Two points charges `Q_(1)` and `Q_(2)` placed at separation `d` in vacuum and force acting between them is `F`. Now a dielectric slab of thickness `d//2` and dielectric constant `K = 4` is placed between them. The new force between the charges will be

Options

A. `(4 F)/(9)`
B. `(2F)/(9)`
C. `(F)/(9)`
D. `(5F)/(9)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the initial force between the charges
The force \( F \) between two point charges \( Q_1 \) and \( Q_2 \) separated by a distance \( d \) in vacuum is given by Coulomb's law:
\[ F = \frac{1}{4 \pi \epsilon_0} \frac{Q_1 Q_2}{d^2} \]

### Step 2: Introduce the dielectric slab
When a dielectric slab of thickness \( \frac{d}{2} \) and dielectric constant \( K = 4 \) is placed between the charges, the effective distance between the charges changes. The dielectric constant modifies the permittivity of the medium between the charges.

### Step 3: Calculate the new effective distance
The dielectric slab occupies half the distance \( d \), leaving two segments of \( \frac{d}{4} \) on either side of the slab. The effective distance \( d' \) can be calculated as:
\[ d' = \frac{d}{4} + \frac{d}{2} + \frac{d}{4} = d \]
However, we need to consider the effect of the dielectric on the force.

### Step 4: Modify the permittivity
With the dielectric slab, the effective permittivity \( \epsilon \) becomes:
\[ \epsilon = K \epsilon_0 = 4 \epsilon_0 \]

### Step 5: Calculate the new force between the charges
The new force \( F' \) when the dielectric is present is given by:
\[ F' = \frac{1}{4 \pi \epsilon} \frac{Q_1 Q_2}{d^2} \]
Substituting \( \epsilon = 4 \epsilon_0 \):
\[ F' = \frac{1}{4 \pi (4 \epsilon_0)} \frac{Q_1 Q_2}{d^2} = \frac{1}{16 \pi \epsilon_0} \frac{Q_1 Q_2}{d^2} \]

### Step 6: Relate the new force to the original force
We can relate \( F' \) to the original force \( F \):
\[ F' = \frac{F}{4} \]
This means the new force \( F' \) is one-fourth of the original force \( F \).

### Final Result
Thus, the new force between the charges after placing the dielectric slab is:
\[ F' = \frac{F}{4} \]