The separation between the plates of a parallel plate capacitor is made double while it remains connected to a cell. 

Let initial capacitance of capacitor is, $C_0=\frac{{ϵ}_{0}A}{d}\mathrm{\&}$ emf of battery is E. As d gets double & V remains constant so, $C=\frac{C_0}{2},V=V_0=Ed,Q=\frac{Q_0}{2},E=\frac{E_0}{2}$

Charge is given to battery so work is done on the battery. Work has been done by an external agent on the plates to move them apart against electric force between the plates.