Two charges are placed a certain distance apart in air. If a brass plate is introduced between them, the force between them will

Options

A. increase
B. decrease
C. remain the same
D. become zero

Explanation

To solve the question about the effect of introducing a brass plate between two charges, we can follow these steps:

### Step 1: Understand the Initial Setup
We have two charges, \( q_1 \) and \( q_2 \), placed a certain distance apart in air. The force between them can be calculated using Coulomb's law, which states that the force \( F \) between two point charges is given by:

\[ F = \frac{1}{4\pi \epsilon_0} \cdot \frac{q_1 q_2}{D^2} \]

where:
- \( \epsilon_0 \) is the permittivity of free space,
- \( D \) is the distance between the charges.

### Step 2: Introduce the Brass Plate
When a brass plate (a conductor) is introduced between the two charges, it affects the electric field between them. Brass, being a metal, has a very high conductivity and can be considered to have an infinite dielectric constant.

### Step 3: Analyze the Effect of the Brass Plate
The introduction of a conductor like brass between the two charges leads to the redistribution of charges on the surface of the brass plate. This redistribution creates an electric field that opposes the electric field created by the two charges.

### Step 4: Calculate the New Force
Since the dielectric constant \( k \) of a conductor is considered to be infinite, the effective permittivity \( \epsilon \) in the region between the charges becomes:

\[ \epsilon = \epsilon_0 \cdot k = \infty \]

This means that the force between the two charges can be expressed as:

\[ F' = \frac{1}{4\pi \epsilon} \cdot \frac{q_1 q_2}{D^2} = \frac{1}{4\pi \cdot \infty} \cdot \frac{q_1 q_2}{D^2} = 0 \]

### Step 5: Conclusion
Thus, when the brass plate is introduced between the two charges, the force between them becomes zero.

### Final Answer
The force between the two charges will be **zero** when a brass plate is introduced between them.

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