JEE MAIN - Physics (2017 - 9th April Morning Slot - No. 26)
A combination of parallel plate capacitors is maintained at a certain potential difference.
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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.
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.
3
4
5
6
Explanation
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
$$ \Rightarrow 3K = 2.4K + 3$$
$$ \Rightarrow 0.6K = 3$$
$$ \Rightarrow K = {3 \over {0.6}} = {{30} \over 6} = 5$$
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