An empty parallel plate capacitor is connected between the terminals of a 8.85-V battery and charges up. The capacitor is then disconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor

The new voltage is 17.7 V.  

Explanation:

Voltage, V = 8.85 V

The spacing is doubled.

When it is disconnected, the charge remains same,

q = C V (1)

where, C is the capacitance, V is the voltage.  

The capacitance is inversely proportional to the distance between the two plates.

So, when the spacing is doubled, the capacitance is halved.

Let the new voltage is V'.

C V = C' V'

C x 8.85 = C/2 x V'

V' = 17.7 V  

i think its