QJEEM18Phc82f: JEE Main 2018 Physics - Class 12 JEEM18Phc82f Solution: A parallel plate capacitor with area 200 cm 2 and separation…

Given area of Parallel plate capacitor, A = 200 cm²

Separation between the plates, d = 1.5 cm

Force of attraction between the plates, F = 25 × 10^–6 N

F = QE

$\Rightarrow$ F = ${{{Q^2}} \over {2A{ \in _0}}}$

[ As E due to parallel plate = ${\sigma \over {2{ \in _0}}} = {Q \over {2A{ \in _0}}}$ ]

Also we know, Q = CV = ${{{ \in _0}AV} \over d}$

$\therefore$ F = ${{{{\left( {{ \in _0}AV} \right)}^2}} \over {{d^2} \times 2A{ \in _0}}}$

$\Rightarrow$ V = d $\sqrt {{{2F} \over {{ \in _0}A}}}$

$\Rightarrow$ V = $1.5 \times {10^{ - 2}}\sqrt {{{2 \times 25 \times {{10}^{ - 6}}} \over {8.85 \times {{10}^{ - 12}} \times 2 \times {{10}^{ - 2}}}}}$

= $1.5 \times {10^{ - 2}}\sqrt {{{25} \over {8.85}}}$ = 250 V