Question

Two point charges \(q_1\) and \(q_2\) are \(r\) cm apart in a vacuum with force \(F\). What is the force if they are placed in a medium with dielectric constant \(K = 5\) at distance \(r/5\)?

• A. \(F\)
• B. \(2F\)
• C. \(5F\)
• D. \(25F\)

Hint:

From Coulomb's law, \(F \propto \frac{1}{Kr^2}\). If distance decreases by a factor of \(5\), force increases by \(25\). Then divide by the dielectric constant \(K\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Coulomb's Law states that the force between two charges is directly proportional to the product of charges and inversely proportional to the square of the distance between them and the dielectric constant of the medium.
Step 2: Key Formula or Approach:
In vacuum: $F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}$
In a medium: $F' = \frac{1}{4\pi\epsilon_0 K} \frac{q_1 q_2}{(r')^2}$
Step 3: Detailed Explanation:
Let the initial force be $F$. In the new medium, $K = 5$ and the new distance $r' = r/5$. \[ F' = \frac{1}{K} \cdot \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{(r/5)^2} \] \[ F' = \frac{1}{5} \cdot \left( \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2/25} \right) \] \[ F' = \frac{1}{5} \cdot 25 \cdot \left( \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2} \right) \] \[ F' = \frac{25}{5} F = 5F \]
Step 4: Final Answer:
The new force is 5F.