The drawing shows the electric potential as a function of dis- tance along the x axis. Determine the magnitude of the electric field in the region (a) A to B, (b) B to C, and (c) C to D. A B 5.0 4.0 C 3.0 2.0 1.0 D 0.40 0.80 x, m Potential, V 0.20 0.60
The drawing shows the electric potential as a function of dis- tance along the x axis. Determine the magnitude of the electric field in the region (a) A to B, (b) B to C, and (c) C to D. A B 5.0 4.0 C 3.0 2.0 1.0 D 0.40 0.80 x, m Potential, V 0.20 0.60
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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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![**Electric Potential and Electric Field Analysis**
*Educational Objective: Understanding the Relationship between Electric Potential and Electric Field*
**Problem Statement:**
The drawing shows the electric potential as a function of distance along the x-axis. Determine the magnitude of the electric field in the regions (a) \(A\) to \(B\), (b) \(B\) to \(C\), and (c) \(C\) to \(D\).
**Graph Analysis:**
The graph presented illustrates the electric potential \( V \) (in volts) as a function of distance \( x \) (in meters) along the x-axis. The potential changes at specific points labeled \( A \), \( B \), \( C \), and \( D \).
- **Region \( A \) to \( B \):**
- \( V_A = 5.0 \) V at \( x = 0.00 \) m
- \( V_B = 5.0 \) V at \( x = 0.20 \) m
- The potential remains constant, indicating that the electric field magnitude, \( E \), is zero in this region.
- **Region \( B \) to \( C \):**
- \( V_B = 5.0 \) V at \( x = 0.20 \) m
- \( V_C = 2.0 \) V at \( x = 0.40 \) m
- The electric potential decreases linearly. The electric field can be calculated as the negative rate of change of potential with respect to distance:
\[
E = -\frac{\Delta V}{\Delta x} = -\frac{(2.0 - 5.0) \ \text{V}}{(0.40 - 0.20) \ \text{m}} = \frac{3.0 \ \text{V}}{0.20 \ \text{m}} = 15 \ \text{V/m}
\]
- **Region \( C \) to \( D \):**
- \( V_C = 2.0 \) V at \( x = 0.40 \) m
- \( V_D = 0.0 \) V at \( x = 0.80 \) m
- The electric potential decreases linearly. The electric field can be calculated similarly](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6746fa6f-1856-443b-addc-ecae5d9da9b9%2F0efe0468-9de1-4d8a-ac37-66b84c30d31d%2F9qec5hb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Electric Potential and Electric Field Analysis**
*Educational Objective: Understanding the Relationship between Electric Potential and Electric Field*
**Problem Statement:**
The drawing shows the electric potential as a function of distance along the x-axis. Determine the magnitude of the electric field in the regions (a) \(A\) to \(B\), (b) \(B\) to \(C\), and (c) \(C\) to \(D\).
**Graph Analysis:**
The graph presented illustrates the electric potential \( V \) (in volts) as a function of distance \( x \) (in meters) along the x-axis. The potential changes at specific points labeled \( A \), \( B \), \( C \), and \( D \).
- **Region \( A \) to \( B \):**
- \( V_A = 5.0 \) V at \( x = 0.00 \) m
- \( V_B = 5.0 \) V at \( x = 0.20 \) m
- The potential remains constant, indicating that the electric field magnitude, \( E \), is zero in this region.
- **Region \( B \) to \( C \):**
- \( V_B = 5.0 \) V at \( x = 0.20 \) m
- \( V_C = 2.0 \) V at \( x = 0.40 \) m
- The electric potential decreases linearly. The electric field can be calculated as the negative rate of change of potential with respect to distance:
\[
E = -\frac{\Delta V}{\Delta x} = -\frac{(2.0 - 5.0) \ \text{V}}{(0.40 - 0.20) \ \text{m}} = \frac{3.0 \ \text{V}}{0.20 \ \text{m}} = 15 \ \text{V/m}
\]
- **Region \( C \) to \( D \):**
- \( V_C = 2.0 \) V at \( x = 0.40 \) m
- \( V_D = 0.0 \) V at \( x = 0.80 \) m
- The electric potential decreases linearly. The electric field can be calculated similarly
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