The energy and capacity of a charged parallel plate capacitor are $U$ and $C$ respectively. Now a dielectric slab of $\in _r = 6$ is inserted in it then energy and capacity becomes  (Assume charge on plates remains constant)
  • A$6\,U,\, 6\,C$
  • B$U,\, C$
  • C$\frac{U}{6}\,,6C$
  • D$U,\, 6\,C$
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