A straight wire 10 mm long carries 28 μC�C distributed uniformly over its length.

**Part A**

Part complete

What is the line charge density on the wire?

Express your answer in microcoulombs per meter.

λ� =

2.8

μC/m�C/mFind the electric field strength 13 cmcm perpendicular from the wire axis, not near either end.

Note: make suitable approximation. Express your answer in newtons per coulomb.

E� =

3.9×105

N/CN/C  Find the electric field strength 470 mm from the wire.

Note: make suitable approximation. Express your answer in newtons per coulomb.

Let's tackle the problem step by step.

### Part A: Find the Line Charge Density
Given:
- Total charge \( Q = 28 \, \mu\text{C} \)
- Length of the wire \( L = 10 \, \text{mm} = 0.01 \, \text{m} \)

The line charge density \( \lambda \) is calculated as:

\[
\lambda = \frac{Q}{L} = \frac{28 \, \mu\text{C}}{0.01 \, \text{m}} = \frac{28 \times 10^{-6} \, \text{C}}{0.01 \, \text{m}} = 2.8 \times 10^{-3} \, \text{C/m} = 2800 \, \mu\text{C/m}
\]

### Part B: Electric Field Strength at 13 cm from the Wire
Given:
- Distance \( r = 13 \, \text{cm} = 0.13 \, \text{m} \)

Using the formula for the electric field due to a long charged wire:

\[
E = \frac{\lambda}{2 \pi \epsilon_0 r}
\]

Substitute the values:

\[
E = \frac{2800 \times 10^{-6} \, \text{C/m}}{2 \pi \times 8.854 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2 \times 0.13 \, \text{m}}
\]

\[
E \approx \frac{2800 \times 10^{-6}}{7.237 \times 10^{-12} \times 0.13}
\]

\[
E \approx \frac{2800 \times 10^{-6}}{9.408 \times 10^{-13}} \, \text{N/C}
\]

\[
E \approx 2.975 \times 10^{6} \, \text{N/C} \approx 3.0 \times 10^{6} \, \text{N/C}
\]

### Part C: Electric Field Strength at 470 mm from the Wire
Given:
- Distance \( r = 470 \, \text{mm} = 0.47 \, \text{m} \)

Using the same formula:

\[
E = \frac{\lambda}{2 \pi \epsilon_0 r}
\]

Substitute the values:

\[
E = \frac{2800 \times 10^{-6} \, \text{C/m}}{2 \pi \times 8.854 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2 \times 0.47 \, \text{m}}
\]

\[
E \approx \frac{2800 \times 10^{-6}}{2.617 \times 10^{-11}} \, \text{N/C}
\]

\[
E \approx 1.071 \times 10^{6} \, \text{N/C} \approx 1.1 \times 10^{6} \, \text{N/C}
\]

### Summary
- **Line charge density \( \lambda \)**: \( 2800 \, \mu\text{C/m} \)
- **Electric field at 13 cm**: \( 3.0 \times 10^{6} \, \text{N/C} \)
- **Electric field at 470 mm**: \( 1.1 \times 10^{6} \, \text{N/C} \)

Would you like more details, or do you have any questions?

Here are five related questions:
1. What would the electric field be at a different distance, such as 1 m, from the wire?
2. How does the electric field strength change as you move further from the wire?
3. How would the electric field strength change if the wire were shorter but had the same total charge?
4. What would happen to the electric field strength if the charge on the wire were doubled?
5. How does the concept of an infinite wire help in making these calculations?

**Tip:** Always ensure that units are consistent throughout your calculations to avoid errors, particularly when converting between units like millimeters and meters.

Learn how to calculate line charge density and electric field strength due to a uniformly charged wire. Solve problems step-by-step with examples including finding electric field strength at specific distances from the wire.

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