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Chapter 3: Problem 47

When the separation between two charges is increased, the electric potential energy of the charges a. increases b. decreases c. remains the same d. may increase or decrease

Short Answer

Expert verified
b. decreases

Step by step solution

01

Understand the Problem

We need to determine how the electric potential energy between two charges changes when the separation between them is increased.
02

Recall the Formula for Electric Potential Energy

The electric potential energy between two point charges is given by the formula:\[ U = \frac{k \cdot q_1 \cdot q_2}{r} \]where \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the separation between the charges.
03

Analyze the Effect of Increasing Separation

As the separation \( r \) between the charges increases, the denominator in the formula \( \frac{k \cdot q_1 \cdot q_2}{r} \) becomes larger, thereby reducing the value of the electric potential energy \( U \).
04

Determine the Result

Since increasing the separation \( r \) leads to a larger denominator and thus a smaller value for \( U \), the electric potential energy decreases.

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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 a fundamental principle in physics that explains the force between two point charges. It is represented in the equation:\[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]Here,
  • \( F \) is the force between the charges.
  • \( k \) is 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.
  • \( r \) is the distance between the centers of the two charges.
Coulomb's Law tells us how the force is either attractive or repulsive. This depends on the sign of the charges. If charges are of opposite signs, the force is attractive; if the same, it is repulsive. Understanding this law is vital for grasping the concept of electric potential energy, which relies on similar components.
Point Charges
In physics, a point charge refers to an idealized object with a certain amount of charge that is located at a single point in space. Imagine the charge is concentrated at one point with no size or volume. There are a couple of key reasons why we use point charges in physics:
  • They simplify calculations as they allow us to use clean mathematical models, such as Coulomb's Law and the formula for electric potential energy.
  • By treating them as point charges, we can understand the basic principles governing the behavior of electric charges without interference from complex shapes or charge distributions.
Real-world objects aren't perfect point charges, but many small charged particles like electrons act similarly. Knowing how point charges work helps build a foundation for more complex electrical phenomena analysis.
Separation Distance Effect
Separation distance between charges plays a critical role in determining electric forces and potential energy. As the distance \( r \) between point charges increases, it affects how strongly they interact.For example:
  • When two charges are close, the force and potential energy are higher because the distance \( r \) is smaller.
  • As the separation distance \( r \) increases, electric potential energy \( U \)
    • Decreases, since the formula \( U = \frac{k \cdot q_1 \cdot q_2}{r} \) shows that increasing \( r \) leads to a larger denominator.
  • Distance and force have an inverse square relationship, meaning the force decreases rapidly as distance grows.
By understanding the separation distance effect, we can determine how systems of charges behave dynamically and how objects with electric charges will influence each other in various configurations.

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Most popular questions from this chapter

There is an infinite straight chain of alternating charges \(q\) and \(-q\). The distance between the two neighboring charges is equal to \(a\). Find the interaction energy of any charge with all the other charges. a. \(-\frac{2 q^{2}}{4 \pi \varepsilon_{0} a}\) b. \(\frac{2 q^{2} \log _{e} 2}{4 \pi \varepsilon_{0} a}\) c. \(-\frac{2 q^{2} \log _{e} 2}{4 \pi \varepsilon_{0} a}\) d. None of these

A hollow metal sphere of radius \(5 \mathrm{~cm}\) is charged such that the potential on its surface is \(10 \mathrm{~V}\). The potential at the center of the sphere is a. \(0 \mathrm{~V}\) b. \(10 \mathrm{~V}\) c. same as at point \(5 \mathrm{~cm}\) away from the surface d. same as at a point \(20 \mathrm{~cm}\) away from the surface

A particle of mass \(m\) carrying charge ' \(q^{\prime}\) is projected with velocity ' \(v^{\prime}\) from point ' \(P\) ' towards an infinite line of charge from a distance ' \(a\) '. Its speed reduces to zero momentarily at point \(Q\) which is at a distance \(a / 2\) from the line of charge. If an贸ther particle with mass \(m\) and charge \(-q^{\prime}\) is projected with the same velocity ' \(v^{\prime}\) from \(P\) towards the line of charge, what will be its speed at \(Q\) ? a. \(\sqrt{2} v\) b. \(\sqrt{3} v\) c. \(\frac{v}{\sqrt{2}}\) d. \(\frac{v}{\sqrt{3}}\)

A point charge \(q\) is placed inside a conducting spherical shell of inner radius \(2 R\) and outer radius \(3 R\) at a distance of \(R\) from the center of the shell. Find the electric potential at the center of the shell. a. \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{2 R}\) b. \(\frac{1}{4 \pi \varepsilon_{0}} \frac{4 q}{3 R}\) c. \(\frac{1}{4 \pi \varepsilon_{0}} \frac{5 q}{6 R}\) d. \(\frac{1}{4 \pi \varepsilon_{0}} \frac{2 q}{3 R}\)

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
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