Question

A parallel plate capacitor with air between the plates has a capacitance of 12 pF (1pF = 10-12F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with s substance of dielectric constant 8 ?

Answer 1

Answer:

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Answer 2

Answer:

192 pF

Explanation:

we know capacitance ,

    C =  $ A          (  $ ---- epsilon)

              d

   when distance is reduced to half and dielectric is introduced new capacitance C' will be:

     C'  = 8 * $ A

                 d/2

   therefore,

  C' = 16C       ( C= 12 pF)

 C' = 192pF.