Two charge spheres separated at a distance d exert a force F on each other. If they are immersed in a liquid of dielectric constant K=2, then the force (if all conditions are same) is

Options

A. `F//2`
B. `F`
C. `2F`
D. `4F`

Answer

To solve the problem, we will use Coulomb's law and the concept of dielectric constant. The force between two point charges in a vacuum (or air) is given by:

\[ F = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q_1 Q_2}{r^2} \]

Where:
- \( F \) is the force between the charges,
- \( Q_1 \) and \( Q_2 \) are the magnitudes of the charges,
- \( r \) is the distance between the charges,
- \( \epsilon_0 \) is the permittivity of free space.

When the charges are placed in a medium with a dielectric constant \( K \), the force between the charges is modified as follows:

\[ F' = \frac{1}{4\pi \epsilon} \cdot \frac{Q_1 Q_2}{r^2} \]

Where \( \epsilon \) is the permittivity of the medium, given by:

\[ \epsilon = K \cdot \epsilon_0 \]

Thus, the modified force in the medium becomes:

\[ F' = \frac{1}{4\pi K \epsilon_0} \cdot \frac{Q_1 Q_2}{r^2} \]

Now, we can express this in terms of the original force \( F \):

1. **Calculate the original force \( F \)**:
\[ F = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q_1 Q_2}{d^2} \]

2. **Calculate the new force \( F' \) in the medium**:
\[ F' = \frac{1}{4\pi K \epsilon_0} \cdot \frac{Q_1 Q_2}{d^2} \]

3. **Substituting \( K = 2 \)**:
\[ F' = \frac{1}{4\pi (2) \epsilon_0} \cdot \frac{Q_1 Q_2}{d^2} \]

4. **Relate \( F' \) to \( F \)**:
\[ F' = \frac{1}{2} \cdot \frac{1}{4\pi \epsilon_0} \cdot \frac{Q_1 Q_2}{d^2} = \frac{F}{2} \]

Thus, the new force \( F' \) when the charges are immersed in a liquid with dielectric constant \( K = 2 \) is:

\[ F' = \frac{F}{2} \]

### Final Answer:
The force when the charges are immersed in the liquid is \( \frac{F}{2} \).