The plates of a parallel-plate capacitor, each of area 0.10 m^2 and separated by 2.0 mm in air, remain connected to a 100 V battery. A dielectric…

Initially $C_0=\varepsilon_0A/d=4.43\times10^{-10}\ \mathrm{F}$. With the slab the series combination of the air gap and the dielectric gives $C=\dfrac{\varepsilon_0A}{(d-t)+t/K}=\dfrac{\varepsilon_0A}{1.0+0.25\ \mathrm{mm}}=\dfrac{\varepsilon_0A}{1.25\ \mathrm{mm}}=7.08\times10^{-10}\ \mathrm{F}$, i.e. $1.6\,C_0$. The potential $V$ is fixed by the battery, so the battery supplies charge $\Delta Q=\Delta C\,V$ and does work $W_b=\Delta C\,V^2$. The stored energy rises by only $\Delta U=\tfrac12\Delta C\,V^2$. Energy conservation, $W_b=\Delta U+W_{field}$, then shows the field does work $W_{field}=W_b-\Delta U=\tfrac12\Delta C\,V^2$ on the slab (the slab is drawn in). Numerically $\Delta C=2.66\times10^{-10}\ \mathrm{F}$, so $W=\tfrac12(2.66\times10^{-10})(10^4)=1.33\ \mathrm{\mu J}$. The slab is pulled in because the capacitor at fixed potential lowers its energy-plus-battery potential energy by increasing capacitance, so the field does positive work.