Question

A 13-

Answer 1

Initially, the capacitors are in parallel, and have the same voltage.

When disconnected from the power source, each plate maintains its amount of charge.

When connected with positive and negative plates in electrical contact, the charge then re-distributes itself. to be equal on both plates.

Initial charge on positive plate and negative plate of capacitor 1 (known size capacitor):
q1ip = C1*Vi
q1in = -C1*Vi

Initial charge on positive plate and negative plate of capacitor 2 (unknown size):
q2i = C2*Vi
q2in = -C2*Vi

Suppose that C1 has its positive plate at the top, when re-connected. Therefore C2 has its negative plate at the top. Thus, the net charge on the top plates:
qtop = q1ip + q2in = Vi*(C1 - C2)

Do the same for the connected bottom plates:
qbottom = q2ip + q1in = Vi*(C2 - C1)

upon redistributing this charge,
q1ft = qtop/2 = Vi*(C1 - C2)/2
q2ft = qtop/2 = Vi*(C1 - C2)/2
q1fb = qbottom/2 = Vi*(C2 - C1)/2
q2fb = qbottom/2 = Vi*(C2 - C1)/2

using q1ft and q1fb, find the final voltage across its terminals. The problem is, that the capacitor isn't net-neutral. In order to use Vf = qf/C1, qf is the charge on one plate, relative to the midpoint chagre q0

midpoint charge:
q0 = (q1ft + q1fb)/2

q0 = (Vi*(C1 - C2)/2 + Vi*(C2 - C1)/2)/2

Simplify, and it turns out that q0 does equal zero.

Thus, qf = | qft |

So,
Vf = Vi*(C1 - C2)/(2*C1)

Solve for C2, the size of our unknown capacitor:
C2 = (Vi - 2*Vf)*C1/Vi

Data:
C1 := 13