A 200µF parallel plate capacitor having plate separation of 5 mm is charged by a 100 V dc source. It remains connected to the source. Using an insulated handle, the distance between the plates is doubled and a dielectric slab of thickness 5 mm and dielectric constant 10 is introduced between the plates. Explain with reason, how the (i) capacitance, (ii) electric field between the plates, (iii) energy density of the capacitor will change?
A 200µF parallel plate capacitor having plate separation of 5 mm is charged by a 100 V dc source. It remains connected to the source. Using an insulated handle, the distance between the plates is doubled and a dielectric slab of thickness 5 mm and dielectric constant 10 is introduced between the plates. Explain with reason, how the (i) capacitance, (ii) electric field between the plates, (iii) energy density of the capacitor will change?
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![A 200µF parallel plate capacitor having plate
separation of 5 mm is charged by a 100 V dc
source. It remains connected to the source. Using
an insulated handle, the distance between the
plates is doubled and a dielectric slab of thickness
5 mm and dielectric constant 10 is introduced
between the plates. Explain with reason, how the
(i) capacitance, (ii) electric field between the plates,
(iii) energy density of the capacitor will change ?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F92ac0be1-7b82-48a6-9fb4-d70435da9f41%2F9b81c192-ec03-4c2e-8c2e-5c075c82a579%2Fzfdvz44_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A 200µF parallel plate capacitor having plate
separation of 5 mm is charged by a 100 V dc
source. It remains connected to the source. Using
an insulated handle, the distance between the
plates is doubled and a dielectric slab of thickness
5 mm and dielectric constant 10 is introduced
between the plates. Explain with reason, how the
(i) capacitance, (ii) electric field between the plates,
(iii) energy density of the capacitor will change ?
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