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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Question
![**Title: Calculating the Electric Field in a Capacitor**
**Objective:**
Determine the magnitude and direction of the electric field for the given capacitor setup.
**Description:**
The diagram illustrates a parallel plate capacitor. The left plate is connected to 150 V, and the right plate is connected to 0 V. The distance between the plates is indicated as 3.0 mm. The spacing between specific points is marked as 1.0 mm segments.
**Analysis:**
1. **Potential Difference (ΔV):**
- The potential difference between the plates is calculated as:
\[
\Delta V = 150 \, \text{V} - 0 \, \text{V} = 150 \, \text{V}
\]
2. **Distance Between Plates (d):**
- The given distance between the plates is 3.0 mm, which can be converted to meters for standard SI units:
\[
d = 3.0 \, \text{mm} = 0.003 \, \text{m}
\]
3. **Electric Field (E):**
- The electric field (E) between the plates of the capacitor can be calculated using the formula:
\[
E = \frac{\Delta V}{d} = \frac{150 \, \text{V}}{0.003 \, \text{m}} = 50,000 \, \text{V/m}
\]
4. **Direction of the Electric Field:**
- The electric field direction is from the positive plate (150 V) to the negative plate (0 V).
**Conclusion:**
- **Magnitude of the Electric Field:** 50,000 V/m
- **Direction:** From left (150 V) to right (0 V)
This analysis of the capacitor helps understand basic electric field concepts and their calculation in electrostatics.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F070e8b65-931e-44e7-bfb4-6d69e5ced655%2Ffe96cb42-e3d4-46f8-8ee4-af7b234fe2e1%2Fpipde1o_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Calculating the Electric Field in a Capacitor**
**Objective:**
Determine the magnitude and direction of the electric field for the given capacitor setup.
**Description:**
The diagram illustrates a parallel plate capacitor. The left plate is connected to 150 V, and the right plate is connected to 0 V. The distance between the plates is indicated as 3.0 mm. The spacing between specific points is marked as 1.0 mm segments.
**Analysis:**
1. **Potential Difference (ΔV):**
- The potential difference between the plates is calculated as:
\[
\Delta V = 150 \, \text{V} - 0 \, \text{V} = 150 \, \text{V}
\]
2. **Distance Between Plates (d):**
- The given distance between the plates is 3.0 mm, which can be converted to meters for standard SI units:
\[
d = 3.0 \, \text{mm} = 0.003 \, \text{m}
\]
3. **Electric Field (E):**
- The electric field (E) between the plates of the capacitor can be calculated using the formula:
\[
E = \frac{\Delta V}{d} = \frac{150 \, \text{V}}{0.003 \, \text{m}} = 50,000 \, \text{V/m}
\]
4. **Direction of the Electric Field:**
- The electric field direction is from the positive plate (150 V) to the negative plate (0 V).
**Conclusion:**
- **Magnitude of the Electric Field:** 50,000 V/m
- **Direction:** From left (150 V) to right (0 V)
This analysis of the capacitor helps understand basic electric field concepts and their calculation in electrostatics.
![The image displays a diagram at the top with two rectangles on either side, representing plates, separated by three spaces each measuring 1.0 mm across, summing up to a total distance of 3.0 mm between the plates.
Below the diagram, a question poses multiple-choice options regarding the electric field strength and direction between the plates:
- ○ 450,000 V/m, to the right
- ○ 50 V/m to the right
- ○ 50,000 V/m, to the left
- ○ 450,000 V/m, to the left
- ○ 50 V/m to the left
- ○ 50,000 V/m, to the right
This setup is typically used to ask about electric field calculations in physics, focusing on concepts like field strength, direction, and distance between charged plates.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F070e8b65-931e-44e7-bfb4-6d69e5ced655%2Ffe96cb42-e3d4-46f8-8ee4-af7b234fe2e1%2Fjgbynz_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The image displays a diagram at the top with two rectangles on either side, representing plates, separated by three spaces each measuring 1.0 mm across, summing up to a total distance of 3.0 mm between the plates.
Below the diagram, a question poses multiple-choice options regarding the electric field strength and direction between the plates:
- ○ 450,000 V/m, to the right
- ○ 50 V/m to the right
- ○ 50,000 V/m, to the left
- ○ 450,000 V/m, to the left
- ○ 50 V/m to the left
- ○ 50,000 V/m, to the right
This setup is typically used to ask about electric field calculations in physics, focusing on concepts like field strength, direction, and distance between charged plates.
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