Two identical parallel plate capacitors of capacitance C and charge Q are placed next to each other as shown in figure. If the two positive by charged plates are connected by a resistor of resistance R .

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A total charge of $Q / 3$ flows through the resistor.

No charge flows through resistor.

Heat loss will be $\frac{Q^{2}}{18 C}$

Heat loss, instead, if the negatively charged plates are connected will be same as when positive plates are connected.

answer is A, D.

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Detailed Solution

Initially, $V_{A C} \neq 0 \Rightarrow$ charge flows through wire
$\frac{Q - q}{A \in_{0}} d = \frac{q}{A \in_{0}} \times 2 d q = \frac{Q}{3} U_{i} = \frac{Q^{2}}{2 C} + \frac{Q^{2}}{2 C} = \frac{Q^{2}}{C} U_{f} = \frac{4 Q^{2}}{18 C} + \frac{Q^{2}}{2 C} + \frac{2 Q^{2}}{9 C} = \frac{17 Q^{2}}{18 C} \Rightarrow H = \frac{Q^{2}}{18 C}$