Questions
A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in
a
Reduction of charge on the plates and increase of potential difference across the plates
b
Increase in the potential difference across the plate, reduction in stored energy, but no change in the charge on the plates
c
Decrease in the potential difference across the plates, reduction in the stored energy, but no change in the charge on the plates
d
None of the above
detailed solution
Correct option is C
Battery in disconnected so Q will be constant as C∝ K. So with introduction of dielectric slab capacitance will increase using Q = CV, V will decrease and using U=Q22C, energy will decrease.Talk to our academic expert!
Similar Questions
The energy of a charged capacitor is given by the expression (q = charge on the conductor and C = its capacity)
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