QJEEM17Phb522: JEE Main 2017 Physics - Class 12 JEEM17Phb522 Solution: A combination of parallel plate capacitors is maintained at a…

Before introducing the slab between the plates of all three capacitors, we have

${C_1} = {{{\varepsilon _0}A} \over 3}$

After introducing the slab between the plates of all three capacitors, we have

${C_1} = K{{{\varepsilon _0}A} \over 3} + {{{\varepsilon _0}A} \over {2.4}}$

As the capacitors are in series, we get

${{{\varepsilon _0}A} \over 3} = {{\left( {{{K{\varepsilon _0}A} \over 3}} \right)\,.\,\left( {{{{\varepsilon _0}A} \over {2.4}}} \right)} \over {\left( {{{K{\varepsilon _0}A} \over 3}} \right) + \left( {{{{\varepsilon _0}A} \over {2.4}}} \right)}}$

Therefore, the dielectric constant of the slab is

$\Rightarrow 3K = 2.4K + 3$

$\Rightarrow 0.6K = 3$

$\Rightarrow K = {3 \over {0.6}} = {{30} \over 6} = 5$