Question

A parallel-plate capacitor is charged by a battery and then the battery is removed and a dielectric of constant K is used to fill the gap between the plates. Inserting the dielectric changes the energy stored by a factor of

Answer 1

Step-by-step explanation:

Battery is removed so its charge stored becomes fixed .

Energy of capacitor E₁ = Q² / 2C

When dielectric is inserted , the capacity of the capacitor is increased to KC .

Charge on it still remains the same ie Q

So its new energy

E₂ = Q² / 2 KC

E₂ / E₁ = 1 / K

E₂ = E₁ / K

Change in energy

= E₂ - E₁

= E₁ / K - E₁

= E₁ [ (1 / K) - 1 ]

Factor by which energy is changed

= [ (1 / K) - 1 ] .