P26_017.PDF

17. The charge initially on the charged capacitor is given by q = C 1 V 0 ,where C 1 = 100pF is the capacitance

and V 0 = 50V is the initial potential diﬀerence. After the battery is disconnected and the second

capacitor wired in parallel to the ﬁrst, the charge on the ﬁrst capacitor is q 1 = C 1 V ,where v =35Vis

the new potential diﬀerence. Since charge is conserved in the process, the charge on the second capacitor

is q 2 = q

V ). The potential diﬀerence across the second capacitor is also V ,sothe

capacitance is

V = V 0 −

C 1 = 50V

−

(100pF) = 3 pF .

q 1 ,where C 2 is the capacitance of the second capacitor. Substituting C 1 V 0 for q and C 1 V

for q 1 ,weobtain q 2 = C 1 ( V 0 −

−

C 2 = q 2

35V