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haridas_mandal

A parallel plate capacitor fitted with a material of dielectric constant 'k' is charged to a certain voltage. The dielectric material removed. Then ;- a)capacitance reduces by a factor 'k'. b) Electric field reduces by a factor 'k'. c) Voltage across the capacitor increases by a factor 'k'. d) charge stored in the capacitor increases by a factor 'k'. options: 1) a and b 2) a and c 3) b and c 4) b and d I know for sure capacitance reduces by a factor 'k' but not getting the combination right. pl help

  • one year ago
  • one year ago

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  1. telltoamit
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    2 assuming battery is also removed....charge remains same...hence voltage and field have to increase K times

    • one year ago
  2. Mashy
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    exactly right.. capacitance decreases.. and hence voltage increases!

    • one year ago
  3. haridas_mandal
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    So option in this case would be serial 2 ( a&c). Thanks

    • one year ago
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