A parallel plate capacitor of capacity ${C_0}$ is charged to a potential ${V_0}$

$(i)$ The energy stored in the capacitor when the battery is disconnected and the separation is doubled ${E_1}$

$(ii)$ The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is ${E_2}.$

Then ${E_1}/{E_2}$ value is

  • A$4$
  • B$1.5$
  • C$2$
  • D$0.25$
Medium
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