Reason (R): Energy of system decreases when dielectric slab enters between plates of charged parallel plate capacitor.
Solution:
For an isolated charged capacitor, charge (Q) is constant. When a dielectric slab enters the capacitor, its capacitance (C) increases C' = KC. Since energy \(U = \frac{Q^2}{2C}\), the energy of the system decreases. A system tends to move towards a state of lower potential energy, so the slab is pulled in.
Leave a Reply