The force between two point charges in vacuum is 15N, if a brass plate is introduced between the two charges, then force between them will -

Options

A. Becomes zero
B. Remains the same
C. Becomes 30 N
D. Becomes 60 N

Answer

To solve the problem, we need to analyze the effect of introducing a brass plate between two point charges on the electrostatic force between them.

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
- We are given that the force between two point charges in vacuum is \( F = 15 \, \text{N} \).
- The force between two point charges can be calculated using Coulomb's Law:
\[ F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2} \]
- Here, \( q_1 \) and \( q_2 \) are the magnitudes of the charges, \( r \) is the distance between them, and \( \epsilon_0 \) is the permittivity of free space.

2. **Introducing the Brass Plate**:
- When a brass plate is introduced between the two charges, it acts as a dielectric material.
- The dielectric constant \( K \) of brass is significantly high (theoretically considered infinite for perfect conductors).

3. **Effect of Dielectric on Force**:
- The presence of a dielectric reduces the effective force between the charges. The new force \( F' \) can be expressed as:
\[ F' = \frac{F}{K} \]
- Since \( K \) (the dielectric constant of brass) is very high (approaching infinity), we can say:
\[ F' = \frac{F}{\infty} = 0 \]

4. **Conclusion**:
- Therefore, when the brass plate is introduced, the force between the two charges effectively becomes zero.

### Final Answer:
The force between the two charges after introducing the brass plate will be \( 0 \, \text{N} \).

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