Given data

• Initial capacitance: $C = 12.5\,\text{pF} = 12.5 \times 10^{-12}\,\text{F}$
• Initial potential difference: $V = 12.0\,\text{V}$
• Dielectric constant: $k = \epsilon_r = 6$
• Battery is disconnected, so charge remains constant.

Initial energy stored in the capacitor

Since initially the capacitor is charged to voltage $V$, $U_i = \frac{1}{2}CV^2$ Substituting, $U_i = \frac{1}{2}(12.5 \times 10^{-12})(12)^2$ $U_i = \frac{1}{2}(12.5 \times 10^{-12})(144)$ $U_i = 6.25 \times 144 \times 10^{-12}$ $U_i = 900 \times 10^{-12}\,\text{J}$

Effect of inserting dielectric after disconnecting battery

When dielectric is inserted fully, $C' = kC = 6C$ Since battery is disconnected, charge $Q$ remains constant. For constant charge, energy is $U = \frac{Q^2}{2C}$ Therefore, when capacitance becomes $6$ times, energy becomes $\frac{1}{6}$ times: $U_f = \frac{U_i}{6} = \frac{900 \times 10^{-12}}{6}$ $U_f = 150 \times 10^{-12}\,\text{J}$

Change in potential energy

$\Delta U = U_f - U_i$ $\Delta U = 150 \times 10^{-12} - 900 \times 10^{-12}$ $\Delta U = -750 \times 10^{-12}\,\text{J}$

So the energy decreases by $750 \times 10^{-12}\,\text{J}$

Integer to be filled

Since the blank is in $\underline{\hspace{2cm}} \times 10^{-12}\,\text{J}$ the required integer is: $750$

Comparison with stored answer

Stored correct answer: $750$

This matches our derived answer.