Electric Potential and Capacitance | Physics JEE previous year questions with Solutions

(2) Initial potential energy of system

$\Rightarrow \frac{1}{2} C V^{2} + \frac{1}{2} (2 C) {(2 V)}^{2}$

$U_{i} = E + 8 E = 9 E$

Common potential difference across capacitors, finally

$\Rightarrow V_{0} = 5 V / 3$

Final potential energy of system

$U_{f} = \frac{1}{2} C V_{0}^{2} + \frac{1}{2} (2 C) V_{0}^{2}$

$U_{f} = \frac{3 C}{2} V_{0}^{2} = \frac{3 C}{2} {(\frac{5 V}{3})}^{2} = \frac{1}{2} C V^{2} (\frac{25}{3}) = \frac{25}{3} E$

Loss of energy

$\Rightarrow U_{i} - U_{f} = 9 E - \frac{25 E}{3} = \frac{(27 - 25) E}{3} = \frac{2 E}{3}$

(1320)

$V = \frac{Q}{C} = \frac{i t}{(\frac{ε_{0} A}{d})} = \frac{i t d}{ε_{0} (π r^{2})}$

$\Rightarrow d = \frac{ε_{0} (π r^{2})}{i} (\frac{V}{t})$

$d = \frac{(9 \times 10^{- 12}) (\frac{22}{7}) {(0.1)}^{2}}{0.15} (7 \times 10^{8}) m = 1320 μm$

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