Two 2.0 cm * 2.0 cm metal electrodes are spaced 1.0 mm apart and connected by wires to the terminals of a 9.0 V battery. a. What are the charge on each electrode and the potential difference between them? The wires are disconnected, and insulated handles are used to pull the plates apart to a new spacing of 2.0 mm. b. What are the charge on each electrode and the potential difference between them?

Answer: a) 31.86 * 10^-12 C=31.86 pC; b) 18 V

Explanation: In order to explain this problem we have to consider the expression a parallel plates capacitor, which is given by:

C=Q/V where C is equal to C=εo*A/d where A and D are the area and the separation between the plates.

also we have

Q=C*V=ε (o*A/d) * V = (8.85*10^-12*0.02*0.02/1*10^-3) * 9=31.86*10^-12 C=31.86pC

Then if the plates apart to a new spacing of 2.0 mm the new capacitance is equal

Cnew=εo*A/2*d so Cnew = Cinitial/2

then Cnew = Q/Vnew (Q is constant after disconnection to the battery)

Finally Vnew = Q / (Cinitial/2) = 2 * (Q/Cinitial) = 2*Vinitial = 2*9=18V