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?

icon
Related questions
Question
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 ?
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 ?
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 4 steps with 4 images

Blurred answer
Similar questions