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A parallel plate capacitor has plates of area A =0.055 m2 separated by distance d =1.60...

A parallel plate capacitor has plates of area A =0.055 m2 separated by distance d =1.60 ✕ 10−5 m and is attached to a battery with potential difference ΔV = 22 V.(The permittivity of free space is εo= 8.85 ✕ 10−12 C2/N·m2).

  1. calculate the capacitance in F if the space between the plates is filled with air
  2. what is the electric field between the plates of the capacitor
  3. suppose the capacitor is replaced with another capacitance C= 1.90 x10^-5 F, what is the energy stored
  4. with the same condition from c, the charging battery is now disconnected and a cermamic slab with dielectric constant K=3.10 is slipped between the plates of the new capacitor. What is the energy stored in the capacitor now?
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