KCET2022PhysicsCapacitance
A parallel place capacitor is charged by connecting a 2 ~V battery across it. It is then disconnected from the battery and a glass slab is introduced between plates. Which of the following pairs of quantities decrease?
Options
- APotential difference and energy stored
- BEnergy stored and capacitance
- CCapacitance and charge
- DCharge and potential difference
Correct answer
A. Potential difference and energy stored
Step-by-step solution
When a charged capacitor is disconnected from the battery and a glass slab (dielectric material) is introduced between the plates, then charge (Q), capacitance (C) , electric potential (V) and stored energy (U) is given as Q=Q₀, C=K C₀, V= V₀ K and U= U₀ K where, K= dielectric constant aligned V & = electric potential U & = potential energy V₀ & = initial potential of charged capacitor U₀ & = Initial energy of charged capacitor aligned From above, we see that V and U decreases after introducing glass plate of diele