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 Online 09-04-2017]
A) 6
B) 4
C) 3
D) 5
Correct Answer: D
Solution :
\[{{C}_{1}}=\frac{{{\varepsilon }_{0}}A}{3}\,\] before \[{{C}_{1}}=\frac{k{{\varepsilon }_{0}}A}{3}\,+\frac{{{\varepsilon }_{0}}A}{2.4}\] after \[\frac{{{\varepsilon }_{0}}A}{3}=\frac{k\frac{{{\varepsilon }_{0}}A}{3}.\,\frac{{{\varepsilon }_{0}}A}{2.4}}{k\frac{{{\varepsilon }_{0}}A}{3}+\frac{{{\varepsilon }_{0}}A}{2.4}}\] \[3k=2.4k+3\] \[0.6\,k=3\Rightarrow \,k=\frac{3}{0.6}\] \[k=\frac{30}{6}=5\]
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