Dielectrics

When a capacitor is maintained at constant voltage (V) by a battery, the energy stored is (U = \frac{1}{2}CV^2). The critical insight is that energy is proportional to capacitance at constant voltage. With a dielectric slab of constant (K = 4) filling half the gap (thickness (d/2)), the system can be treated as two capacitors in series: one with dielectric of thickness (d/2) and one vacuum gap of (d/2). The equivalent capacitance with dielectric is higher than the vacuum capacitance (C_0 = \varepsilon_0 A/d). Specifically, (\frac{1}{C_{with}} = \frac{d/2}{K\varepsilon_0 A} + \frac{d/2}{\varepsilon_0 A} = \frac{d}{2\varepsilon_0 A}\left(\frac{1}{4} + 1\right) = \frac{5d}{8\varepsilon_0 A}), giving (C_{with} = \frac{8\varepsilon_0 A}{5d} = 1.6 C_0). When the dielectric is removed, (C = C_0), which is smaller. At constant (V), energy (U = \frac{1}{2}CV^2) decreases. The energy released goes into the work done in pulling out the dielectric (the slab is pulled into a region of lower energy). Option 'energy remains same' is incorrect because energy depends on capacitance, which changes. Option 'energy increases' is incorrect because removing the dielectric reduces capacitance. Option 'doubles' is incorrect because no physical basis exists for a factor-of-two change here. Plausibility check: pulling out a dielectric reduces (C) and therefore reduces stored energy at constant (V).