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Two identical rings \(P\) and \(Q\) of radius \(0.1 \mathrm{~m}\) are mounted coaxially at a distance \(0.5 \mathrm{~m}\) apart. The charges on the two rings are 2 and \(4 \mu \mathrm{C}\), respectively. The work done in transferring a charge of \(5 \mu \mathrm{C}\) from the center of \(P\) to that of \(Q\) is a. \(1.28 \mathrm{~J}\) b. \(0.72 \mathrm{~J}\) c. \(0.144 \mathrm{~J}\) d. \(2.24 \mathrm{~J}\)

Short Answer

Expert verified
The work done in transferring the charge is approximately \(2.24 \mathrm{~J}\).

Step by step solution

01

Understanding the Concept of Electric Potential

The work done in moving a charge in an electric field is given by the change in electric potential energy, which is the product of the charge and the potential difference between the starting and ending points. The potential at a point due to a charged ring is calculated considering the geometry and distribution of charge on the ring.
02

Electric Potential at the Center of a Charged Ring

The electric potential at the center of ring due to its charge is given by the formula: \[ V = \frac{kQ}{\sqrt{R^2 + x^2}} \] where \( k \) is Coulomb's constant \(8.99 \times 10^9 \mathrm{~N~m^2/C^2}\), \( Q \) is the charge on the ring, \( R \) is the radius of the ring, and \( x \) is the distance of the center from the ring along the axis. For point \( P\), \( x = 0 \), hence \( V_P = \frac{k(2 \times 10^{-6})}{0.1} \).
03

Calculate Potential at the Center of Ring Q

Similarly, calculate the potential at the center of ring \( Q \) due to both rings \( P \) and \( Q \). For the influence of \( P \) on \( Q \), we calculate \[ V' = \frac{k(2 \times 10^{-6})}{0.5} \]. For ring \( Q \)'s own contribution: \[ V_Q = \frac{k(4 \times 10^{-6})}{0.1} \]. Total \( V_Q = V' + V_Q \).
04

Potential Difference Calculation

Calculate the potential difference between the center of ring \( Q \) and \( P \): \( \Delta V = V_Q - V_P \). Substitute the calculated potentials from steps 2 and 3.
05

Calculate the Work Done

The work done in moving a charge \( q \) from \( P \) to \( Q \) is \[ W = q \Delta V \]. Substitute \( q = 5 \times 10^{-6} \mathrm{~C} \) and \( \Delta V \) from previous calculations to find \( W \).
06

Final Calculation

Substitute into the expression \[W = 5 \times 10^{-6} \](\(V_Q - V_P\)) to compute \(W\). Using given values and constants, calculate the numerical final result.

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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 fundamental to understanding electric force and electric fields. It defines the force between two charges. This law states that the force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. The mathematical expression for Coulomb’s Law is: \[ F = k \frac{|q_1 q_2|}{r^2} \]Where:
  • \( F \) is the force between the charges (in newtons)
  • \( k \) is the Coulomb's constant, approximately \(8.99 \times 10^9 \; \text{N m}^2/\text{C}^2\)
  • \( q_1 \) and \( q_2 \) are the magnitudes of the charges (in coulombs)
  • \( r \) is the distance between the charges (in meters)
Coulomb’s Law helps us calculate the electric fields created by the charged objects like rings or spheres, which then impact how other charges move around those objects. For systems of multiple charges, like in our original exercise, understanding how forces and potentials combine is crucial.
Electric Potential Energy
Electric potential energy is energy that results from the positions of charges in an electric field. It is closely related to work done, as moving a charge in or against an electric field involves work. In simple terms, electric potential energy is the ability to perform work due to electric forces. The formula to calculate electric potential energy ( \( U \)) between two point charges is:\[ U = k \frac{q_1 q_2}{r} \]Where:
  • \( U \) is the electric potential energy (in joules)
  • \( k \) is Coulomb's constant
  • \( q_1 \) and \( q_2 \) are the electrical charges
  • \( r \) is the distance between the charges
In our exercise, transferring a charge between the centers of two rings changes the system's electric potential energy, as the positions of charges involved in the field are altered. The ability to store energy in an electric field by relocating charges is crucial in understanding electric circuits and systems.
Work Done in Electric Fields
When we talk about work done with charges, we're referring to the energy required to move a charge from one position to another in an electric field. Work done in electric fields is related to electric potential difference and is expressed as:\[ W = q \Delta V \]Where:
  • \( W \) is the work done (in joules)
  • \( q \) is the charge being moved (in coulombs)
  • \( \Delta V \) is the potential difference (in volts) between two positions
The potential difference, \( \Delta V \), describes how much energy per coulomb is stored or required to move a charge from one point to another. In the case of the exercise, it determines how much work is needed to transfer a charge from the center of one ring to another, considering both fields created by the rings. Understanding work in this sense enables us to analyze energy transfer processes in physics and engineering contexts, including how many components in electrical systems work and interact.

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Most popular questions from this chapter

Four identical charges are placed at the points \((1,0,0)\), \((0,1,0),(-1,0,0)\) and \((0,-1,0)\). Then, a. the potential at the origin is zero b. the electric field at the origin is not zero c. the potential at all points on the \(z\) -axis, other than the origin, is zero d. the field at all points on the \(z\) -axis, other than the origin, acts along the \(z\) -axis

A positive charge is moved from a low potential point \(A\) to a high potential point \(B\). Then, the electric potential energy of the system a. increases b. decreases c. will remain the same d. nothing definite can be predicted

Two point charges \(Q\) and \(-Q / 4\) placed along \(x\) -axis are separated by a distance \(r\). Take \(-Q / 4\) as origin and it is placed right of \(Q\). Then, potential is zero a. at \(x=r / 3\) only b. at \(x=-r / 5\) only c. both at \(x=r / 3\) and at \(x=-r / 5\) d. there exist two points on the axis where electric field is zero

If a charge is moved against the coulomb force of an electric field, then a. positive work is done by the electric field b. energy is used from some outside source which does positive work c. the strength of the field is decreased d. the energy of the system is decreased

Let \(V_{0}\) be the potential at the origin in an electric field \(\vec{E}=E_{x} \hat{i}+E_{y} \hat{j}\). The potential at the point \((x, y)\) is a. \(V_{0}-x E_{x}-y E_{y}\) b. \(V_{0}+x E_{x}+y E_{y}\) c. \(x E_{x}+y E_{y}-V_{0}\) d. \(\left(\sqrt{x^{2}+y^{2}}\right) \sqrt{E_{x}^{2}+E_{y}^{2}}-V_{0}\)

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