Two charges are at a distance `d` apart. If a copper plate (conducting medium) of thickness `d//2` is placed between them , the effictive force will be

Options

A. `2 F`
B. `F//2`
C. `0`
D. `sqrt(2) F`

Answer

To solve the problem of determining the effective force between two charges when a copper plate (a conducting medium) of thickness \( \frac{d}{2} \) is placed between them, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Charges and Distance**:
Let the two point charges be \( q_1 \) and \( q_2 \), and they are separated by a distance \( d \).

2. **Introduce the Conducting Medium**:
A copper plate of thickness \( \frac{d}{2} \) is placed between the two charges. This means that the distance between the charges is effectively reduced due to the presence of the conducting plate.

3. **Understanding the Effect of a Conducting Medium**:
When a conducting medium is introduced between two charges, it affects the electric field and the force between the charges. The conducting medium will polarize and create an induced charge on its surfaces.

4. **Use of the Electric Force Formula**:
The electric force \( F \) between two point charges in a medium is given by:
\[ F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2} \cdot \frac{1}{k} \]
where \( k \) is the dielectric constant of the medium.

5. **Dielectric Constant of a Conductor**:
For a perfect conductor, the dielectric constant \( k \) tends to infinity. This means that the electric field inside the conductor is zero.

6. **Calculate the Effective Force**:
Since the dielectric constant \( k \) approaches infinity, the effective force \( F \) between the two charges becomes:
\[ F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{d^2} \cdot \frac{1}{\infty} = 0 \]

7. **Conclusion**:
Therefore, the effective force between the two charges when a copper plate of thickness \( \frac{d}{2} \) is placed between them is **zero**.