College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
expand_more
expand_more
format_list_bulleted
Related questions
Concept explainers
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.](https://content.bartleby.com/qna-images/question/070e8b65-931e-44e7-bfb4-6d69e5ced655/fe96cb42-e3d4-46f8-8ee4-af7b234fe2e1/pipde1o_processed.jpeg)
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.
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.
Expert Solution
arrow_forward
Step 1
Step by stepSolved in 2 steps with 2 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Similar questions
- A power supply is connected in series to a 3 mF capacitor. What voltage does the power supply need to generate so that when the capacitor is fully charged, it has 5 J of energy stored in it? 3333.33 V 262.32 V 57.73 V None of the above 40.82 Varrow_forwardGiven a 3.00 μF capacitor, a 6.00 μF capacitor, and a 2.00 V battery, find the charge on each capacitor if you connect them in the following ways. (a) in series across the battery 3.00 uF capacitor 6.00 μF capacitor μC μC (b) in parallel across the battery 3.00 μF capacitor 6.00 μF capacitor JC μCarrow_forwardTwo capacitors with capacitances of 1.5 μF and 0.25 μF, respectively, are connected in parallel. The system is connected to a 40-V battery. What electrical potential energy is stored in the 1.5-μF capacitor? O 1.1E-4 J O2.0E-4 J O 1.2E-3 J 1.1E-3 Jarrow_forward
- What is the equivalent capacitance of 8 equal 400 nF capacitors in series? ○ 5 nF ○ 3200 nF ○ 32 nF ○ 50 nFarrow_forwardCan you explain how the voltage drop across capacitor C2 is equal to 2V?arrow_forwarda个4 3 of 3 + Automatic Zoom In the figure, V = 9.0 Volt, C1 = C2 = 30.µF, and C3 = C4 = 15µF. (a) What is the charge on capacitor 4? (b) What is the voltage across capacitor 1? C4 V C3 C2 ENGarrow_forward
- A parrallel plate capacitor consisting of two 30.0 cm x 30.0 cm square conductors are separated by a 3.00 mm piece of mica (k=5.00) is charged by a connecting it to an 18.0 V battery. What is the capacitance of this capacitor? What is the charge on the capacitor? If the mica is removed and replaced with glass (k=7.00), what is the potential difference of the plates now? What is the charge on the capacitor now with the glass?arrow_forwardThe capacitance of a parallel-plate capacitor can be increased by: O increasing the charge O decreasing the charge O increasing the plate separation O decreasing the plate separation O decreasing the plate areaarrow_forwardFind the following. (In the figure use C1 = 37.80 µF and C2 = 31.80 µF.) (a) the equivalent capacitance of the capacitors in the figure above µF(b) the charge on each capacitor on the right 37.80 µF capacitor µC on the left 37.80 µF capacitor µC on the 31.80 µF capacitor µC on the 6.00 µF capacitor µC (c) the potential difference across each capacitor on the right 37.80 µF capacitor V on the left 37.80 µF capacitor V on the 31.80 µF capacitor V on the 6.00 µF capacitor Varrow_forward
- You measure the voltage across two parallel capacitors to be 38 V. If they have a separation distance 0f 3.5 mm, what is the strength of the electric field in N/C? a. 1.09 x 10^4 N/C b. 1.33 x 10^4 N/C c. 9.21 x 10^-5 N/C d. 722 N/C e. None of the abovearrow_forward1arrow_forwardK Solve for the indicated variable. Assume all variables are positive. QV₁ QV₂ T. L T₂ ; for T₁ (Simplify your answer.)arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- College PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage Learning
University Physics (14th Edition)PhysicsISBN:9780133969290Author:Hugh D. Young, Roger A. FreedmanPublisher:PEARSON
Introduction To Quantum MechanicsPhysicsISBN:9781107189638Author:Griffiths, David J., Schroeter, Darrell F.Publisher:Cambridge University Press 
Physics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
Lecture- Tutorials for Introductory AstronomyPhysicsISBN:9780321820464Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina BrissendenPublisher:Addison-Wesley
College Physics: A Strategic Approach (4th Editio...PhysicsISBN:9780134609034Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart FieldPublisher:PEARSON
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
University Physics (14th Edition)
Physics
ISBN:9780133969290
Author:Hugh D. Young, Roger A. Freedman
Publisher:PEARSON
Introduction To Quantum Mechanics
Physics
ISBN:9781107189638
Author:Griffiths, David J., Schroeter, Darrell F.
Publisher:Cambridge University Press
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:9780321820464
Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:Addison-Wesley
College Physics: A Strategic Approach (4th Editio...
Physics
ISBN:9780134609034
Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:PEARSON