At position x=0 the voltage is 100V. At x=0.5m the voltage is OV. What is the direction and strength of the average x-component of the electric field in that region?
At position x=0 the voltage is 100V. At x=0.5m the voltage is OV. What is the direction and strength of the average x-component of the electric field in that region?
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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At position x=0 the voltage is 100V. At x=0.5m the voltage is 0V. What is the direction and strength of the average x-component of the electric field in that region?
![**Problem Statement: Electric Field Calculation**
At position \( x = 0 \), the voltage is \( 100V \). At \( x = 0.5m \), the voltage is \( 0V \). What is the direction and strength of the average x-component of the electric field in that region?
***Explanation:***
To determine the direction and strength of the electric field (\( \mathbf{E} \)), we can use the relationship between electric field and potential difference, \( \Delta V \), over a distance, \( \Delta x \):
\[ E_x = -\frac{\Delta V}{\Delta x} \]
Where:
- \( \Delta V \) is the change in voltage,
- \( \Delta x \) is the change in position.
Given data:
- \( V(x=0) = 100V \)
- \( V(x=0.5m) = 0V \)
- \( x_2 - x_1 = 0.5m - 0m = 0.5m \)
Now we can calculate:
\[ \Delta V = V(x=0.5m) - V(x=0) = 0V - 100V = -100V \]
\[ E_x = -\frac{\Delta V}{\Delta x} = -\frac{-100V}{0.5m} = \frac{100V}{0.5m} = 200 \, V/m \]
Thus, the electric field strength is \( 200 \, V/m \).
***Direction:***
The negative sign in the equation \( E_x = -\frac{\Delta V}{\Delta x} \) indicates that the electric field points in the direction of decreasing potential. Here, since voltage decreases from \( 100V \) to \( 0V \) as \( x \) increases from \( 0 \) to \( 0.5m \), the electric field points in the positive x-direction.
Therefore, the average x-component of the electric field in the region is \( 200 \, V/m \) in the positive x-direction.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F225ea937-ec91-4f60-983d-e6a97001b794%2F073b2a4b-3a1f-4d3e-a13a-ab2b2418dc9c%2Fzyjafm_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement: Electric Field Calculation**
At position \( x = 0 \), the voltage is \( 100V \). At \( x = 0.5m \), the voltage is \( 0V \). What is the direction and strength of the average x-component of the electric field in that region?
***Explanation:***
To determine the direction and strength of the electric field (\( \mathbf{E} \)), we can use the relationship between electric field and potential difference, \( \Delta V \), over a distance, \( \Delta x \):
\[ E_x = -\frac{\Delta V}{\Delta x} \]
Where:
- \( \Delta V \) is the change in voltage,
- \( \Delta x \) is the change in position.
Given data:
- \( V(x=0) = 100V \)
- \( V(x=0.5m) = 0V \)
- \( x_2 - x_1 = 0.5m - 0m = 0.5m \)
Now we can calculate:
\[ \Delta V = V(x=0.5m) - V(x=0) = 0V - 100V = -100V \]
\[ E_x = -\frac{\Delta V}{\Delta x} = -\frac{-100V}{0.5m} = \frac{100V}{0.5m} = 200 \, V/m \]
Thus, the electric field strength is \( 200 \, V/m \).
***Direction:***
The negative sign in the equation \( E_x = -\frac{\Delta V}{\Delta x} \) indicates that the electric field points in the direction of decreasing potential. Here, since voltage decreases from \( 100V \) to \( 0V \) as \( x \) increases from \( 0 \) to \( 0.5m \), the electric field points in the positive x-direction.
Therefore, the average x-component of the electric field in the region is \( 200 \, V/m \) in the positive x-direction.
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