Two infinitely long parallel wires having linear charge densities `lambda_(1)` and `lambda_(2)` respectively are placed at a distance of `R` metres. The force per unit length on either wire will be `(k= 1/(4piepsilon_(0)))`

Options

A. `k(2lambda_(1)lambda_(2))/(R^(2))`
B. `k(2lambda_(1)lambda_(2))/(R)`
C. `k(lambda_(1)lambda_(2))/(R^(2))`
D. `k(lambda_(1)lambda_(2))/(R)`

Explanation

To solve the problem of finding the force per unit length between two infinitely long parallel wires with linear charge densities \(\lambda_1\) and \(\lambda_2\) separated by a distance \(R\), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
We have two infinitely long parallel wires. Wire 1 has a linear charge density \(\lambda_1\) and wire 2 has a linear charge density \(\lambda_2\). The distance between the two wires is \(R\).

2. **Determine the Charge on a Length of Wire**:
Consider a segment of length \(L\) of wire 2. The total charge \(Q\) on this segment can be calculated as:
\[ Q = \lambda_2 \cdot L \]

3. **Calculate the Electric Field due to Wire 1**:
The electric field \(E\) produced by an infinitely long wire with linear charge density \(\lambda_1\) at a distance \(R\) from the wire is given by:
\[ E = \frac{1}{2 \pi \epsilon_0} \cdot \frac{\lambda_1}{R} \]
Here, \(\epsilon_0\) is the permittivity of free space.

4. **Substitute \(K\) into the Electric Field Expression**:
We know that \(k = \frac{1}{4 \pi \epsilon_0}\). Therefore, we can express \(\epsilon_0\) in terms of \(k\):
\[ \epsilon_0 = \frac{1}{4 \pi k} \]
Substituting this into the electric field expression gives:
\[ E = \frac{\lambda_1}{2 \pi \epsilon_0 R} = \frac{2 k \lambda_1}{R} \]

5. **Calculate the Force on Wire 2**:
The force \(F\) on wire 2 due to the electric field \(E\) created by wire 1 is given by:
\[ F = Q \cdot E = (\lambda_2 \cdot L) \cdot \left(\frac{2 k \lambda_1}{R}\right) \]
Simplifying this gives:
\[ F = \frac{2 k \lambda_1 \lambda_2 L}{R} \]

6. **Find the Force per Unit Length**:
To find the force per unit length \(f\), we divide the total force \(F\) by the length \(L\):
\[ f = \frac{F}{L} = \frac{2 k \lambda_1 \lambda_2 L}{R \cdot L} = \frac{2 k \lambda_1 \lambda_2}{R} \]

### Final Answer:
The force per unit length on either wire is given by:
\[ f = \frac{2 k \lambda_1 \lambda_2}{R} \]