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MHT CET202613 April 2026Evening ShiftPhysicsCapacitanceActual

A parallel plate capacitor with air between the plate has a capacitance of 15 pF. The separation between the plates becomes twice and the space between them is filled with a medium of dielectric constant 3.5 . Then the capacitance becomes x/4 pF. The value of x is

Options

  1. A105
  2. B109
  3. C111
  4. D115

Correct answer

A. 105

Step-by-step solution

The initial capacitance of the parallel plate capacitor with air is given by: C₁ = ₀ A d = 15 pF When the separation between the plates is doubled ( d' = 2d ) and the space is filled with a dielectric of constant K = 3.5 , the new capacitance becomes: C₂ = K ₀ A d' = 3.5 ₀ A 2d Substituting the value of ₀ A d from the initial condition: C₂ = 3.5 2 15 = 7 4 15 = 105 4 pF Given that the new capacitance is x 4 pF , we can compare the two expressions: x 4 = 105 4 x = 105 Answer: 105

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