CapacitanceMedium

Question

On a parallel plate capacitor following operations can be performed.
P - connect the capacitor to a battery of emf V
Q - disconnect the battery
R - reconnect the battery with polarity reversed
S - insert a dielectric slab in the capacitor

Options

A.In PQR (perform P, then Q, then R), the stored electric energy remains unchanged and no thermal energy is developed   
B.The charge appearing on the capacitor is greater after the action PSQ then after the action PQS
C.The electric energy stored in the capacitor is greater after the action SPQ then after the action PQS
D.The electric field in the capacitor after the action PS is the same as that after SP

Solution

In PQS process charge on capacitor is Q = CV
In PSQ process charge on capacitor is Q′ = KCV
Electric energy stored in PQS is = CV2
Electric energy stored in PSQ is = KCV2
UPSQ > UPQS
Electric field in PS is E =
Electric field in SP is E =
EPS  = ESP

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