A parallel-plate capacitor having plate area 400 cm2 and separation between the plates 1.0 mm is connected to a power supply of 100V. A dielectric slab of thickness 1.0 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?
Given,
Area, A = 400cm2 = 400 × 10–4m2
distance between plates d = 1cm = 1× 10–3m
voltage V = 100V
Thickness t = 0.5 = 5 × 10–4m
dielectric constant, k = 5
The capacitance of the capacitor without the dielectric slab is given by
Where,
A= Plate Area
d= separation between the plates,
∈0 = Permittivity of free space = 8.854 × 10-12 m-3 kg-1 s4 A2
When the dielectric slab is inserted, the capacitance becomes
where, t is the thickness of the slab
