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Question

A capacitor is charged by a battery and then the battery is disconnected. A dielectric slab is introduced between the plates. The result is
  1. P.d between the plates increases, charge on the plate decreases
  2. P.d between the plates decreases, charge remains same
  3. P.d increases, charge remain constant
  4. P.d decreases, charge increases

A
P.d between the plates decreases, charge remains same
B
P.d between the plates increases, charge on the plate decreases
C
P.d increases, charge remain constant
D
P.d decreases, charge increases
Solution
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As dielectric slab does not affect charges:
V=QC=QdKϵoA

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