Answer
(a) $U_{parallel} = 4U_{series}$
(b) $Q_{parallel} = 2Q_{series}$
(c) $E_{parallel} = 2E_{series}$
Work Step by Step
(a) In parallel both capacitors have the same potential, therefore their total energy stored is
$$U_{parallel} = CV^2$$
While in series, each capacitor has the half value of the total potential $V/2 $, therefore, their stored energy in series is given by
$$U_{series} = \dfrac{1}{4} CV^2$$
Let us take the ratio between both energies
$$\dfrac{U_{parallel}}{U_{series}} = \dfrac{CV^2}{\dfrac{1}{4} CV^2} = 4 $$
$$\therefore \boxed{U_{parallel} = 4U_{series}}$$
(b) In parallel, each capacitor has its own charge and the total charge for their combination is
$$Q_{parallel} = 2Q$$
While in series, both capacitors have the same charge even for their combination and is given by
$$Q_{series} = Q$$
Let us take the ratio between both charges
$$\dfrac{Q_{parallel}}{Q_{series}} = \dfrac{2Q}{Q} = 2 $$
$$\therefore \boxed{Q_{parallel} = 2Q_{series}}$$
(c) The electric field depends on the potential of the capacitor and is given in the form
$$E = \dfrac{V}{d}$$
In parallel, both have the same potential, therefore the same electric field for their combination
$$E_{parallel} = \dfrac{V}{d}$$
While in series, each capacitor has half of the potential, therefore, the electric field for their combination in series
$$E_{series} = \dfrac{V}{2} d$$
Let us take the ratio between both charges
$$\dfrac{E_{parallel}}{E_{series}} = \dfrac{ \dfrac{V}{d}}{ \dfrac{V}{2} d} = 2 $$
$$\therefore \boxed{E_{parallel} = 2E_{series}}$$
