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Q.

The capacitance of an air capacitor is 15 μF the separation between the parallel plates is 6 mm. A copper plate of 3 mm thickness is introduced symmetrically between the plates. The capacitance now becomes 

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a

22.5 μF

b

7.5 μF

c

30 μF

d

5 μF

answer is D.

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Detailed Solution

By using Cair=ε0Ad, Cmedium=ε0Ad−t+tK

For K=∞ Cmedium =ε0Ad−t

⇒CmCa=dd−t⇒Cm15=66−3⇒Cm=30 μF

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