In parallel plate capacitor if battery is connected and dielectric slab is inserted then..#sdphysics

When we introduce a dielectric slab between plates of capacitor then its capacitance increases. What happen if we introduce a dielectric slab between plates of capacitor and battery is disconnected/connected. When battery is connected and dielectric slab of dielectric constant K is introduced in between the plates: Potential difference between plates remains constant.(V’=V) Capacitance increases K times. (C=KC0) Electric field E remains the same . E=Vd as V and d are constant. Hence E remains same. Charge becomes K times.(q’=Kq0) C=KC0 ⇒ q’V=Kq0V (as potential is same) q’=Kq0 Energy stored becomes K times. U = 1/2CV2 = 1/2KC0V2 U=KU0 When battery is removed and dielectric slab of dielectric constant K is introduced between plates: Potential difference between plates become 1/k times.(V’=V/K) Capacitance increases K times.(C=KC0) Charge remains same because charge is conserved. Electric field E becomes 1/K times. E’=V’d =V/K d #sdphysics #electrostatics #electrostatic_potential_and_capacitance #class12 #cbseboard