What is the magnitude of the electric field at the point (2.60 i -6.20j+ 2.20 k) m if the electric potential is given by V = 1.50xyz2, where V is in volts and x, y, and z are in meters?
What is the magnitude of the electric field at the point (2.60 i -6.20j+ 2.20 k) m if the electric potential is given by V = 1.50xyz2, where V is in volts and x, y, and z are in meters?
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)...
Related questions
Question
Please type the answer instead of hand writting thanks
![### Electric Field Magnitude Calculation
**Problem Statement:**
What is the magnitude of the electric field at the point \( (2.60 \hat{i} - 6.20 \hat{j} + 2.20 \hat{k}) \) m if the electric potential is given by \( V = 1.50xyz^2 \), where \( V \) is in volts and \( x \), \( y \), and \( z \) are in meters?
**Input Form:**
- Number: [________]
- Units: [No units ▼] (Dropdown)
**Additional Resources:**
- eTextbook and Media
**Solution Outline:**
1. **Determine the Partial Derivatives:**
- Calculate the partial derivatives of \( V \) with respect to \( x \), \( y \), and \( z \) to find the electric field components.
2. **Evaluate at Given Point:**
- Plug in the values of \( x = 2.60 \), \( y = -6.20 \), and \( z = 2.20 \) into the derived expressions from step 1.
3. **Calculate the Magnitude:**
- Use the electric field components to calculate the magnitude of the electric field vector. This involves computing the square root of the sum of the squared components.
**Details:**
To find the components of the electric field \( \vec{E} \):
\[ \vec{E} = -\nabla V \]
Where:
\[ \frac{\partial V}{\partial x} = 1.50yz^2 \]
\[ \frac{\partial V}{\partial y} = 1.50xz^2 \]
\[ \frac{\partial V}{\partial z} = 3.00xyz \]
These partial derivatives will be evaluated at the point \( (2.60, -6.20, 2.20) \).
**Electric Field Components at the Given Point:**
Calculate the field components:
\[ E_x = - \left.\frac{\partial V}{\partial x}\right|_{(2.60, -6.20, 2.20)} \]
\[ E_y = - \left.\frac{\partial V}{\partial y}\right|_{(2.60, -6.20, 2.20)} \]
\[ E_z = - \left.\frac{\partial](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe462800b-ac16-4590-bafb-82d68c005ec5%2F00b3cd0e-435f-45f2-9bd4-24c1e270bcc8%2Fx0v2c8v_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Electric Field Magnitude Calculation
**Problem Statement:**
What is the magnitude of the electric field at the point \( (2.60 \hat{i} - 6.20 \hat{j} + 2.20 \hat{k}) \) m if the electric potential is given by \( V = 1.50xyz^2 \), where \( V \) is in volts and \( x \), \( y \), and \( z \) are in meters?
**Input Form:**
- Number: [________]
- Units: [No units ▼] (Dropdown)
**Additional Resources:**
- eTextbook and Media
**Solution Outline:**
1. **Determine the Partial Derivatives:**
- Calculate the partial derivatives of \( V \) with respect to \( x \), \( y \), and \( z \) to find the electric field components.
2. **Evaluate at Given Point:**
- Plug in the values of \( x = 2.60 \), \( y = -6.20 \), and \( z = 2.20 \) into the derived expressions from step 1.
3. **Calculate the Magnitude:**
- Use the electric field components to calculate the magnitude of the electric field vector. This involves computing the square root of the sum of the squared components.
**Details:**
To find the components of the electric field \( \vec{E} \):
\[ \vec{E} = -\nabla V \]
Where:
\[ \frac{\partial V}{\partial x} = 1.50yz^2 \]
\[ \frac{\partial V}{\partial y} = 1.50xz^2 \]
\[ \frac{\partial V}{\partial z} = 3.00xyz \]
These partial derivatives will be evaluated at the point \( (2.60, -6.20, 2.20) \).
**Electric Field Components at the Given Point:**
Calculate the field components:
\[ E_x = - \left.\frac{\partial V}{\partial x}\right|_{(2.60, -6.20, 2.20)} \]
\[ E_y = - \left.\frac{\partial V}{\partial y}\right|_{(2.60, -6.20, 2.20)} \]
\[ E_z = - \left.\frac{\partial
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 3 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.Recommended textbooks for you
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
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