Problem 47 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

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

Step 1: 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 \( U_{13} \):\[ U_{13} = \frac{(8.99 \times 10^9)(4.00 \times 10^{-9})(2.00 \times 10^{-9})}{0.10} = 7.192 \times 10^{-7} \text{ J} \].

Step 2: 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 \( U_{23} \):\[ U_{23} = \frac{(8.99 \times 10^9)(-3.00 \times 10^{-9})(2.00 \times 10^{-9})}{0.10} = -5.394 \times 10^{-7} \text{ J} \].

Step 3: 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 \( U_{23} \):\[ U = U_{13} + U_{23} = 7.192 \times 10^{-7} \text{ J} - 5.394 \times 10^{-7} \text{ J} = 1.798 \times 10^{-7} \text{ J} \].

Step 4: 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:\[ \frac{q_1}{x} = -\frac{q_2}{0.20-x} \].

Step 5: 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 \( x \):\[ 4.00(0.20 - x) = 3.00x \].\[ 0.80 - 4.00x = 3.00x \].\[ 0.80 = 7.00x \].\[ x = \frac{0.80}{7.00} = 0.1143 \text{ m} \text{ or } 11.43 \text{ cm} \].

Key Concepts

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.