A parallel plate capacitor is first charged using a battery and then isolated. A dielectric slab is then introduced between the plates.
How does energy vary in this case?
Answer & explanation
Correct answer: option 2
When the capacitor is kept at a voltage, it gains charge.
Now when the system is isolated, the charge present on capacitor cannot change because of law of conservation of charge.
∴ Charge always remains constant in isolated systems.
initial capacitance < final capacitance and so from Conservation of charge, Energy $U = \frac{Q^2}{2C} $ decreases.