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A parallel plate capacitor is first charged and then a dielectric slab is introduced between the plates after disconnecting it from the battery. The quantity that remains unchanged is
  • a)
    charge
  • b)
    energy
  • c)
    potential
  • d)
    capacitance
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
A parallel plate capacitor is first charged and then a dielectric slab...
Charge on plate remain same and other options will change energy change by formula q2 /2c since capacitance increases by putting slab between plates so energy will decrease potential difference across plates also decreases because net electric field become decrease
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A parallel plate capacitor is first charged and then a dielectric slab is introduced between the plates after disconnecting it from the battery. The quantity that remains unchanged isa)chargeb)energyc)potentiald)capacitanceCorrect answer is option 'A'. Can you explain this answer?
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