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 plates are connected by a resistor of resistance $R$ .

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

Heat loss will be $\frac{Q^{2}}{6 C}$

No charge flows through resistor.

A total charge of $\frac{Q}{3}$ flows through the resistor.

answer is A, C, D.

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Detailed Solution

Initially, $V_{A C} \neq 0 \Rightarrow$ charge flows through wire

$\frac{Q - q}{A \in_{0}} d = \frac{q}{A \in_{0}} \times 2 d q = \frac{Q}{3} U_{i} = \frac{Q^{2}}{2 C} + \frac{Q^{2}}{2 C} = \frac{Q^{2}}{C} U_{f} = \frac{4 Q^{2}}{18 C} + \frac{Q^{2}}{2 C} + \frac{Q^{2}}{9 C} = \frac{5 Q^{2}}{6 C} \Rightarrow H = \frac{Q^{2}}{18 C}$