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 (∊_r = 6) is inserted between the plates. The change in its potential energy after inserting the dielectric slab is _______× 10^–12 J.

The correct answer is 750.

Step 1: Understand the initial state of the capacitor
Initially, the capacitor is charged by a battery. The parameters given are:
- Capacitance, $C_0=\; 12.5 \text{pF}=\; 12.5\times {10}^{-12} \text{F}$
- Potential difference, $V_0=\; 12.0 \text{V}$

The charge stored on the capacitor initially is:
$Q=C_0{V}_0$
$Q=12.5\times {10}^{-12}\times 12.0=150\times {10}^{-12} \text{C}$

The initial electrostatic potential energy stored in the capacitor ($U_i$) is:
$U_i=\frac{1}{2}{C}_0{V}_0^2$
$U_i=\frac{1}{2}\times 12.5\times {10}^{-12}\times {\left(12.0\right)}^2$
$U_i=0.5\times 12.5\times {10}^{-12}\times 144$
$U_i=900\times {10}^{-12} \text{J}$

Step 2: Understand the final state of the capacitor
The battery is now disconnected, which means the charge $Q$ on the plates remains constant:
$Q=150\times {10}^{-12} \text{C}$

A dielectric slab of relative permittivity (${\epsilon }_r=6$) is inserted, so the new capacitance ($C_f$) becomes:
$C_f={\epsilon }_r{C}_0=6\times 12.5 \text{pF}=75.0 \text{pF}$

The final potential energy ($U_f$) is given by:
$U_f=\frac{Q^2}{2C_f}=\frac{U_i}{{\epsilon }_r}$
$U_f=\frac{900\times {10}^{-12}}{6}$

Step 3: Calculate the change in potential energy
The change in potential energy ($\Delta U$) is:
$\Delta U=U_i-U_f$
$\Delta U=900\times {10}^{-12}-150\times {10}^{-12}$
$\Delta U=750\times {10}^{-12} \text{J}$

Thus, the magnitude of the change in its potential energy after inserting the dielectric slab is 750 × 10^-12 J.