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
5 μF
b
7.5 μF
c
22.5 μF
d
30 μF
answer is D.
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Detailed Solution
By using Cair=ε0Ad, Cmedium=ε0Ad−t+tKFor K=∞ Cmedium =ε0Ad−t⇒CmCa=dd−t⇒Cm15=66−3⇒Cm=30 μF
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