Question
Find the electric field at a distance of 5 cm from a long, straight wire with a linear charge density of 2 μC/m. Provide the formula used to calculate the electric field and the numerical value of the electric field.
Answers
GPT-4o mini
To calculate the electric field generated by a long, straight wire with a given linear charge density, we use the following formula for the electric field \( E \) at a distance \( r \) from the wire:
\[
E = \frac{\lambda}{2 \pi \epsilon_0 r}
\]
Where:
- \( E \) is the electric field (in N/C),
- \( \lambda \) is the linear charge density (in C/m),
- \( \epsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \),
- \( r \) is the distance from the wire (in m).
Given:
- \( \lambda = 2 \, \mu\text{C/m} = 2 \times 10^{-6} \, \text{C/m} \)
- \( r = 5 \, \text{cm} = 0.05 \, \text{m} \)
Now substituting the values into the formula:
\[
E = \frac{2 \times 10^{-6}}{2 \pi (8.85 \times 10^{-12}) (0.05)}
\]
Calculating the denominator:
\[
E = \frac{2 \times 10^{-6}}{2 \pi (8.85 \times 10^{-12}) (0.05)} \approx \frac{2 \times 10^{-6}}{2.785 \times 10^{-12}} \approx \frac{2 \times 10^{-6}}{2.785 \times 10^{-12}} \approx 719.42 \, \text{N/C}
\]
Thus, the electric field at a distance of 5 cm from the wire is approximately:
\[
E \approx 719.42 \, \text{N/C}
\]
So the electric field at a distance of 5 cm from the long, straight wire with a linear charge density of \( 2 \, \mu \text{C/m} \) is approximately **719.42 N/C**.
\[
E = \frac{\lambda}{2 \pi \epsilon_0 r}
\]
Where:
- \( E \) is the electric field (in N/C),
- \( \lambda \) is the linear charge density (in C/m),
- \( \epsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \),
- \( r \) is the distance from the wire (in m).
Given:
- \( \lambda = 2 \, \mu\text{C/m} = 2 \times 10^{-6} \, \text{C/m} \)
- \( r = 5 \, \text{cm} = 0.05 \, \text{m} \)
Now substituting the values into the formula:
\[
E = \frac{2 \times 10^{-6}}{2 \pi (8.85 \times 10^{-12}) (0.05)}
\]
Calculating the denominator:
\[
E = \frac{2 \times 10^{-6}}{2 \pi (8.85 \times 10^{-12}) (0.05)} \approx \frac{2 \times 10^{-6}}{2.785 \times 10^{-12}} \approx \frac{2 \times 10^{-6}}{2.785 \times 10^{-12}} \approx 719.42 \, \text{N/C}
\]
Thus, the electric field at a distance of 5 cm from the wire is approximately:
\[
E \approx 719.42 \, \text{N/C}
\]
So the electric field at a distance of 5 cm from the long, straight wire with a linear charge density of \( 2 \, \mu \text{C/m} \) is approximately **719.42 N/C**.