A parallel-plate capacitor having plate area 400cm2 and separatio… - Vidyadip

Question

A parallel-plate capacitor having plate area 400cm² and separation between the plates 1.0mm is connected to a power supply of 100V. A dielectric slab of thickness 1.0mm and dielectric constant 5.0 is inserted into the gap:

Find the increase in electrostatic energy.
If the power supply is now disconnected and the dielectric slab is taken out, find the further increase in energy.
Why does the energy increase in inserting the slab as well as in taking it out?

✓

Answer

A = 400cm² = 4 × 10^-2m²
d = 1cm = 1 × 10^-3m
V = 160V
t = 0.5 = 5 × 10^-4m
k = 5
$\text{C}=\frac{\in_0\text{A}}{\text{d}-\text{t}+\frac{\text{t}}{\text{k}}}$
$=\frac{8.85\times10^{-12}\times4\times10^{-2}}{10^{-3}-5\times10^{-4}+\frac{5\times10^{-4}}{5}}$
$=\frac{35.4\times10^{-4}}{10^{-3}-0.5}$

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