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Three identical metal plates of area S are at equal distances b as shown. Initially metal plate A is uncharged, while metal plates B and C have respective charges $+ Q_{0} a n d - Q_{0}$ initially as shown. Metal plates A and C are connected by switch K through a resistor of resistance R. The key K is closed at time t = 0. Then

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The magnitude of current in amperes through the resistor at any later time t is $\frac{Q_{0} b}{R S ε_{0}} e^{\frac{- b t}{R S ε_{0}}}$

The total heat dissipated through resistor of resistance R is $\frac{Q_{0}^{2} b}{4 S ε_{0}}$

The magnitude of current in amperes through the resistor at any later time t is $\frac{Q_{0} b}{R S ε_{0}} e^{\frac{- 2 b t}{R S ε_{0}}}$

The total heat dissipated through resistor of resistance R is $\frac{Q_{0}^{2} b}{2 S ε_{0}}$

answer is B, C.

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Detailed Solution

$\frac{Q_{0} + q}{C} + i R + \frac{q}{C} = 0 \frac{Q_{0} + 2 q}{2 C R} = \frac{d q}{d t} q = \frac{Q_{0}}{2} (1 - e^{\frac{- 2 t}{R c}}) i = \frac{Q_{0}}{R C} e^{\frac{- 2 t}{R C}}$

In steady state change on both capacitor is $\frac{Q_{0}}{2}$ ; Hence heat produced

$\frac{Q_{0}^{2}}{2 C} - \frac{(\frac{Q_{0}}{2})}{2 C} \times 2 = \frac{Q_{0}^{2}}{4 S ε_{0}} b$

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