Question

a parallel plate condenser has initially air medium between the plates .If a slab of dielectric constant 5 having thickness half the distance of separation between the plates is introduced, the percentage increase in its capacity is

Answer 1

Answer:

The answer is 66.7%

Explanation:

According to the problem the slab of dielectric constant 5 having thickness half the distance of separation.

Let the initial capacitance= Co

Therefore,

Co=ε0a/r [ a is the area and r is the diameter]

New  capacitance:

C=ε0a/(r−s+s/k)

  =ε0a/(r−r/2+r/2/5)

   =ε0a/(r−r/2+r/10)

 =ε0a/(10r/10−5r/10+r/10)

 =ε0a/6r/10

 =10/6 x ε0a/r

 =10/6 x Co

Percentage increase in capacity = C−Co/Co ×100

                                                    =10/6Co−Co/Co×100

                                                    =4/6×100

                                                       =66.7%