Two similar point charges `q_1 and q_2` are placed at a distance r apart in the air. The force between them is `F_1` . A dielectric slab of thickness `t(ltr)` and dielectric constant K is placed between the charges . Then the force between the same charge . Then the fore between the same charges is `F_2` . The ratio is

Options

A. 1
B. K
C. `[(r-t+tsqrtK)/r]^2`
D. `[r/(r-t+tsqrtK)]^2`

Answer

To solve the problem, we need to find the ratio of the forces \( F_1 \) and \( F_2 \) between two similar point charges \( q_1 \) and \( q_2 \) when a dielectric slab is introduced between them.

### Step-by-Step Solution:

1. **Understanding the Initial Force \( F_1 \)**:
The force between two point charges in air (or vacuum) is given by Coulomb's law:
\[ F_1 = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2} \]
Here, \( \epsilon_0 \) is the permittivity of free space, and \( r \) is the distance between the charges.

2. **Introducing the Dielectric Slab**:
When a dielectric slab of thickness \( t \) and dielectric constant \( K \) is placed between the charges, the effective distance \( r' \) between the charges changes. The effective distance can be approximated as:
\[ r' = r - t(1 - \frac{1}{\sqrt{K}}) \]
This accounts for the reduction in the effective distance due to the presence of the dielectric slab.

3. **Calculating the New Force \( F_2 \)**:
The new force \( F_2 \) when the dielectric is present is given by:
\[ F_2 = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{(r - t(1 - \frac{1}{\sqrt{K}}))^2} \]

4. **Finding the Ratio \( \frac{F_1}{F_2} \)**:
To find the ratio of the two forces, we can write:
\[ \frac{F_1}{F_2} = \frac{\frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}}{\frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{(r - t(1 - \frac{1}{\sqrt{K}}))^2}} \]
Simplifying this gives:
\[ \frac{F_1}{F_2} = \frac{(r - t(1 - \frac{1}{\sqrt{K}}))^2}{r^2} \]

5. **Final Expression**:
Thus, the final expression for the ratio of the forces is:
\[ \frac{F_1}{F_2} = \left( \frac{r - t(1 - \frac{1}{\sqrt{K}})}{r} \right)^2 \]