A parallel-plate air capacitor is made by using two plates 10 cm square, spaced 4.7 mm apart. It is connected to a 15 V battery. If the plates remain connected to the battery and the plates are pulled apart to a separation of 9.9 mm.
What is the new capacitance? What is the new charge on each plate?
C = Q/V = (permittivity of freespace)(A/D)
The Attempt at a Solution
I attempted to solve for the new capacitance by simply substituting in the new separation value and it isn't working. Why?