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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### Physics Problem - Charge Distribution
#### Question:
If \( Q_{1} = 3.2 \, \mu C \), what is \( Q_{2} \)?
#### Diagram Description:
The diagram shows two spherical conductors connected by a wire. The radius \( r_{1} \) of the first sphere (Q1) is 0.8 cm, and the radius \( r_{2} \) of the second sphere (Q2) is 1.4 cm.
A thin wire connects these two spheres, allowing charge to redistribute between the spheres.
#### Answer Choices:
- \(2.55 \, \mu C\)
- \(9.8 \, \mu C\)
- \(5.6 \, \mu C\)
- \(3.2 \, \mu C\)
#### Solution Explanation:
To determine \( Q_{2} \), we utilize the principle of conservation of charge and the fact that the potential on both spheres must be equal (since they are connected by a conducting wire).
1. **Conservation of Charge:**
The total charge \( Q \) is the sum of the charges on both spheres:
\[
Q_{total} = Q_{1} + Q_{2}
\]
2. **Equal Potential:**
The potential \( V \) on each sphere is given by:
\[
V = \frac{Q}{r}
\]
Since the potential must be equal on both spheres:
\[
\frac{Q_{1}}{r_{1}} = \frac{Q_{2}}{r_{2}}
\]
Substituting the given values:
\[
\frac{3.2 \, \mu C}{0.8 \, cm} = \frac{Q_{2}}{1.4 \, cm}
\]
3. **Solving for \( Q_{2} \):**
\[
\frac{3.2}{0.8} = \frac{Q_{2}}{1.4}
\]
\[
4 = \frac{Q_{2}}{1.4}
\]
\[
Q_{2} = 4 \times 1.4 = 5.6 \, \mu C
\]
Thus, \( Q_{2} \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdb529b68-3495-4a4f-837a-ddb8b824decf%2F18aba7d2-d131-407b-9a84-9358b61ac01a%2Fpauuxho_processed.jpeg&w=3840&q=75)
Transcribed Image Text:---
### Physics Problem - Charge Distribution
#### Question:
If \( Q_{1} = 3.2 \, \mu C \), what is \( Q_{2} \)?
#### Diagram Description:
The diagram shows two spherical conductors connected by a wire. The radius \( r_{1} \) of the first sphere (Q1) is 0.8 cm, and the radius \( r_{2} \) of the second sphere (Q2) is 1.4 cm.
A thin wire connects these two spheres, allowing charge to redistribute between the spheres.
#### Answer Choices:
- \(2.55 \, \mu C\)
- \(9.8 \, \mu C\)
- \(5.6 \, \mu C\)
- \(3.2 \, \mu C\)
#### Solution Explanation:
To determine \( Q_{2} \), we utilize the principle of conservation of charge and the fact that the potential on both spheres must be equal (since they are connected by a conducting wire).
1. **Conservation of Charge:**
The total charge \( Q \) is the sum of the charges on both spheres:
\[
Q_{total} = Q_{1} + Q_{2}
\]
2. **Equal Potential:**
The potential \( V \) on each sphere is given by:
\[
V = \frac{Q}{r}
\]
Since the potential must be equal on both spheres:
\[
\frac{Q_{1}}{r_{1}} = \frac{Q_{2}}{r_{2}}
\]
Substituting the given values:
\[
\frac{3.2 \, \mu C}{0.8 \, cm} = \frac{Q_{2}}{1.4 \, cm}
\]
3. **Solving for \( Q_{2} \):**
\[
\frac{3.2}{0.8} = \frac{Q_{2}}{1.4}
\]
\[
4 = \frac{Q_{2}}{1.4}
\]
\[
Q_{2} = 4 \times 1.4 = 5.6 \, \mu C
\]
Thus, \( Q_{2} \)
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