A combination of parallel plate capacitors is maintained at a certain potential difference When a $3\, mm$ thick slab is introduced between all the plates, in order to maintain the same potential difference, the distance between the plates is increased by $2.4\, mm$. Find the dielectric constant of the slab.
JEE MAIN 2017, Diffcult
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Before introducing a slab capacitance of plates
$C_{1}=\frac{\varepsilon_{0} A}{3}$
If a slab of dielectric constant $K$ is introduced between plates then
$C=\frac{K \varepsilon_{0} A}{d}$ then $C'_1 = \frac{{{\varepsilon _0}A}}{{2.4}}$
$\mathrm{C}_{1}$ and ${C'}_{1}$ are in series hence,
$\frac{{{\varepsilon _0}A}}{3} = \frac{{{\text{k}}\frac{{{\varepsilon _0}A}}{3} \cdot \frac{{{\varepsilon _0}A}}{{2.4}}}}{{{\text{k}}\frac{{{\varepsilon _0}A}}{3} + \frac{{{\varepsilon _0}{\text{A}}}}{{2.4}}}}$
$3\, k=2.4 \,k+3$
$0.6\, \mathrm{k}=3$
Hence, the dielectric constant of slap is given by,
$k=\frac{30}{6}=5$
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