/*! 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 1 A point charge \(q_{1}\) is held... [FREE SOLUTION] | 91Ó°ÊÓ

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A point charge \(q_{1}\) is held stationary at the origin. A second charge \(q_{2}\) is placed at point \(a\), and the electric potential energy of the pair of charges is \(+5.4 \times 10^{-8} \mathrm{~J}\). When the second charge is moved to point \(b\), the electric force on the charge does \(-1.9 \times 10^{-8} \mathrm{~J}\) of work. What is the electric potential energy of the pair of charges when the second charge is at point \(b\) ?

Short Answer

Expert verified
The electric potential energy at point b is \(7.3 \times 10^{-8} \mathrm{~J}\).

Step by step solution

01

Understand Electric Potential Energy at Point a

First, note that the electric potential energy at point \(a\) is given as \(U_a = +5.4 \times 10^{-8} \mathrm{~J}\). This energy is specific to when the second charge, \(q_2\), is at point \(a\) relative to \(q_1\).
02

Calculate Work Done by Electric Force

The electric force does work on the charge \(q_2\) as it moves from point \(a\) to point \(b\). Given that this work is \(W = -1.9 \times 10^{-8} \mathrm{~J}\), it contributes to the change in electric potential energy between points \(a\) and \(b\).
03

Determine the Potential Energy at Point b Using Work-Energy Principle

The principle of conservation of energy tells us that the change in electric potential energy between two points is equal to the negative of the work done by the electric force when moving between those points. Thus, \( \Delta U = U_b - U_a = -W \), where \(W\) is the work done by the electric force. Substitute the given values: \( \Delta U = U_b - 5.4 \times 10^{-8} = -(-1.9 \times 10^{-8})\).
04

Solve for Electric Potential Energy at Point b

Now solve for \( U_b \) using \( \Delta U = U_b - U_a = -W \). Plug in the numbers: \( U_b = 5.4 \times 10^{-8} + 1.9 \times 10^{-8} = 7.3 \times 10^{-8} \mathrm{~J}\). This represents the electric potential energy of the pair of charges when charge \( q_2 \) is at point \( b \).

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Key Concepts

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

Conservation of Energy
The concept of the conservation of energy is all about understanding that energy cannot just disappear or appear out of nowhere. Instead, it can only change forms or transfer from one system to another. In the context of our problem, we are dealing with electric potential energy, which is the energy stored in the positions of charged particles.
When the second charge moves from point \(a\) to point \(b\), the conservation of energy tells us that the total energy before and after the charge's movement remains the same. The initial electric potential energy and the work done by the electric force are all part of this energy balance.
In our scenario, as the charge \(q_2\) moves from point \(a\) to point \(b\), the work done on the charge by the electric force changes the system's electric potential energy. It's crucial to consider this very principle in calculations, as it ensures that all forms of energy are conserved across the transformation.
Work-Energy Principle
The work-energy principle explains how work done by forces results in a change in energy. Simply put, work can modify the energy state of a system.
In our problem, this principle is evident when the electric force does work on the charge \(q_2\) as it moves from point \(a\) to point \(b\). The work done has been given as \(-1.9 \times 10^{-8} \mathrm{~J}\).
This principle allows us to understand that:
  • The amount of work done is equivalent to the change in electric potential energy.
  • This change in energy determines how the electric potential energy at point \(b\) will differ from the initial state at point \(a\).
By applying the work-energy principle, we recognize that any work performed (negative or positive) directly translates to a change in the system's energy.
Electric Force
An electric force is a fundamental interaction between charges, and it can do work on charges, causing changes in their energy states. Electric force is central to this problem since it brings about the movement of the charge \(q_2\) and helps us determine the final electric potential energy.
The electric force can exert a push or pull on the charge, depending on the nature of the charges involved. Importantly, it has the ability to do work as the charge moves, influencing the potential energy of the system.
As given in the problem, the work done by this force is negative, \(-1.9 \times 10^{-8} \mathrm{~J}\), which signifies that energy is released by the electric field. The negative sign suggests that the energy of the system decreases in that direction or is transferred elsewhere. This interaction is what leads to the change in potential energy that we calculated using the conservation of energy and work-energy principle.

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Most popular questions from this chapter

A positive charge \(q\) is fixed at the point \(x=0, y=0\), and a negative charge \(-2 q\) is fixed at the point \(x=a, y=0\). (a) Show the positions of the charges in a diagram. (b) Derive an expression for the potential \(V\) at points on the \(x\)-axis as a function of the coordinate \(x\). Take \(V\) to be zero at an infinite distance from the charges. (c) At which positions on the \(x\)-axis is \(V=0 ?\) (d) Graph \(V\) at points on the \(x\)-axis as a function of \(x\) in the range from \(x=-2 a\) to \(x=+2 a .\) (e) What does the answer to part (b) become when \(x \gg a\) ? Explain why this result is obtained.

The electric potential \(V\) in a region of space is given by $$ V(x, y, z)=A\left(x^{2}-3 y^{2}+z^{2}\right) $$ where \(A\) is a constant. (a) Derive an expression for the electric field \(\overrightarrow{\boldsymbol{E}}\) at any point in this region. (b) The work done by the field when a \(1.50-\mu \mathrm{C}\) test charge moves from the point \((x, y, z)=\) \((0,0,0.250 \mathrm{~m})\) to the origin is measured to be \(6.00 \times 10^{-5} \mathrm{~J}\). Determine \(A\). (c) Determine the electric field at the point \((0,0\), \(0.250 \mathrm{~m}\) ). (d) Show that in every plane parallel to the \(x z\)-plane the equipotential contours are circles. (e) What is the radius of the equipotential contour corresponding to \(V=1280 \mathrm{~V}\) and \(\mathrm{y}=2.00 \mathrm{~m}\) ?

A positive charge \(+q\) is located at the point \(x=0\), \(y=-a\), and a negative charge \(-q\) is located at the point \(x=0\), \(y=+a .\) (a) Derive an expression for the potential \(V\) at points on the \(y\)-axis as a function of the coordinate \(y\). Take \(V\) to be zero at an infinite distance from the charges. (b) Graph \(V\) at points on the \(y\)-axis as a function of \(y\) over the range from \(y=-4 a\) to \(y=+4 a\). (c) Show that for \(y>a\), the potential at a point on the positive \(y\)-axis is given by \(V=-\left(1 / 4 \pi \epsilon_{0}\right) 2 q a / y^{2}\). (d) What are the answers to parts (a) and (c) if the two charges are interchanged so that \(+q\) is at \(y=+a\) and \(-q\) is at \(y=-a\) ?

A uniform surface charge of density \(\sigma\) is given to quarter infinite non conducting plane in first quardrant of \(x-y\) plane. Find the \(z\)-component of the electric field at the point \((0,0, z)\). Hence or otherwise find the potential difference between the points \((0,0, d) \&(0,0,2 d)\).

A total electric charge of \(3.50 \mathrm{nC}\) is distributed uniformly over the surface of a metal sphere with a radius of \(24.0 \mathrm{~cm}\). If the potential is zero at a point at infinity, find the value of the potential at the following distances from the center of the sphere: (a) \(48.0 \mathrm{~cm}\); (b) \(24.0 \mathrm{~cm} ;\) (c) \(12.0 \mathrm{~cm}\)

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