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Electrostatic Potential and Capacitance — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsElectrostatic Potential and Capacitance5 Marks
Question
A parallel-plate capacitor having plate area $400cm^2$ 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:
  1. Find the increase in electrostatic energy.
  2. If the power supply is now disconnected and the dielectric slab is taken out, find the further increase in energy.
  3. Why does the energy increase in inserting the slab as well as in taking it out?
✓

Answer

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

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