Each of the two long parallel threads carries a uniform `lambda` per unit length. The threads are separated by a distance l. Find the maximum magnitude of the electric field strenght in the symmetry plane of this system located between the threads.

Options

A. `E_(max)=lambda//2piepsilon_(0)l`
B. `E_(max)=lambda//3piepsilon_(0)l`
C. `E_(max)=lambda//piepsilon_(0)`
D. `E_(max)=lambda//piepsilon_(0)l`

Answer

To find the maximum magnitude of the electric field strength in the symmetry plane located between two long parallel threads carrying a uniform linear charge density (λ), we can follow these steps:

### Step 1: Understanding the Setup
We have two long parallel threads separated by a distance \( l \), each carrying a uniform charge density \( \lambda \) per unit length. The symmetry plane is located exactly in the middle of the two threads.

### Step 2: Electric Field Due to a Single Thread
The electric field \( E \) due to a single long charged thread at a distance \( r \) from it is given by the formula:
\[ E = \frac{\lambda}{2 \pi \epsilon_0 r} \]
where \( \epsilon_0 \) is the permittivity of free space.

### Step 3: Determine the Position in the Symmetry Plane
In the symmetry plane, the distance from each thread to the midpoint is \( \frac{l}{2} \).

### Step 4: Calculate the Electric Field Contributions
1. **Electric Field from the First Thread**: The electric field \( E_1 \) at the midpoint due to the first thread (to the right) is:
\[ E_1 = \frac{\lambda}{2 \pi \epsilon_0 \left(\frac{l}{2}\right)} = \frac{\lambda}{\pi \epsilon_0 l} \]

2. **Electric Field from the Second Thread**: The electric field \( E_2 \) at the midpoint due to the second thread (to the left) is:
\[ E_2 = \frac{\lambda}{2 \pi \epsilon_0 \left(\frac{l}{2}\right)} = \frac{\lambda}{\pi \epsilon_0 l} \]

### Step 5: Direction of the Electric Fields
- The electric field due to the first thread points away from the thread (to the right).
- The electric field due to the second thread also points away from the thread (to the left).

### Step 6: Net Electric Field in the Symmetry Plane
Since both electric fields are in the same direction (away from the threads), the total electric field \( E_{total} \) at the midpoint is:
\[ E_{total} = E_1 + E_2 = \frac{\lambda}{\pi \epsilon_0 l} + \frac{\lambda}{\pi \epsilon_0 l} = \frac{2\lambda}{\pi \epsilon_0 l} \]

### Step 7: Conclusion
The maximum magnitude of the electric field strength in the symmetry plane located between the threads is:
\[ E_{max} = \frac{2\lambda}{\pi \epsilon_0 l} \]

### Final Answer
The maximum magnitude of the electric field strength in the symmetry plane is:
\[ E_{max} = \frac{2\lambda}{\pi \epsilon_0 l} \]

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