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Chapter 23: Problem 47

A point charge \(q_1 =\) 4.00 nC is placed at the origin, and a second point charge \(q_2 = -\)3.00 nC is placed on the \(x\)-axis at \(x = +\)20.0 cm. A third point charge \(q_3 =\) 2.00 nC is to be placed on the \(x\)-axis between \(q_1\) and \(q_2\) . (Take as zero the potential energy of the three charges when they are infinitely far apart.) (a) What is the potential energy of the system of the three charges if \(q_3\) is placed at \(x = +\)10.0 cm? (b) Where should \(q_3\) be placed to make the potential energy of the system equal to zero?

Short Answer

Expert verified
(a) 1.798 x 10^-7 J, (b) q3 should be at 11.43 cm from q1.

Step by step solution

01

Calculate the potential energy due to q1 and q3

The formula for the potential energy between two point charges is given by \[ U = \frac{k \cdot q_1 \cdot q_3}{r} \]where \( k = 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2 \) is Coulomb's constant, \( q_1 = 4.00 \times 10^{-9} \text{ C} \), \( q_3 = 2.00 \times 10^{-9} \text{ C} \), and \( r = 10.0 \times 10^{-2} \text{ m} \). Calculate \( U_{13} \):\[ U_{13} = \frac{(8.99 \times 10^9)(4.00 \times 10^{-9})(2.00 \times 10^{-9})}{0.10} = 7.192 \times 10^{-7} \text{ J} \].
02

Calculate the potential energy due to q2 and q3

For the interaction between \( q_2 \) and \( q_3 \), use the same formula. Here, \( q_2 = -3.00 \times 10^{-9} \text{ C} \) and the distance is \( 10.0 \text{ cm} \), or \( r = 10.0 \times 10^{-2} \text{ m} \).Calculate \( U_{23} \):\[ U_{23} = \frac{(8.99 \times 10^9)(-3.00 \times 10^{-9})(2.00 \times 10^{-9})}{0.10} = -5.394 \times 10^{-7} \text{ J} \].
03

Calculate total potential energy of the system at x = +10.0 cm

The total potential energy is the sum of the potential energies from all pairwise interactions. Since there is no direct interaction between \( q_1 \) and \( q_2 \) as they are fixed, only add \( U_{13} \) and \( U_{23} \):\[ U = U_{13} + U_{23} = 7.192 \times 10^{-7} \text{ J} - 5.394 \times 10^{-7} \text{ J} = 1.798 \times 10^{-7} \text{ J} \].
04

Set up equation for zero potential energy

We need to find the position \( x \) where the potential energy is zero. The equation becomes:\[ \frac{k \cdot q_1 \cdot q_3}{x} + \frac{k \cdot q_2 \cdot q_3}{(0.20 - x)} = 0 \].Divide by \( k \cdot q_3 \) and simplify:\[ \frac{q_1}{x} = -\frac{q_2}{0.20-x} \].
05

Solve for x

Substitute \( q_1 = 4.00 \times 10^{-9} \text{ C} \) and \( q_2 = -3.00 \times 10^{-9} \text{ C} \):\[ \frac{4.00 \times 10^{-9}}{x} = \frac{3.00 \times 10^{-9}}{0.20 - x} \].Solve for \( x \):\[ 4.00(0.20 - x) = 3.00x \].\[ 0.80 - 4.00x = 3.00x \].\[ 0.80 = 7.00x \].\[ x = \frac{0.80}{7.00} = 0.1143 \text{ m} \text{ or } 11.43 \text{ cm} \].

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Key Concepts

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

Potential Energy
Potential energy in the context of electrostatics refers to the work done in bringing charges from infinity to their respective positions in a configuration, without any acceleration.
It helps us understand how charges interact and the stability of a configuration.
The potential energy between two point charges depends on:
  • The magnitudes of the charges.
  • The distance between them.
  • The medium in which the charges are placed, usually characterized by Coulomb's constant.
When charges are of opposite signs, such as in our exercise, the potential energy is negative, indicating that work is gained when bringing the charges together from infinity. This is due to the attractive nature of the force.
In our particular example, calculating potential energy involved considering interactions between all pairs: (1) first and third, and (2) second and third charges.
The total potential energy of the system is the sum of these individual potentials.
Coulomb's Law
Coulomb's Law is the fundamental principle that quantifies the electrostatic force between two point charges.
This law states that the magnitude of the force (\( F \)) between two point charges is directly proportional to the product of the absolute values of the charges, and inversely proportional to the square of the distance between them. It is mathematically expressed as:
  • \( F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \)
where \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between them.
In our exercise, we used this law to calculate potential energy, noting that potential energy equates analogous features with force over a distance.
This gives potential energy the form:
  • \( U = \frac{k \cdot q_1 \cdot q_2}{r} \)
This relationship highlights key dynamics:
  • A greater distance between charges decreases potential energy, aligning with the observation that it takes more effort to bring distant charges together.
  • A larger product of charge magnitudes increases the potential energy, reflecting a stronger interaction field.
Point Charges
Point charges are idealized representations of charges, where all the charge volume is assumed to be concentrated at a single point.
This simplifies calculating electric fields and forces between them, especially with analytical methods involving Coulomb's Law.
Real-world charges may be spread across an area or volume, but assuming point charges helps simplify complex calculations and provides accurate estimations in many scenarios.
In the exercise undertaken, the charges involved were considered point charges which could be conveniently used to find the total potential energy.
Point charge calculations are especially useful in understanding electrostatic interactions by focusing on the essence of interactions between separate cores of charge without considering the complexities of their finite sizes.

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Most popular questions from this chapter

A point charge \(q_1 = +\)5.00 \(\mu\)C is held fixed in space. From a horizontal distance of 6.00 cm, a small sphere with mass 4.00 \(\times 10^{-3}\) kg and charge \(q2 = +\)2.00 \(\mu\)C is fired toward the fixed charge with an initial speed of 40.0 m\(/\)s. Gravity can be neglected. What is the acceleration of the sphere at the instant when its speed is 25.0 m\(/\)s?

A total electric charge of 3.50 nC is distributed uniformly over the surface of a metal sphere with a radius of 24.0 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 cm; (b) 24.0 cm; (c) 12.0 cm.

Two stationary point charges \(+\)3.00 nC and \(+\)2.00 nC are separated by a distance of 50.0 cm. An electron is released from rest at a point midway between the two charges and moves along the line connecting the two charges. What is the speed of the electron when it is 10.0 cm from the \(+\)3.00-nC charge?

(a) An electron is to be accelerated from 3.00 \(\times 10^6\) m\(/\)s to 8.00 \(\times 10^6\) m\(/\)s. Through what potential difference must the electron pass to accomplish this? (b) Through what potential difference must the electron pass if it is to be slowed from 8.00 \(\times 10^6\) m\(/\)s to a halt?

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 \(x = -2a\) to \(x = +2a\). (e) What does the answer to part (b) become when \(x \gg a\)? Explain why this result is obtained.
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