Force between two identical charges placed at a distance of `r` in vacume 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. F
B. `(3)/(5) F`
C. `(4)/(9) F`
D. `(F)/(4)`

Explanation

To solve the problem of finding the new force between two identical charges when a dielectric slab is inserted between them, we can follow these steps:

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

### Step 2: Insert the Dielectric Slab
When a dielectric slab with a dielectric constant \( K = 4 \) is inserted between the two charges, the effective distance between the charges changes. The thickness of the slab is \( \frac{r}{2} \), which means that the remaining distance outside the slab is also \( \frac{r}{2} \).

### Step 3: Calculate the New Force
The effective force between the charges when a dielectric is present is given by:
\[ F' = \frac{1}{4 \pi \epsilon_0 K} \frac{q_1 q_2}{d^2} \]
where \( d \) is the effective distance between the charges.

In this case, the effective distance \( d \) is the distance outside the dielectric slab, which is \( \frac{r}{2} \). Therefore, we can substitute \( K \) and \( d \) into the equation:
\[ F' = \frac{1}{4 \pi \epsilon_0 \cdot 4} \frac{q_1 q_2}{\left( \frac{r}{2} \right)^2} \]

### Step 4: Simplify the Expression
Now we simplify the expression:
\[ F' = \frac{1}{4 \pi \epsilon_0 \cdot 4} \frac{q_1 q_2}{\frac{r^2}{4}} = \frac{1}{4 \pi \epsilon_0 \cdot 4} \cdot \frac{4 q_1 q_2}{r^2} \]
\[ F' = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q_1 q_2}{r^2} \]

### Step 5: Relate to Original Force
Since the original force \( F \) was:
\[ F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2} \]
we can see that:
\[ F' = \frac{F}{4} \]

### Conclusion
Thus, the new force between the charges when the dielectric slab is inserted is:
\[ F' = \frac{F}{4} \]