Two charges are at distance (d) apart in air Coulomb force between them is F. If a dielectric material of dielectric constant (K) is placed between them, the coulomb force now becomes.

Options

A. `F//K`
B. FK
C. `F//K^(2)`
D. `K^(2)F`

Answer

To solve the problem, we need to understand how the introduction of a dielectric material affects the Coulomb force between two charges. Here’s a step-by-step solution:

### Step 1: Understand the Coulomb's Law
Coulomb's law states that the force \( F \) between two point charges \( Q_1 \) and \( Q_2 \) separated by a distance \( d \) in a vacuum (or air, which is approximately the same) is given by the formula:
\[ F = \frac{1}{4 \pi \epsilon_0} \frac{Q_1 Q_2}{d^2} \]
where \( \epsilon_0 \) is the permittivity of free space.

### Step 2: Introduce the Dielectric Constant
When a dielectric material with dielectric constant \( K \) is introduced between the charges, the effective force \( F' \) between the charges is reduced by a factor of \( K \). The modified formula becomes:
\[ F' = \frac{1}{4 \pi \epsilon} \frac{Q_1 Q_2}{d^2} \]
where \( \epsilon = K \epsilon_0 \) is the permittivity of the dielectric material.

### Step 3: Relate the Forces
Since \( \epsilon = K \epsilon_0 \), we can express the modified force \( F' \) in terms of the original force \( F \):
\[ F' = \frac{1}{K} \left( \frac{1}{4 \pi \epsilon_0} \frac{Q_1 Q_2}{d^2} \right) \]
This shows that:
\[ F' = \frac{F}{K} \]

### Step 4: Conclusion
Thus, the new Coulomb force \( F' \) when a dielectric is placed between the charges is:
\[ F' = \frac{F}{K} \]

### Final Answer
The Coulomb force now becomes \( \frac{F}{K} \).

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