A capacitor has a capacitance C and is charged using a battery. After disconnecting the battery the two plates are separated until they are twice as far apart as originally. What happens to the voltage across the capacitor? I think answer should be voltage increases . As battery has been disconnected q will remain same . Distance between the plates is increasing , capacitance should decrease. ##Q##= CΔV As Q is constant and C is decreasing , V should increase. But according to http://www.bama.ua.edu/~tmewes/PH-106/2008/clicker%20questions/CH26%20Clicker.pdf [Broken] Go to page 10 Voltage goes down . It is wrong . And formula is also wrong. Is that correct? I have gone through all 14 pages of this pdf and found them right except this one , but I want to confirm are all other 13 pages correct?