Force between two identical charges placed at a distance of `r` in vacuum is `F`.Now a slab of dielectric constant 4 is inserted between these two charges . If the thickness of the slab is `r//2`, then the force between the charges will becomes

Options

A. `3/5F`
B. `4/9F`
C. `F/4`
D. `F/2`

Answer

To solve the problem, we need to determine the force between two identical charges when a dielectric slab is inserted between them. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the initial force between the charges
The force \( F \) between two identical charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) in vacuum is given by Coulomb's law:
\[ F = \frac{1}{4\pi \epsilon_0} \cdot \frac{q_1 \cdot q_2}{r^2} \]

### Step 2: Introduce the dielectric slab
When a dielectric slab with a dielectric constant \( K = 4 \) is inserted between the charges, it affects the electric field and thus the force between the charges. The thickness of the slab is given as \( \frac{r}{2} \).

### Step 3: Calculate the new distance between the charges
The total distance between the charges after inserting the slab is the sum of the thickness of the slab and the remaining distance outside the slab. Since the slab occupies \( \frac{r}{2} \), the remaining distance on either side of the slab is also \( \frac{r}{2} \).

Thus, the new effective distance \( d \) between the charges is:
\[ d = \frac{r}{2} + \frac{r}{2} \cdot \sqrt{K} = \frac{r}{2} + \frac{r}{2} \cdot \sqrt{4} = \frac{r}{2} + \frac{r}{2} \cdot 2 = \frac{r}{2} + r = \frac{3r}{2} \]

### Step 4: Calculate the new force with the dielectric
The new force \( F' \) between the charges when the dielectric is present is given by:
\[ F' = \frac{1}{4\pi \epsilon_0} \cdot \frac{q_1 \cdot q_2}{d^2} \]
Substituting \( d = \frac{3r}{2} \):
\[ F' = \frac{1}{4\pi \epsilon_0} \cdot \frac{q_1 \cdot q_2}{\left(\frac{3r}{2}\right)^2} = \frac{1}{4\pi \epsilon_0} \cdot \frac{q_1 \cdot q_2}{\frac{9r^2}{4}} = \frac{4}{9} \cdot \frac{1}{4\pi \epsilon_0} \cdot \frac{q_1 \cdot q_2}{r^2} \]

### Step 5: Relate the new force to the original force
From the original force \( F \):
\[ F = \frac{1}{4\pi \epsilon_0} \cdot \frac{q_1 \cdot q_2}{r^2} \]
Thus, we can express the new force \( F' \) in terms of \( F \):
\[ F' = \frac{4}{9} F \]

### Final Answer
The force between the charges after inserting the dielectric slab becomes:
\[ F' = \frac{4}{9} F \]