Two infinitely long rods carry equal linear denisty `lambda` each. They are perpendicular to each other and they are in different planes and separated by a distance `d`. Find the electrostatic force on once rod due to the other

Explanation

To find the electrostatic force on one infinitely long rod due to another infinitely long rod that is perpendicular to it and separated by a distance \( d \), we can follow these steps:

### Step 1: Understand the Configuration
We have two infinitely long rods, each carrying a linear charge density \( \lambda \). One rod is placed along the \( y \)-axis (YZ plane), and the other rod is placed along the \( x \)-axis (YX plane). The distance between the two rods is \( d \).

### Step 2: Identify the Electric Field Due to One Rod
The electric field \( E \) due to an infinitely long charged rod at a distance \( r \) from it can be expressed as:
\[ E = \frac{2k\lambda}{r} \]
where \( k \) is Coulomb's constant (\( k = \frac{1}{4\pi\epsilon_0} \)).

### Step 3: Calculate the Electric Field at the Location of the Other Rod
Since we want to find the force on the rod along the \( y \)-axis due to the rod along the \( x \)-axis, we will calculate the electric field at a distance \( d \) from the \( x \)-axis rod:
\[ E = \frac{2k\lambda}{d} \]

### Step 4: Calculate the Force on the Rod
The force \( F \) on a segment of the rod due to the electric field can be calculated using:
\[ F = qE \]
where \( q \) is the charge on the segment of the rod. For a small length \( dx \) of the rod along the \( y \)-axis, the charge \( dq \) is given by:
\[ dq = \lambda \, dy \]
Thus, the force on this segment is:
\[ dF = dq \cdot E = \lambda \, dy \cdot \frac{2k\lambda}{d} \]
This simplifies to:
\[ dF = \frac{2k\lambda^2}{d} \, dy \]

### Step 5: Integrate to Find the Total Force
To find the total force \( F \) on the entire rod along the \( y \)-axis, we integrate \( dF \) over the entire length of the rod:
\[ F = \int_{-\infty}^{\infty} dF = \int_{-\infty}^{\infty} \frac{2k\lambda^2}{d} \, dy \]
Since the rod is infinitely long, the integral diverges, but we can express the total force in terms of the linear charge density:
\[ F = \frac{2k\lambda^2}{d} \cdot L \]
where \( L \) is the length of the rod considered. For practical purposes, we can consider the force per unit length.

### Step 6: Substitute \( k \) and Simplify
Substituting \( k = \frac{1}{4\pi\epsilon_0} \):
\[ F = \frac{2 \cdot \frac{1}{4\pi\epsilon_0} \lambda^2}{d} \cdot L = \frac{\lambda^2}{2\pi\epsilon_0 d} \cdot L \]

### Final Result
The electrostatic force per unit length on one rod due to the other is:
\[ F = \frac{\lambda^2}{2\pi\epsilon_0 d} \]