Download the app

Questions  

Plates of a parallel plate capacitor of capacitance 4μF is connected to the terminals of  a battery of 12 volt. Now the battery is disconnected from the capacitor and the plates of the capacitor are connected to the plates of a  1μF parallel plate uncharged capacitor. If a dielectric slab of dielectric constant k = 2 is introduced into the space between the plates of 1μF capacitor, The potential difference across the combination of capacitors is

a
7.5 volt
b
4 volt
c
6 volt
d
8 volt

detailed solution

Correct option is D

Q = Charge on the combination of capacitors =4 x 12 μC=48μCCe = Equivalent capacitance =4+1 x 2μF=6μF∴V'=QCe=486volt=8 volt

Talk to our academic expert!

+91

Are you a Sri Chaitanya student?


Similar Questions

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


phone icon
whats app icon