question_answer

A parallel plate capacitor with area \[200c{{m}^{2}}\] and separation between the plates\[1.5cm\], is connected across a battery of emf V. If the force of attraction between the plates is \[25\times {{10}^{-6}}N\],  the value of \[V\] is approximately:                                 [JEE Online 15-04-2018 (II)] \[\left( {{\in }_{0}}=8.85\times {{10}^{-12}}\frac{{{C}^{2}}}{N.{{m}^{2}}} \right)\]

A)                     \[150V\]                              

B) \[100V\]              

C) \[250V\]              

D)          \[300V\]

Correct Answer: C

Solution :

\[A=200c{{m}^{2}}\] \[d=1.5cm\] \[F=25\times {{10}^{-6}}N\] \[\because E=\frac{\sigma }{2{{\in }_{o}}}=\frac{Q}{2A{{\in }_{o}}}\] \[F=QE\] \[F=\frac{{{Q}^{2}}}{2A{{\in }_{o}}}\] But \[Q=CV=\frac{{{\in }_{o}}A(V)}{d}\] \[\therefore F=\frac{{{({{\in }_{o}}AV)}^{2}}}{{{d}^{2}}\times 2A{{\in }_{o}}}\] \[=\frac{{{({{\in }_{o}}A)}^{2}}\times {{V}^{2}}}{{{d}^{2}}\times 2\times (A{{\in }_{o}})}\] \[=\frac{({{\in }_{o}}A){{V}^{2}}}{{{d}^{2}}\times 2}\] \[25\times {{10}^{-6}}=\frac{(8.85\times {{10}^{-12}})\times (200\times {{10}^{-4}})\times {{V}^{2}}}{2.25\times {{10}^{-4}}\times 2}\] \[V=\sqrt{\frac{25\times {{10}^{-6}}\times 2.25\times {{10}^{-4}}\times 2}{8.85\times {{10}^{-12}}\times 200\times {{10}^{-4}}}}\] Here, on solving, \[v\approx 250V\]