The ratio of the force between two small conducting spheres of equal charge in (a) a medium of dielectric constant 2, and (b) air is respectively

Options

A. `1:4`
B. `4:1`
C. `1:2`
D. `2:1`

Explanation

To find the ratio of the force between two small conducting spheres of equal charge in (a) a medium of dielectric constant 2, and (b) air, we can use Coulomb's law, which states that the electric force \( F \) between two point charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) in a medium with dielectric constant \( \kappa \) is given by:

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

### Step 1: Identify the forces in both media
1. **Force in air**: The dielectric constant of air is approximately 1. Thus, the force \( F_{\text{air}} \) between the two charges in air is:
\[ F_{\text{air}} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q^2}{r^2} \]
where \( q \) is the charge on each sphere.

2. **Force in medium with dielectric constant 2**: The force \( F_{\text{medium}} \) in a medium with dielectric constant \( \kappa = 2 \) is:
\[ F_{\text{medium}} = \frac{1}{4 \pi \epsilon_0 \cdot 2} \cdot \frac{q^2}{r^2} \]

### Step 2: Write the ratio of the forces
Now, we can find the ratio of the forces in the two different media:
\[ \text{Ratio} = \frac{F_{\text{medium}}}{F_{\text{air}}} \]

Substituting the expressions we found:
\[ \text{Ratio} = \frac{\frac{1}{4 \pi \epsilon_0 \cdot 2} \cdot \frac{q^2}{r^2}}{\frac{1}{4 \pi \epsilon_0} \cdot \frac{q^2}{r^2}} \]

### Step 3: Simplify the ratio
The \( \frac{q^2}{r^2} \) and \( \frac{1}{4 \pi \epsilon_0} \) terms cancel out:
\[ \text{Ratio} = \frac{1}{2} \]

### Final Answer
Thus, the ratio of the force between the two small conducting spheres of equal charge in a medium of dielectric constant 2 to the force in air is:
\[ \text{Ratio} = \frac{1}{2} \]