Two very long, uniformly charged lines are situated near each other with linear charge densities and distances given in the illustration. What is the correct expression for the electric field at point P in the illustration? P -λ A. B. C. D. E. 3 d 3. d -22 3 d -1. 27 d TE d 2d 2
Two very long, uniformly charged lines are situated near each other with linear charge densities and distances given in the illustration. What is the correct expression for the electric field at point P in the illustration? P -λ A. B. C. D. E. 3 d 3. d -22 3 d -1. 27 d TE d 2d 2
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Transcribed Image Text:**Determining the Electric Field at a Point between Uniformly Charged Lines**
Two very long, uniformly charged lines are situated near each other with linear charge densities and distances given in the illustration below. What is the correct expression for the electric field at point P in the illustration?
### Illustration Description:
- There are two parallel, horizontally oriented, uniformly charged lines.
- The upper line has a linear charge density of \(-\lambda\).
- The lower line has a linear charge density of \(+\lambda\).
- Vertical distance between the upper line and point P is \(d\).
- Distance between point P and the lower line is \(2d\).
### Diagram Explanation:
In the provided diagram:
- Point P is marked above the middle section between the two lines.
- Point P is located a distance "d" from the upper line and "2d" from the lower line.
- The electric field contributions from each line need to be considered to determine the net electric field at P.
### Question and Answer Options:
What is the correct expression for the electric field at point P in the illustration?
A. \( \frac{-\lambda}{3 \pi \epsilon_0 d} \hat{j} \)
B. \( \frac{\lambda}{3 \pi \epsilon_0 d} \hat{j} \)
C. \( \frac{-2 \lambda}{3 \pi \epsilon_0 d} \hat{j} \)
D. \( \frac{-\lambda}{2 \pi \epsilon_0 d} \hat{j} \)
E. \( \frac{\lambda}{\pi \epsilon_0 d} \hat{j} \)
### Solution Approach:
To determine the correct expression for the electric field at point P:
1. Calculate the electric field due to each charged line at point P.
2. The electric field due to a line charge at a perpendicular distance \(r\) is given by \(E = \frac{\lambda}{2 \pi \epsilon_0 r}\).
3. Superpose the electric fields from both lines at point P, keeping in mind the directions.
By applying these principles, the correct expression can be determined based on the linear charge densities and the distances.
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