0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

A parallel-plate capacitor having plate area $$400 cm^2$$ and separation between the plates $$1.0 mm$$ is connected to a power supply of $$100 V.$$ A dielectric slab of thickness 0.5 mm and dielectric constant 5.0 is inserted into the gap.
a) Find the increase in electrostatic energy.
b) If the power supply is now disconnected and the dielectric slab is taken out, find the further increase in energy.
c) Why does the energy increase in inserting the slab as Well as in taking it out ?

Solution
Verified by Toppr


Was this answer helpful?
1
Similar Questions
Q1
A parallel-plate capacitor having plate area 400 cm2 and separation between the plates 1⋅0 mm is connected to a power supply of 100 V. A dielectric slab of thickness 0⋅5 mm and dielectric constant 5⋅0 is inserted into the gap. (a) Find the increase in electrostatic energy. (b) If the power supply is now disconnected and the dielectric slab is taken out, find the further increase in energy. (c) Why does the energy increase in inserting the slab as well as in taking it out?
View Solution
Q2
67.A parallel plate capacitor having plate area 400cm2 and separation between the plates 1mm is connected to a power supply of 100V.A dielectric slab of thickness 0.5 mm and dielectric constant 5 is inserted into the gap . if the power supply is disconnected and the dielectric slab is taken out , find the increase in energy
View Solution
Q3
A parallel plate capacitor of capacitance 100 μF is connected to a power supply of 200 V. A dielectric slab of dielectric constant 5 is now inserted into the gap between the plates. The workdone by power supply will be:
View Solution
Q4

The plates of a parallel plate capacitor are charged to 200 V and then, the charging battery is disconnected. Now, a dielectric slab of dielectric constant 5 and thickness 4 mm is inserted between the capacitor plates. To maintain the original capacity, the increase in the separation between the plates of the capacitor is:


View Solution
Q5
146. A parallel plate capacitor of capacitance C is connected to a power supply V . A dielectric slab of dielectric constant k= 5 is now inserted into the gap between the plates to occupy the entire region between the plates . Work done by the supply is
View Solution