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Manipal MET2017PhysicsCapacitance

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 these two charges. If the thickness of the slab is r / 2 , then the force between the charges will become

Options

  1. AF
  2. B3 5 F
  3. C4 9 F
  4. DF 4

Correct answer

D. F 4

Step-by-step solution

From Coulomb's law the force (F) between two charges is F= 1 4 ₀ k q^2 r^2 First case : k=1F= 1 4 ₀ q^2 r^2 ...(i) Second case : k=4F^ = 1 4 ₀ 4 q^2 r^2 ....(ii) Dividing Eq. (i) by Eq. (ii), we have F F^ =4 F^ = F 4

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