A parallel plate capacitor of capacitance $$12.5$$ pF is charged by a battery connected between its plates to potential difference of $$12.0$$ V. The battery is now disconnected and a dielectric slab ($$\epsilon_r = 6$$) is inserted between the plates. The change in its potential energy after inserting the dielectric slab is _____ $$ 10^{-12}$$ J.

Initial capacitance of the parallel-plate capacitor
$$C = 12.5\ \text{pF} = 12.5 \times 10^{-12}\ \text{F}$$

The plates are first charged by a battery to a potential difference
$$V = 12.0\ \text{V}$$

Because the battery is disconnected before inserting the dielectric, the charge $$Q$$ on the plates remains constant.

Step 1 : Calculate the initial charge
Using $$Q = C V$$,

$$Q = \left(12.5 \times 10^{-12}\right)\!\text{F}\;\left(12\ \text{V}\right) = 150 \times 10^{-12}\ \text{C}$$

Step 2 : Initial electrostatic energy
For a charged capacitor, the energy stored is
$$U = \frac{1}{2} C V^{2}$$

$$U_i = \tfrac{1}{2}\left(12.5 \times 10^{-12}\right)\!\text{F}\;\left(12\ \text{V}\right)^{2}$$
$$U_i = \tfrac{1}{2}\left(12.5 \times 10^{-12}\right)\!\left(144\right)$$
$$U_i = 9.0 \times 10^{-10}\ \text{J}$$

Step 3 : Capacitance after inserting the dielectric
Relative permittivity of the slab $${\epsilon_r} = 6$$. New capacitance
$$C' = \epsilon_r\,C = 6 \times 12.5\ \text{pF} = 75\ \text{pF}$$
$$C' = 75 \times 10^{-12}\ \text{F}$$

Step 4 : Final electrostatic energy
Charge stays the same, so use $$U = \dfrac{Q^{2}}{2C}$$:

$$U_f = \frac{\left(150 \times 10^{-12}\ \text{C}\right)^{2}}{2 \left(75 \times 10^{-12}\ \text{F}\right)}$$
$$U_f = \frac{2.25 \times 10^{-20}}{150 \times 10^{-12}}$$
$$U_f = 1.5 \times 10^{-10}\ \text{J}$$

Step 5 : Change in potential energy
$$\Delta U = U_f - U_i = 1.5 \times 10^{-10}\ \text{J} \;-\; 9.0 \times 10^{-10}\ \text{J}$$
$$\Delta U = -7.5 \times 10^{-10}\ \text{J}$$

The negative sign shows that the energy decreases. Expressing the magnitude in units of $$10^{-12}\ \text{J}$$:

$$|\Delta U| = 7.5 \times 10^{-10}\ \text{J} = 750 \times 10^{-12}\ \text{J}$$

Hence, the change in potential energy is $$\mathbf{750} \times 10^{-12}\ \text{J}$$ (decrease).