A parallel-plate capacitor of area A and separation d is charged to V and disconnected. A dielectric slab K is inserted fully. If Q, E, W denote charge, electric field and work done in insertion, choose the incorrect relation.
Options

Question

A parallel-plate capacitor of area A and separation d is charged to V and disconnected. A dielectric slab K is inserted fully. If Q, E, W denote charge, electric field and work done in insertion, choose the incorrect relation.

Options

G1 Physics · Accepted Answer

After disconnection Q remains ε₀AV/d. Final field is E=Q/(Kε₀A)=V/(Kd). The energy decrease, equal to work gained by insertion, is Ui−Uf=[ε₀AV²/(2d)](1−1/K). Therefore option B is the incorrect relation. Correct answer: B.

A parallel-plate capacitor of area A and separation d is charged to V and disconnected. A dielectric slab K is inserted fully. If Q,E,W denote charge, electric field and work done in insertion, choose the incorrect relation.