A parallel plate capacitor of capacitance $200 \,\mu {F}$ is connected to a battery of $200 \, {V} .$ A dielectric slab of dielectric constant $2$ is now inserted into the space between plates of capacitor while the battery remain connected. The change in the electrostatic energy in the capacitor will be ......$ J.$
  • A$400$
  • B$0.4$
  • C$40$
  • D$4$
JEE MAIN 2021, Diffcult
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