A parallel-plate capacitor having plate area 400 cm^2 and separat… - Vidyadip

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 $100V.$ A dielectric slab of thickness $1.0\ mm$ 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

$A = 400\ cm^2 = 4 \times 10^{-2}m^2$
$d = 1\ cm = 1 \times 10^{-3}m$
$V = 160V$
$t = 0.5 = 5 \times 10^{-4}m$
$k = 5$
$\text{C}=\frac{\epsilon_0\text{A}}{\text{d}-\text{t}+\frac{\text{t}}{\text{k}}}$
$=\frac{8.85\times10^{-12}\times4\times10^{-2}}{10^{-3}-5\times10^{-4}+\frac{5\times10^{-4}}{5}}$
$=\frac{35.4\times10^{-4}}{10^{-3}-0.5}$

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