### Electric Field Calculation for Linear Charge Density**Problem Statement:**The entire y-axis is covered with a uniform linear charge density of \( \lambda = 2.2 \, \text{nC/m} \). Determine the magnitude of the electric field on the x-axis at \( x = 5.5 \, \text{m} \).Round your answer to 1 decimal place.**Detailed Explanation:**1. **Understanding the Setup:** - You have a uniform linear charge distribution along the y-axis. - A linear charge density (\(\lambda\)) implies the charge per unit length along the y-axis. - You are asked to find the electric field at a point on the x-axis, \( x = 5.5 \, \text{m} \).2. **Relevant Formulas:** - The electric field (\(E\)) due to a line of charge can be determined using the formula: \[ E = \frac{\lambda}{2 \pi \epsilon_0 r} \] where: - \(\lambda\) is the linear charge density - \(r\) is the perpendicular distance from the line of charge to the point where the field is being calculated - \(\epsilon_0\) is the permittivity of free space (\(8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2\))3. **Applying to the Given Problem:** - Given \(\lambda = 2.2 \, \text{nC/m}\) or \(2.2 \times 10^{-9} \, \text{C/m}\) - Distance \(r = 5.5 \, \text{m}\)4. **Calculation:** Substitute the values into the formula to find \(E\): \[ E = \frac{2.2 \times 10^{-9} \, \text{C/m}}{2 \pi (8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2)(5.5 \, \text{m})} \] Simplify the expression step-by-step: \[

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The entire y axis is covered with a uniform linear charge density 2.2 nC/m. Determine the magnitude of the electric field on the x axis at x= 5.5 m.

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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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Electric Charges and Fields

One of the fundamental properties is the electromagnetic property. Charge cannot be destroyed by any process and this contributes formally to the law of charge conservation.

Question

### Electric Field Calculation for Linear Charge Density

**Problem Statement:**
The entire y-axis is covered with a uniform linear charge density of \( \lambda = 2.2 \, \text{nC/m} \). Determine the magnitude of the electric field on the x-axis at \( x = 5.5 \, \text{m} \).

Round your answer to 1 decimal place.

**Detailed Explanation:**
1. **Understanding the Setup:**
- You have a uniform linear charge distribution along the y-axis.
- A linear charge density (\(\lambda\)) implies the charge per unit length along the y-axis.
- You are asked to find the electric field at a point on the x-axis, \( x = 5.5 \, \text{m} \).

2. **Relevant Formulas:**
- The electric field (\(E\)) due to a line of charge can be determined using the formula:
\[
E = \frac{\lambda}{2 \pi \epsilon_0 r}
\]
where:
- \(\lambda\) is the linear charge density
- \(r\) is the perpendicular distance from the line of charge to the point where the field is being calculated
- \(\epsilon_0\) is the permittivity of free space (\(8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2\))

3. **Applying to the Given Problem:**
- Given \(\lambda = 2.2 \, \text{nC/m}\) or \(2.2 \times 10^{-9} \, \text{C/m}\)
- Distance \(r = 5.5 \, \text{m}\)

4. **Calculation:**
Substitute the values into the formula to find \(E\):
\[
E = \frac{2.2 \times 10^{-9} \, \text{C/m}}{2 \pi (8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2)(5.5 \, \text{m})}
\]
Simplify the expression step-by-step:
\[

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