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Since the battery is disconnected, charge $Q$ is constant:
$Q = C_0 V_0 = 12 \times 10^{-12} \times 10 = 120\ \text{pC}$
Initial energy: $U_i = \frac{Q^2}{2C_0} = \frac{(120\times10^{-12})^2}{2\times12\times10^{-12}} = 600\ \text{pJ}$
New capacitance: $C = KC_0 = 6.5 \times 12 = 78\ \text{pF}$
Final energy: $U_f = \frac{Q^2}{2C} = \frac{(120)^2 \times 10^{-24}}{2\times78\times10^{-12}} \approx 92.3\ \text{pJ}$
Work done by capacitor on slab $= U_i - U_f = 600 - 92.3 \approx 508\ \text{pJ}$
Given that $st = \sqrt{10}$, compare:
Quantity A: $s^2$
Quantity B: $\dfrac{10}{t^2}$
From $st = \sqrt{10}$, squaring both sides: $(st)^2 = 10$, so $s^2 t^2 = 10$.
Therefore $s^2 = \dfrac{10}{t^2}$.
Quantity A $= s^2 = \dfrac{10}{t^2} =$ Quantity B.
The two quantities are equal.
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