The electric field 0.300 m from a very long uniform line of charge is 8.40\times 10^(2)(N)/(C).

How much charge is contained in a 2.00 cm section of the line?

To find how much charge is contained in a 2.00 cm section of the line, we will use the relation for the electric field produced by a very long line of charge. The electric field \( E \) at a distance \( r \) from a line of charge with linear charge density \( \lambda \) is given by the formula:

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

Where:
- \( E \) is the electric field (in N/C),
- \( \lambda \) is the linear charge density (in C/m),
- \( r \) is the distance from the line of charge (in meters),
- \( \epsilon_0 \) is the permittivity of free space, \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2 \).

### Step 1: Rearrange the formula to solve for \( \lambda \):
\[
\lambda = 2 \pi \epsilon_0 r E
\]

### Step 2: Substitute the given values:
- \( E = 8.40 \times 10^2 \, \text{N/C} \),
- \( r = 0.300 \, \text{m} \),
- \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2 \).

\[
\lambda = 2 \pi (8.85 \times 10^{-12}) (0.300) (8.40 \times 10^2)
\]

### Step 3: Calculate \( \lambda \):
\[
\lambda = 2 \pi (8.85 \times 10^{-12}) (0.300) (8.40 \times 10^2) = 1.401 \times 10^{-8} \, \text{C/m}
\]

### Step 4: Calculate the charge in a 2.00 cm section:
The charge contained in a section of length \( L = 2.00 \, \text{cm} = 0.0200 \, \text{m} \) is:

\[
Q = \lambda L = (1.401 \times 10^{-8}) (0.0200)
\]

\[
Q = 2.80 \times 10^{-10} \, \text{C}
\]

### Final Answer:
The charge contained in a 2.00 cm section of the line is \( 2.80 \times 10^{-10} \, \text{C} \).

Would you like further details or have any questions?

Here are 5 related questions to consider:
1. What is the electric field at a distance of 0.500 m from the same line of charge?
2. How does the electric field vary with distance for a line of charge?
3. What would be the charge in a 1.00 cm section of the same line?
4. How would the electric field change if the charge density were doubled?
5. How does the electric field of a line of charge compare to that of a point charge?

**Tip:** Remember that for line charges, the electric field decreases with \( 1/r \), unlike point charges, where it decreases with \( 1/r^2 \).

The electric field 0.300 m from a very long uniform line of charge is 8.40×10^(2) N/C. How much charge is contained in a 2.00 cm section of the line?

This page explains how to calculate the charge contained in a 2.00 cm section of a very long line of charge using electric field data. The electric field is given as 8.40×10² N/C at a distance of 0.300 m, and the solution applies key formulas in electromagnetism and linear charge density. Suitable for advanced high school or college-level students.

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