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 $1/r^2$.