Two infinitely long straight conductors each having a charge density lamda are arranged parallel to each other. The spearation between them is d. What happens to the force per unit length on each conductor, when the separation between them is doubled?

Options

A. remains same
B. doubled
C. halved
D. becomes 1/4 th

Explanation

To solve the problem, we need to analyze the force per unit length between two infinitely long straight conductors with charge density \( \lambda \) and separated by a distance \( d \). We will then determine what happens to this force when the separation is doubled to \( 2d \).

### Step 1: Determine the Electric Field due to One Conductor
The electric field \( E \) created by an infinitely long straight conductor with charge density \( \lambda \) 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 2: Calculate the Electric Field at the Location of the Other Conductor
For two parallel conductors, we consider the electric field due to one conductor at the position of the other. If the distance between the conductors is \( d \), the electric field at the location of one conductor due to the other is:
\[ E = \frac{\lambda}{2 \pi \epsilon_0 d} \]

### Step 3: Calculate the Force per Unit Length on One Conductor
The force \( F \) on a length \( L \) of one conductor due to the electric field created by the other conductor can be calculated using:
\[ F = Q \cdot E \]
where \( Q \) is the charge on the length \( L \) of the conductor. The charge \( Q \) can be expressed as:
\[ Q = \lambda L \]
Thus, the force becomes:
\[ F = \lambda L \cdot \frac{\lambda}{2 \pi \epsilon_0 d} = \frac{\lambda^2 L}{2 \pi \epsilon_0 d} \]
To find the force per unit length \( \frac{F}{L} \):
\[ \frac{F}{L} = \frac{\lambda^2}{2 \pi \epsilon_0 d} \]

### Step 4: Analyze the Effect of Doubling the Separation
Now, if the separation \( d \) is doubled to \( 2d \), we need to recalculate the force per unit length. The new electric field at the location of one conductor due to the other will be:
\[ E' = \frac{\lambda}{2 \pi \epsilon_0 (2d)} = \frac{\lambda}{4 \pi \epsilon_0 d} \]
The new force \( F' \) on a length \( L \) of one conductor is:
\[ F' = Q \cdot E' = \lambda L \cdot \frac{\lambda}{4 \pi \epsilon_0 d} = \frac{\lambda^2 L}{4 \pi \epsilon_0 d} \]
Thus, the new force per unit length \( \frac{F'}{L} \) is:
\[ \frac{F'}{L} = \frac{\lambda^2}{4 \pi \epsilon_0 d} \]

### Step 5: Compare the Two Forces
Now we can compare the original force per unit length \( \frac{F}{L} \) and the new force per unit length \( \frac{F'}{L} \):
\[ \frac{F'}{L} = \frac{1}{2} \cdot \frac{F}{L} \]
This shows that when the separation between the conductors is doubled, the force per unit length on each conductor is halved.

### Final Answer
When the separation between the two conductors is doubled, the force per unit length on each conductor is halved.