inside the wire due to its charge is zero, because the wire can be considered as a continuous charge distribution with charge \( q \). According to Gauss's Law, the enclosed charge by a Gaussian surface containing the point charge is zero because the wire is a line charge outside this surface.
03

Assess Electrostatic Force

Since there is no electric field inside the wire at any point beyond the location of the point charge (which is at the center), the net electrostatic force on the point charge \( Q \) due to the wire's charge distribution is also zero. The forces from symmetrical parts of the charge distribution on opposite sides of the point charge cancel each other out.

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Key Concepts

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

Point Charge
A point charge is a fundamental concept in electrostatics. It represents a charged particle with negligible size, acting as a source of an electric field. Consider a point charge with a magnitude \( Q \). This charge creates an electric field around it that radiates outward in all directions. The electric field \( E \) due to a point charge can be calculated using the formula:
  • \( E = \frac{kQ}{r^2} \)
where \( k \) is Coulomb's constant and \( r \) is the distance from the charge.

Point charges are idealized concepts that help us simplify complex problems, such as the interaction of charges in space. They allow us to model the behavior of charges without worrying about the shape or size of the actual objects involved.
Gauss's Law
Gauss's Law is a powerful tool in electrostatics. It connects the electric field penetrating a closed surface to the charge enclosed by that surface. Mathematically, Gauss's Law is expressed as:
  • \( \oint E \, dA = \frac{Q_{enc}}{\varepsilon_0} \)
where \( \oint E \, dA \) is the electric flux through a closed surface, \( Q_{enc} \) is the total charge enclosed, and \( \varepsilon_0 \) is the permittivity of free space.

In the context of the original exercise, Gauss's Law helped to realize that the electric field within the wire, due to its own charge distribution, is zero. Since the wire's charge lies outside a Gaussian surface enclosing the point charge \( Q \), it does not contribute to the net flux. Hence, the point charge at the center experiences no net electric field from the wire.
Symmetry in Electric Field Distribution
Symmetry in electric field distribution plays a crucial role in analyzing electrostatic problems. Symmetry helps simplify these problems by showing when forces cancel out.
  • Symmetry tells us that if a charge distribution is uniform and symmetrical, the electric fields it generates can cancel out.
  • In the case of the simple circular wire charged uniformly, every small segment of the wire generates an electric field.

These fields, however, cancel out due to symmetry, especially across the central point where a point charge is placed. The symmetrically opposite segments of the wire exert equal and opposite forces on the charge.

Therefore, due to symmetric distribution, these forces cancel out perfectly, leading to a net zero electrostatic force on the point charge from the wire.

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Most popular questions from this chapter

An electron of mass \(m\) and charge \(e\) leaves the lower plate of a parallel plate capacitor of length \(L\), with an initial velocity \(v_{0}\) making an angle \(\alpha\) with the plate and comeout of the capacitor making an angle \(\beta\) to the plate. The electric field intensity between the plates: (a) \(E:=\frac{m v_{0}^{2} \cos ^{2} \alpha}{e L}(\tan \alpha+\tan \beta)\) (b) \(E=\frac{m v_{0}^{2} \cos ^{2} \alpha}{c L}(\tan \alpha-\tan \beta)\) (c) \(E=\frac{m v_{0}^{7} \cos ^{2} \alpha}{e L}(\tan \beta-\tan \alpha)\) (d) none of the above

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Two charged particles of charge \(+2 q\) and \(+q\) have masses \(m\) and \(2 m\) respectively. They are kept in uniform electric field and allowed to move for the same time. The ratio of their kinetic energies is: (a) \(1: 8\) (b) \(16: 1\) (c) \(2: 1\) (d) \(3: 1\)

A dimensionless body having a physical quantity varies as \(1 / r^{2}\), where \(r\) is distance from the body. This physical quantity may be : (a) gravitational potential (b) electric field (c) gravitational fiot, (d) none of the aboue

An electron is projected with velocity \(10^{7} \mathrm{~m} / \mathrm{s}\) at an angle \(\theta\left(=30^{\circ}\right)\) with horizontal in a region of uniform electric field of \(5000 \mathrm{~N} / \mathrm{C}\) vertically upwards. The maximum distance covered by an electron in vertical direction above its initial level is : (a) \(14.2 \mathrm{~mm}\) (b) \(15 \mathrm{~mm}\) (c) \(12.6 \mathrm{~mm}\) (d) \(14.2 \mathrm{~cm}\)
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