Two charges are placed at a distance apart if a glas slab is placed between them force between them will

Options

A. be zero
B. increase
C. decrease
D. remains the same

Answer

To solve the problem of how the force between two charges changes when a glass slab is placed between them, we can 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 \( R \) in a vacuum is given by Coulomb's law:
\[ F = \frac{1}{4 \pi \epsilon_0} \frac{Q_1 Q_2}{R^2} \]
where \( \epsilon_0 \) is the permittivity of free space.

**Hint:** Recall that Coulomb's law describes the electrostatic force between two charges and depends on the distance between them and the medium in which they are placed.

### Step 2: Introduce the glass slab
When a glass slab (or any dielectric medium) is placed between the two charges, it affects the permittivity of the medium. The effective permittivity \( \epsilon \) of the medium becomes:
\[ \epsilon = K \epsilon_0 \]
where \( K \) is the dielectric constant of the glass slab (and \( K > 1 \)).

**Hint:** Remember that the dielectric constant \( K \) indicates how much the electric field is reduced in the medium compared to a vacuum.

### Step 3: Calculate the new force with the glass slab
With the glass slab in place, the new force \( F' \) between the charges is given by:
\[ F' = \frac{1}{4 \pi \epsilon} \frac{Q_1 Q_2}{R^2} = \frac{1}{4 \pi K \epsilon_0} \frac{Q_1 Q_2}{R^2} \]

**Hint:** Notice how the presence of the dielectric medium modifies the formula for the force by introducing the dielectric constant \( K \).

### Step 4: Compare the forces
Since \( K > 1 \), we can see that:
\[ F' = \frac{F}{K} \]
This shows that the new force \( F' \) is less than the original force \( F \):
\[ F' < F \]

**Hint:** Think about how the dielectric constant affects the electric field and the resulting force between the charges.

### Conclusion
Thus, when a glass slab is placed between the two charges, the force between them **decreases**.

**Final Answer:** The force between the two charges decreases.