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

Question

The force between two identical charges placed at a distance of r in a vacuum is F . Now a slab of dielectric constant 4 is inserted between these two charges. If the thickness of the slab is r 2 , then the force between the charges will become, 3 5 F, 4 9 F, F 4, F 2

MARKS App · Accepted Answer

F = 1 4 π ε 0 q 1 q 2 r 2 For a dielectric of dielectric constant K between the charges, the effective separation in air r eff is given by 1 4 π ε 0 q 1 q 2 r eff 2 = 1 4 π ε 0 q 1 q 2 K r 2 ⇒ r eff = K r ∴ F ′ = 1 4 π ε 0 q 1 q 2 r 2 + K r 2 2 ⇒ F ′ F = 1 1 2 + 4 2 2 = 4 9

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