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
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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%
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