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KCET 2023 Physics Solved Paper

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Question : 49

Total: 60

A parallel plate capacitor of capacitance C1 with a dielectric slab in between its plates is connected to a battery. It has a potential difference V1 across its plates. When the dielectric slab is removed, keeping the capacitor connected to the battery, the new capacitance and potential difference are C2 and V2 respectively. Then,

V1<V2,C1>C2
V1=V2,C1>C2
V1=V2,C1<C2
V1>V2,C1>C2

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