College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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What is electrostatic force between two +1.0 C charges that are 0.5 m apart? (Coulomb's constant is 9.0 x 10^9 Nm^2/C^2; e= -1.602 x 10^-19 C)
![**CP Physics B - GHCHS 20-21**
**Question 30:** What is the electrostatic force between two +1.0 C charges that are 0.5 m apart? (Coulomb's constant is \(9.0 \times 10^9 \, \text{Nm}^2/\text{C}^2\); \(e = -1.602 \times 10^{-19} \, \text{C}\))
**Answer Choices:**
- \( \mathbf{2.0 \times 10^9 \, \text{N}} \)
- \( 3.6 \times 10^{10} \, \text{N} \)
- \( 1.8 \times 10^{10} \, \text{N} \)
- \( 4.0 \times 10^9 \, \text{N} \)
**Explanation of Concept:**
To solve for the electrostatic force (\( F \)) between two charges, we use Coulomb's Law, which is given by the formula:
\[
F = k \frac{q_1 q_2}{r^2}
\]
where
- \( k \) is Coulomb's constant (\( 9.0 \times 10^9 \, \text{Nm}^2/\text{C}^2 \)),
- \( q_1 \) and \( q_2 \) are the magnitudes of the charges (both are \( 1.0 \, \text{C} \) in this case),
- \( r \) is the separation distance between the charges ( \( 0.5 \, \text{m} \) in this case).
Substituting the values into the equation:
\[
F = (9.0 \times 10^9) \frac{(1.0)(1.0)}{(0.5)^2}
\]
Simplifying the expression within the parentheses:
\[
F = (9.0 \times 10^9) \frac{1}{0.25} = (9.0 \times 10^9) \times 4 = 3.6 \times 10^{10} \, \text{N}
\]
Therefore, the correct answer is:
**\( 3.6 \times 10^{10} \,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F37797b3e-93b1-4c2c-98d5-fb6f936a5e8e%2F3cc5fe66-e52b-45b9-b02c-418c75cc868e%2Flsw9nmo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**CP Physics B - GHCHS 20-21**
**Question 30:** What is the electrostatic force between two +1.0 C charges that are 0.5 m apart? (Coulomb's constant is \(9.0 \times 10^9 \, \text{Nm}^2/\text{C}^2\); \(e = -1.602 \times 10^{-19} \, \text{C}\))
**Answer Choices:**
- \( \mathbf{2.0 \times 10^9 \, \text{N}} \)
- \( 3.6 \times 10^{10} \, \text{N} \)
- \( 1.8 \times 10^{10} \, \text{N} \)
- \( 4.0 \times 10^9 \, \text{N} \)
**Explanation of Concept:**
To solve for the electrostatic force (\( F \)) between two charges, we use Coulomb's Law, which is given by the formula:
\[
F = k \frac{q_1 q_2}{r^2}
\]
where
- \( k \) is Coulomb's constant (\( 9.0 \times 10^9 \, \text{Nm}^2/\text{C}^2 \)),
- \( q_1 \) and \( q_2 \) are the magnitudes of the charges (both are \( 1.0 \, \text{C} \) in this case),
- \( r \) is the separation distance between the charges ( \( 0.5 \, \text{m} \) in this case).
Substituting the values into the equation:
\[
F = (9.0 \times 10^9) \frac{(1.0)(1.0)}{(0.5)^2}
\]
Simplifying the expression within the parentheses:
\[
F = (9.0 \times 10^9) \frac{1}{0.25} = (9.0 \times 10^9) \times 4 = 3.6 \times 10^{10} \, \text{N}
\]
Therefore, the correct answer is:
**\( 3.6 \times 10^{10} \,
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