question_answer

A parallel plate capacitor is of area \[6\text{ }c{{m}^{2}}\] and a separation 3 mm. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constants \[{{K}_{1}}=10,\text{ }{{K}_{2}}=12\] and\[{{K}_{3}}=14\]. The dielectric constant of a material which when fully inserted in above capacitor, gives same capacitance would be - [JEE Main Online Paper (Held On 10-Jan-2019 Morning]

A) 12                                            

B) 36

C) 14                                

D)   4

Correct Answer: A

Solution :

\[{{C}_{eq}}\,=\,\frac{{{\varepsilon }_{0}}}{d}\left( \frac{A}{3}\times 10+\frac{A}{3}\times 12+\frac{A}{3}\times 14 \right)\] If whole slab is replaced by single dielectric K then \[C\,\,=\,\,\frac{K\,{{\in }_{0}}\,A}{d}\] Now \[\frac{K\,{{\in }_{0}}\,A}{d}=\,\frac{{{\in }_{0}}A}{d}\left[ \frac{10}{3}+\frac{12}{3}+\frac{14}{3} \right]\] \[K=12\]