(d) How about E distribution on the plane right in the middle of two infinite lines with one charged with Q and the other -Q and separated by d?

To determine the electric field (\(\mathbf{E}\)) distribution on the plane that lies in the middle between two infinite lines, one charged with \(+Q\) and the other with \(-Q\), and separated by a distance \(d\), we can follow these steps:

### Setup
- Let the infinite line with charge \(+Q\) be located at \(x = -\frac{d}{2}\).
- Let the infinite line with charge \(-Q\) be located at \(x = \frac{d}{2}\).
- The plane of interest is at \(x = 0\), equidistant from both lines.

### Electric Field from a Single Infinite Line
The electric field due to an infinite line of charge is given by:

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

Where:
- \(\lambda\) is the linear charge density.
- \(r\) is the perpendicular distance from the line to the point where the field is being calculated.
- \(\epsilon_0\) is the vacuum permittivity.
- \(\hat{r}\) is the unit vector pointing from the line towards the point of interest.

### Superposition Principle
Since electric fields are vectors, the total electric field at any point on the plane is the vector sum of the fields due to each line.

1. **Electric Field from the Positive Line at \(x = -\frac{d}{2}\):**
   - The electric field due to the line at \(x = -\frac{d}{2}\) on the plane \(x = 0\) will point away from the positive charge.
   - Magnitude: 
     \[
     E_+ = \frac{\lambda}{2 \pi \epsilon_0 \left(\frac{d}{2}\right)} = \frac{2 \lambda}{\pi \epsilon_0 d}
     \]
   - Direction: Along the positive \(x\)-axis (towards the right).

2. **Electric Field from the Negative Line at \(x = \frac{d}{2}\):**
   - The electric field due to the line at \(x = \frac{d}{2}\) on the plane \(x = 0\) will point towards the negative charge.
   - Magnitude:
     \[
     E_- = \frac{\lambda}{2 \pi \epsilon_0 \left(\frac{d}{2}\right)} = \frac{2 \lambda}{\pi \epsilon_0 d}
     \]
   - Direction: Along the negative \(x\)-axis (towards the left).

### Total Electric Field on the Plane at \(x = 0\)
Since the two fields are equal in magnitude but opposite in direction, they cancel each other out in the \(x\)-direction. However, the \(y\)-components will add up:

1. **Vertical Components:**
   - Each field has a vertical component of \(E_+\) and \(E_-\), and since the fields are symmetric, these components will add up.
   - Let \(\theta\) be the angle between the line connecting the point of interest to the line charge and the horizontal axis. Then,
   \[
   \sin \theta = \frac{y}{\sqrt{\left(\frac{d}{2}\right)^2 + y^2}}
   \]
   - The total vertical component of the field:
     \[
     E_y = 2 \times E_+ \times \sin \theta = \frac{4 \lambda y}{\pi \epsilon_0 d \sqrt{ \left(\frac{d}{2}\right)^2 + y^2 }}
     \]

### Final Answer
The electric field distribution on the plane midway between the two lines is:

\[
\mathbf{E}(x=0, y) = \frac{4 \lambda y}{\pi \epsilon_0 d \sqrt{ \left(\frac{d}{2}\right)^2 + y^2 }} \hat{y}
\]

- The field is directed purely along the \(y\)-axis.
- The magnitude of the field increases as you move away from the center ( \(y = 0\) ).

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Would you like more details on this solution, or do you have any other questions?

Here are 5 questions related to this topic:
1. How does the electric field change if the distance between the lines is increased?
2. What happens to the field distribution if one of the line charges is doubled?
3. How would the field change if the plane of interest were not at the midpoint between the lines?
4. How does the electric field behave at points very far from the charged lines?
5. What is the potential associated with the electric field in this setup?

**Tip:** Remember, the superposition principle is essential when dealing with multiple sources of electric fields—simply add the vectors!

How about E distribution on the plane right in the middle of two infinite lines with one charged with Q and the other -Q and separated by d?

Explore the electric field distribution in the plane midway between two infinite lines, one charged with +Q and the other with -Q, separated by distance d. This solution uses the superposition principle and key electromagnetism formulas to calculate the field directed along the y-axis.

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