College Physics
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Publisher: Cengage Learning
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![**Title: Calculating Energy for Point Charges in an Equilateral Triangle**
**Problem Statement:**
How much energy is necessary to place three +2.0-µC point charges at the vertices of an equilateral triangle of side 2.0 cm if they started out extremely far away?
**Given:**
- Charge of each point charge: \( +2.0 \, \mu C \)
- Side length of the equilateral triangle: \( 2.0 \, \text{cm} \)
- Constant \( k = \frac{1}{4\pi\varepsilon_0} = 9.0 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \)
---
**Explanation:**
The problem requires calculating the electric potential energy needed to assemble three like charges at specific positions forming an equilateral triangle. Given the nature of electrostatic forces, work must be done against the repulsive forces to bring each charge from infinity to its position.
**Solution Steps:**
1. **Convert Units:**
- Convert the side length from cm to meters: \( 2.0 \, \text{cm} = 0.02 \, \text{m} \).
2. **Use the Formula for Potential Energy:**
- The potential energy \( U \) of a system of point charges is given by:
\[
U = k \sum_{i < j} \frac{q_i q_j}{r_{ij}}
\]
- For three charges, consider all pairs:
\[
U = k \left( \frac{q_1 q_2}{r} + \frac{q_2 q_3}{r} + \frac{q_3 q_1}{r} \right)
\]
- Here, \( r = 0.02 \, \text{m} \) and \( q_1 = q_2 = q_3 = +2.0 \times 10^{-6} \, \text{C} \).
3. **Substitute Values:**
- Calculate each pair’s energy and sum them up:
\[
U = 9.0 \times 10^9 \cdot \left( \frac{(2.0 \times 10^{-6})^](https://content.bartleby.com/qna-images/question/af5d5edb-7877-4fa8-9f1a-7a2f9163b8ac/4c7cf3df-0f67-45cb-8707-4d488cc6ccd2/0dy9ei_processed.jpeg)
Transcribed Image Text:**Title: Calculating Energy for Point Charges in an Equilateral Triangle**
**Problem Statement:**
How much energy is necessary to place three +2.0-µC point charges at the vertices of an equilateral triangle of side 2.0 cm if they started out extremely far away?
**Given:**
- Charge of each point charge: \( +2.0 \, \mu C \)
- Side length of the equilateral triangle: \( 2.0 \, \text{cm} \)
- Constant \( k = \frac{1}{4\pi\varepsilon_0} = 9.0 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \)
---
**Explanation:**
The problem requires calculating the electric potential energy needed to assemble three like charges at specific positions forming an equilateral triangle. Given the nature of electrostatic forces, work must be done against the repulsive forces to bring each charge from infinity to its position.
**Solution Steps:**
1. **Convert Units:**
- Convert the side length from cm to meters: \( 2.0 \, \text{cm} = 0.02 \, \text{m} \).
2. **Use the Formula for Potential Energy:**
- The potential energy \( U \) of a system of point charges is given by:
\[
U = k \sum_{i < j} \frac{q_i q_j}{r_{ij}}
\]
- For three charges, consider all pairs:
\[
U = k \left( \frac{q_1 q_2}{r} + \frac{q_2 q_3}{r} + \frac{q_3 q_1}{r} \right)
\]
- Here, \( r = 0.02 \, \text{m} \) and \( q_1 = q_2 = q_3 = +2.0 \times 10^{-6} \, \text{C} \).
3. **Substitute Values:**
- Calculate each pair’s energy and sum them up:
\[
U = 9.0 \times 10^9 \cdot \left( \frac{(2.0 \times 10^{-6})^
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