Physics - Electrostatics Question with Solution | TestHub

The capacitance of air capacitor

$C_0=\frac{A{\epsilon }_0}{d}=\mathrm{3 }\mathrm{\mu }\mathrm{F}$...(i)

When a dielectric of permittivity${\epsilon }_r$or dielectric constant$K$is introduced between the plates of the capacitor, then its capacitance

$C =\frac{KA{\epsilon }_0}{d}=15\mathrm{ }\mathrm{\mu }\mathrm{F}$...(ii)

Dividing Eq. (ii) by Eq. (i)

$\frac{C}{C_0}= \frac{\frac{KA{\epsilon }_0}{d}}{\frac{A{\epsilon }_0}{d}}=\frac{15}{3}$

$\therefore$$K=5$

Permittivity of the medium

${\epsilon }_{}= {\epsilon }_{0}K$

$=8.854\times {10}^{-12}\times 5$

$=44.27\times {10}^{-12}$

$=0.44\times {10}^{-\mathrm{10 }}{\mathrm{C}}^{\mathrm{2 }}{\mathrm{N}}^{-\mathrm{1 }}{\mathrm{m}}^{-2}$