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When two point charges are a distance \(R\) apart, their potential energy is \(-2.0 \mathrm{J} .\) How far far (in terms of \(R )\) should they be from each other so that their potential energy is \(-6.0 \mathrm{J} ?\)

Short Answer

Expert verified
The charges should be \( \frac{R}{3} \) apart for their potential energy to be \(-6.0 \) J.

Step by step solution

01

Understanding the Relationship

The potential energy between two point charges is given by the formula \( U = \frac{k \cdot q_1 \cdot q_2}{r} \), where \( U \) is the potential energy, \( k \) is the Coulomb's constant, \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between the charges. This means the potential energy is inversely proportional to the distance between the charges.
02

Setting Up the Equation

Given that the potential energy at a distance \( R \) is \(-2.0\) J, we can express this as \( U_1 = \frac{kq_1q_2}{R} = -2.0 \). We need to find the new distance \( r \) for which the potential energy \( U_2 = -6.0 \) J.
03

Relating the Two Situations

We know that \( \frac{kq_1q_2}{R} = -2.0 \) and we want \( \frac{kq_1q_2}{r} = -6.0 \). By dividing the second equation by the first, we get:\[ \frac{\frac{kq_1q_2}{r}}{\frac{kq_1q_2}{R}} = \frac{-6.0}{-2.0} \].
04

Simplifying and Solving for r

The division leads to:\[ \frac{R}{r} = 3 \].This implies \( r = \frac{R}{3} \). The two charges should be \( \frac{R}{3} \) apart for their potential energy to be \(-6.0 \) J.

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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 of electrostatics that describes the force between two point charges. It tells us that the electrostatic force () between two charges () is directly proportional to the product of the magnitudes of the charges () and inversely proportional to the square of the distance () between them. The law is expressed mathematically as:
  • \( F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \)
where \( F \) is the electrostatic force, \( k \) is Coulomb's constant (), \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between the charges. It is important in understanding electrostatic interactions and is similar in form to the law of gravitation, which describes gravitational forces.
Potential Energy
Potential energy in electrostatics refers to the stored energy between two charges due to their positions. Just like a stretched spring has potential energy, two point charges also accumulate potential energy. This is given by the formula:
  • \( U = \frac{k \cdot q_1 \cdot q_2}{r} \)
where \( U \) is the potential energy, \( q_1 \) and \( q_2 \) are the charges, \( r \) is the distance between them, and \( k \) is Coulomb's constant. Unlike kinetic energy, which is associated with movement, potential energy in this context is due to configuration. A critical point is that the sign of potential energy can be positive or negative, depending on whether the charges are like or unlike. Like charges store positive potential energy, while unlike charges (one positive, one negative) store negative potential energy, as in the given problem. Their potential energy becomes more negative as they come closer.
Distance Between Charges
The distance between the charges is a key factor in calculating both the force and potential energy between two point charges. In the context of this exercise, changing the distance changes the potential energy. The equation \( U = \frac{k \cdot q_1 \cdot q_2}{r} \) shows that the potential energy is inversely related to the distance; this leads us to the concept of 'inverse proportionality'. When the distance \( r \) changes, it directly affects the amount of potential energy stored between the charges. To manage the potential energy from \(-2.0 \) J to \(-6.0 \) J, we see that the charges need to be closer to each other, as their distance shrinks from \( R \) to \( \frac{R}{3} \), as derived in the solution.
Inverse Proportionality
Inverse proportionality is a relationship where one value increases as another decreases, and vice versa. In electrostatics, potential energy illustrates this beautifully. As seen in the formula \( U = \frac{k \cdot q_1 \cdot q_2}{r} \), potential energy \( U \) is inversely proportional to the distance \( r \). This means if the distance between charges decreases, the potential energy increases (or becomes more negative).This principle of inverse proportionality is crucial when solving problems related to charge interactions over varying distances. For instance, going from a potential energy of \(-2.0\) J to \(-6.0\) J signals that the potential energy level is increasing in magnitude, necessitating that the distance between charges viz., \( r \), must decrease. Therefore, by understanding and applying this idea, we can predict and calculate the behavior of charged particles in different scenarios.

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Most popular questions from this chapter

A potential difference of 4.75 \(\mathrm{kV}\) is established between parallel plates in air. If the air becomes ionized (and hence electrically conducting) when the electric field exceeds \(3.00 \times 10^{6} \mathrm{V} / \mathrm{m},\) what is the minimum separation the plates can have without ionizing the air?

A 10.0\(\mu \mathrm{F}\) parallel-plate capacitor with circular plates is connected to a 12.0 \(\mathrm{V}\) battery. (a) What is the charge on each plate? (b) How much charge would be on the plates if their separation were doubled while the capacitor remained connected to the battery? (c) How much charge would be on the plates if the capacitor were connected to the 12.0 \(\mathrm{V}\) battery after the radius of each plate was doubled without changing their separation?

Two very large charged parallel metal plates are 10.0 \(\mathrm{cm}\) apart and produce a uniform electric field of \(2.80 \times 10^{6} \mathrm{N} / \mathrm{C}\) between them. A proton is fired perpendicular to these plates with an initial speed of 5.20 \(\mathrm{km} / \mathrm{s}\) , starting at the middle of the negative plate and going toward the positive plate. How much work has the electric field done on this proton by the time it reaches the positive plate?

A parallel-plate air capacitor is made from two plates 0.200 m square, spaced 0.800 \(\mathrm{cm}\) apart. It is connected to a \(120-\mathrm{V}\) battery. (a) What is the capacitance? (b) What is the charge on each plate? (c) What is the electric field between the plates? (d) What is the energy stored in the capacitor? (e) If the battery is disconnected and then the plates are pulled apart to a separation of \(1.60 \mathrm{cm},\) what are the answers to parts (a), (b), \((c),\) and \((d) ?\)

How far from a \(-7.20 \mu \mathrm{C}\) point charge must a \(+2.30 \mu \mathrm{C}\) point charge be placed in order for the electric potential energy of the pair of charges to be \(-0.400 \mathrm{J} ?\) (Take the energy to be zero when the charges are infinitely far apart.)

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