Consider a parallel plate capacitor that has been charged from a battery and then disconnected from it. Now the separation between the plates is doubled, then the

Initially when the capacitor is charged to a charge $Q$

$Q=CV$ where capacitance is $C=\frac{A\epsilon }{d}$

when the separation is increased after the battery is disconnected, 

Charge stays the same. 

Capacitance is halved. Thus potential difference between the plates is doubled. 

Energy stored in capacitor is 

$E=\frac{1}{2}C{V}^2$

is also doubled . 

Therefore external work is needed to increase the separation.

Field between the plates does not alter.