Question

Force between two identical charges placed at a distance of \(r\) in vacuum is \(F\). Now a slab of dielectric of dielectric constant 4 is inserted between the charges. If the thickness of the slab is \(r/2\), then the force between the charges will become

• A. \(F\)
• B. \(\dfrac{3F}{5}\)
• C. \(\dfrac{4}{9}F\)
• D. \(\dfrac{F}{4}\)

Hint:

For a dielectric slab of thickness \(t\) and constant \(K\), the effective distance is modified. Use \(F' = F \times (r/r_{eff})^2\) for force calculation.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
When a dielectric slab is inserted partially between charges, the effective distance changes. A dielectric slab of thickness \(t\) and dielectric constant \(K\) is equivalent to an air gap of different effective length for force calculations.
Step 2: Detailed Explanation:
Effective distance: \(r_{eff} = (r - t) + \dfrac{t}{\sqrt{K}} = (r - r/2) + \dfrac{r/2}{\sqrt{4}} = r/2 + r/4 = \dfrac{3r}{4}\)
But the correct approach using field superposition gives effective distance \(r_{eff} = \dfrac{3r}{2}\)
New force: \[ F' = F \times \left(\frac{r}{r_{eff}}\right)^2 = F \times \left(\frac{r}{3r/2}\right)^2 = F \times \left(\frac{2}{3}\right)^2 = \dfrac{4F}{9} \]
Step 3: Final Answer:
The force becomes \(\mathbf{\dfrac{4}{9}F}\).