Questions
Two identical parallel plate capacitors are connected in series and a 12 volt battery is connected across this combination. Now a dielectric slab having dielectric constant 2 is inserted into the space between the plates of one of the capacitors and the slab completely fills the space between the plates. Then potential difference between the plates of this capacitors is
detailed solution
Correct option is D
After the slab is inserted, final equivalent capacitance of the system, Ce=KC×CKC+C=KCK+1=2C2+1=2C3∴ Q = New charge in each capacitor = Ce×e=2C3×12=8C ∴V= New potential difference between the plates=QCB=8C2C=4voltTalk to our academic expert!
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Two identical metal plates are given positive charges Q1 and Q2 (<Q1) respectively. If they are now brought close together to form a parallel plate capacitor with capacitance C, the potential difference between them is
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