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Chapter 24: Problem 39

Over a thin ring of radius \(R\) a charge \(q\) is distributed non-uniformly. The work done of the force field in displacing a point charge \(q^{\prime}\) from centre of the ring to infinity is: (a) \(\frac{q q^{\prime}}{4 \pi \varepsilon_{0} R}\) (b) \(\frac{q q^{\prime}}{2 \pi \varepsilon_{0} R}\) (c) \(\frac{q q^{\prime}}{\pi \varepsilon_{0} R}\) (d) none of these

Short Answer

Expert verified
(a) \(\frac{q q'}{4\pi \varepsilon_0 R}\)

Step by step solution

01

Understand the Problem

We need to find the work done in moving a point charge \( q' \) from the center of a ring to infinity when the charge \( q \) is non-uniformly distributed on the ring with radius \( R \). The key is to find the electric potential difference between the center of the ring and infinity.
02

Conceptualize Electric Potential

The electric potential \( V \) due to a charge distribution is given by integrating the electric potential due to each infinitesimal charge element. Since the potential at infinity is zero, we only need the potential at the center of the ring.
03

Calculate Electric Potential at Center

For a ring with a charge \( q \) uniformly distributed, the electric potential \( V \) at the center is given by: \[ V = \frac{k_e q}{R} \] where \( k_e = \frac{1}{4\pi \varepsilon_0} \) is Coulomb's constant. As the distribution is non-uniform, the formula remains unchanged as it applies to the entire charge on the ring.
04

Calculate Work Done

The work done \( W \) in moving charge \( q' \) from the center to infinity is given by the change in potential energy, which is: \[ W = q' \times V = q' \times \frac{k_e q}{R} = \frac{q q'}{4\pi \varepsilon_0 R} \]
05

Choose the Correct Answer

Compare the calculated work \( \frac{q q'}{4\pi \varepsilon_0 R} \) with the given options. It matches option (a).

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Key Concepts

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

Electric Potential
When we talk about electric potential, we're discussing the amount of electric potential energy per unit charge at a point in an electric field. This concept helps us understand how charges interact with each other in space.
The formula for electric potential \( V \) due to a point charge \( Q \) is \( V = \frac{k_e Q}{r} \), where \( k_e \) is Coulomb's constant (\( \frac{1}{4\pi \varepsilon_0} \)) and \( r \) is the distance from the point charge. For a distribution of charge, as in the exercise, the potential is the sum of contributions from each infinitesimal part of the charge. The potential at infinity is often considered zero, providing a reference point.
Knowing the potential at a particular point allows us to determine the work needed to move a charge between two points in an electric field. For any fixed points, work done is the product of electric potential difference and charge moved, emphasizing the role of potential in calculating work.
Non-uniform Charge Distribution
Charges are not always spread out equally across an object, leading to what is known as non-uniform charge distribution. This distribution can significantly affect the potential created by the charge in space.
In the exercise, a charge \( q \) is non-uniformly distributed over a ring. Despite this complexity, for symmetrical configurations like a ring, the formula for potential at a specific point remains similar to that of uniform distributions. This is because, as far as potential at the ring's center is concerned, the entire ring seems like a single charge at a specific distance \( R \) due to the symmetry.
Understanding non-uniform distributions is crucial because they represent real-world scenarios, where charge does not always distribute evenly. Calculating the effects of such distributions involves integrating the contributions from all differential elements that make up the total charge.
Coulomb's Law
Coulomb's law describes the electrostatic interaction between electrically charged particles. It states that the force between two point charges \( q_1 \) and \( q_2 \) is proportional to the product of their charges and inversely proportional to the square of the distance between them.
Mathematically, it's expressed as \( F = \frac{k_e q_1 q_2}{r^2} \), where \( F \) is the force, \( r \) is the distance between charges, and \( k_e \) is Coulomb's constant. This law is key to understanding how forces act in an electric field and helps in understanding the calculations for electric potential and work done.
In the context of the exercise, Coulomb's law underlies the formula for potential \( V \) used to calculate work. It provides a foundational understanding of how point charges interact, guiding us through the steps needed to solve real-world problems, like moving a charge within an electric field formed by a charge distribution.

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Most popular questions from this chapter

A point charge \(q=2 \times 10^{-7} \mathrm{C}\) is placed at the centre of a spherical cavity of radius \(3 \mathrm{~cm}\) in a metal piece. Points \(a\) and \(b\) are situated at distances \(1.5 \mathrm{~cm}\) and \(4.5 \mathrm{~cm}\) respectively from the centre of cavity. The electric intensities at \(a\) and \(b\) are : (a) \(8 \times 10^{6} \mathrm{~N} / \mathrm{C}\) and zero (b) zero and zero (c) zero and \(8.9 \times 10^{5} \mathrm{~N} / \mathrm{C}\) (d) none of the above

The electric field in a region is given by \(\mathrm{E}=\alpha x \hat{\mathbf{1}}\), where \(\alpha=\) constant of proper dimensions. What should be the charge contained inside a cube bounded by the surface, \(x=1, x=2 l, y=0, y=l, z=0, z=l ?\) (a) \(\frac{\varepsilon_{0} l^{3}}{\alpha}\) (b) \(\alpha \varepsilon_{0} l^{3}\) (c) \(\frac{\alpha \varepsilon_{0}}{l^{3}}\) (d) \(2 \alpha \varepsilon_{0} l^{3}\)

Four charges, all of the same magnitude are placed at the four corners of a square. At the centre of the square, the potential is \(V\) and the field is \(E\). By suitable choices of the signs of the four charges, which of the following can be obtained? (a) \(V=0, E=0\) (b) \(V=0, E \neq 0\) (c) \(V \neq 0, E=0\) (d) None of these

A point charge \(Q\) is placed at the centre of a hemisphere. The electric flux passing through flat surface of hemisphere is : Q (a) \(\varepsilon_{0}\) (b) zero (c) \(\frac{Q}{2 \varepsilon_{0}}\) (d) none of these

Mark the correct option(s): (a) Electric tield must be conservative (b) Electric field must be non-conservative (c) Electric field may be non-conservative (d) The potential throughout the region bounded by a closed surface containing no charge may be variable
See all solutions

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