Effect of dielectrics in capacitors

When a dielectric of relative permittivity epsilon_r is introduced to fill the space between an already-charged parallel-plate capacitor's plates, the outcome depends on whether the battery is left connected or disconnected beforehand. If the battery is disconnected first, the charge Q on the plates is trapped and cannot change; since the polarised dielectric reduces the net field between the plates (section 1.7.5), the voltage V = Ed also decreases, by the same factor epsilon_r, so the new capacitance C = Q/V = epsilon_r C0 increases; because the stored energy U = Q^2/(2C) has C in its denominator with Q fixed, the energy actually decreases when the dielectric is inserted this way -- physically, the dielectric slab is drawn inward by the field and would do positive work on its way in, so an external agent must instead do work to pull it back out, an amount of extra work exactly equal to the drop in stored energy, as demonstrated numerically in Example 1.21. If instead the battery remains connected throughout, the voltage V is held fixed by the battery, so instead of V decreasing, extra charge Q flows in from the battery to compensate the dielectric's field-reducing effect, again giving the new capacitance C = epsilon_r C0; but now, with V constant and Q (and hence C) increasing, the stored energy U = (1/2)CV^2 actually increases in this case, since the battery supplies the additional energy needed as the extra charge flows in. Table 1.2 summarises these two contrasting outcomes side by side. In either case the new capacitance is always C = epsilon_r C0, larger than the original vacuum or air-filled capacitance C0 by the dielectric's relative permittivity (dielectric co …