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Question

Before introducing a slab capacitance of plates

$C_{1}=\frac{\varepsilon_{0} A}{3}$

If a slab of dielectric constant $K$ is introduced between p

Vedclass · Accepted Answer

$5$

Vedclass · Answer

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,

$\frac{{{\varepsilon _0}A}}{3} = \frac{{{	ext{k}}\frac{{{\varepsilon _0}A}}{3} \cdot \frac{{{\varepsilon _0}A}}{{2.4}}}}{{{	ext{k}}\frac{{{\varepsilon _0}A}}{3} + \frac{{{\varepsilon _0}{	ext{A}}}}{{2.4}}}}$

$3\, k=2.4 \,k+3$

$0.6\, \mathrm{k}=3$

Hence, the dielectric constant of slap is given by,

$k=\frac{30}{6}=5$

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