Questions
A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect?
detailed solution
Correct option is B
Due to dielectric insertion, new capacitance C2=CK Initial energy stored in capacitor, U1=q22C Final energy stored in capacitor, U2=q22KCChange in energy stored, ΔU=U2-U1ΔU=q22C1K-1=12CV21K-1New potential difference between plates V'=qCK=VKTalk to our academic expert!
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A parallel plate, air filled capacitor has capacitance C. When a dielectric material of dielectric constant 4 is filled so that half of the space between plates is occupied by it, then percentage increase in the capacitance is equal to
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