Question:medium

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

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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.
Updated On: Apr 8, 2026
  • \(F\)
  • \(\dfrac{3F}{5}\)
  • \(\dfrac{4}{9}F\)
  • \(\dfrac{F}{4}\)
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The Correct Option is C

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