If the field strength is E = 9 V/m a distance of 1 m from the charge, what is the field strength E a distance of 3 m from the charge? • View Available Hint(s) E = .56 V/m

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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### Part C

#### Question:
The magnitude of the electric field 1 m away from the positive charge is ______ the magnitude of the electric field 2 m away.

#### Answer Options:
- two times
- four times
- equal to
- one-half
- one-quarter

#### Selected Answer:
- four times

#### Explanation: 
The magnitude of the field decreases more quickly than the inverse of the distance from the charge. The magnitude of the electric field is proportional to the inverse of the distance squared (\(E \propto 1/r^2\)), where \(r\) is the distance from the charge. You should verify this by looking at the field strength 3 or 4 meters away. This is consistent with Coulomb's law, which states that the magnitude of the force between two charged particles is \(F = k \frac{Q_1 Q_2}{r^2}\).

### Part D

#### Question:
If the field strength is \(E = 9 \, \text{V/m}\) at a distance of 1 m from the charge, what is the field strength \(E\) at a distance of 3 m from the charge?

#### Response:
- Input field with the value: \[ E = 0.56 \, \text{V/m} \]

#### Explanation:
To determine the electric field strength at a different distance, you can use the inverse square law. The relationship between the field strengths at different distances \(r_1\) and \(r_2\) can be expressed as:
\[ \frac{E_1}{E_2} = \left(\frac{r_2}{r_1}\right)^2 \]
\[ \frac{9 \, \text{V/m}}{E_2} = \left(\frac{3 \, \text{m}}{1 \, \text{m}}\right)^2 \]
\[ \frac{9 \, \text{V/m}}{E_2} = 9 \]
\[ E_2 = \frac{9 \, \text{V/m}}{9} = 1 \, \text{V/m} \]

So, the field strength \(E\) at a distance of 3 m from the charge is \(0.56 \, \text{V/m}\).

### Graphs and Diagrams: 
There are no specific graphs
Transcribed Image Text:### Part C #### Question: The magnitude of the electric field 1 m away from the positive charge is ______ the magnitude of the electric field 2 m away. #### Answer Options: - two times - four times - equal to - one-half - one-quarter #### Selected Answer: - four times #### Explanation: The magnitude of the field decreases more quickly than the inverse of the distance from the charge. The magnitude of the electric field is proportional to the inverse of the distance squared (\(E \propto 1/r^2\)), where \(r\) is the distance from the charge. You should verify this by looking at the field strength 3 or 4 meters away. This is consistent with Coulomb's law, which states that the magnitude of the force between two charged particles is \(F = k \frac{Q_1 Q_2}{r^2}\). ### Part D #### Question: If the field strength is \(E = 9 \, \text{V/m}\) at a distance of 1 m from the charge, what is the field strength \(E\) at a distance of 3 m from the charge? #### Response: - Input field with the value: \[ E = 0.56 \, \text{V/m} \] #### Explanation: To determine the electric field strength at a different distance, you can use the inverse square law. The relationship between the field strengths at different distances \(r_1\) and \(r_2\) can be expressed as: \[ \frac{E_1}{E_2} = \left(\frac{r_2}{r_1}\right)^2 \] \[ \frac{9 \, \text{V/m}}{E_2} = \left(\frac{3 \, \text{m}}{1 \, \text{m}}\right)^2 \] \[ \frac{9 \, \text{V/m}}{E_2} = 9 \] \[ E_2 = \frac{9 \, \text{V/m}}{9} = 1 \, \text{V/m} \] So, the field strength \(E\) at a distance of 3 m from the charge is \(0.56 \, \text{V/m}\). ### Graphs and Diagrams: There are no specific graphs
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