The capacitors in the figure below are initially uncharged and are connected, as

Question

The capacitors in the figure below are initially uncharged and are connected, as in the diagram, with switch S open. The applied potential difference is Vab = +280 V.

(a) What is the potential difference Vcd?
V

(b) What is the potential difference across each capacitor after switch S is closed?
3.00 µF capacitor between a and d
V
6.00 µF capacitor between d and b
V
3.00 µF capacitor between b and c
V
6.00 µF capacitor between c and a
V

(c) How much charge flowed through the switch when it was closed?
C

Explanation / Answer

Lets' denote the capacitors. C1 is between ad, C2 is between db, C3 is between ac, C4 is between cb. Their

charges are Q1,Q2,Q3,Q4, potential differences are U1,U2,U3,U4.

1. Calculation of Vcd.

As you know, charges of two connected in series capacitores are equal.

Q1 = Q2.
C1*U1 = C2*U2
U1+U2 = V
U2 = V-U1
C1*U1 = C2*(V-U1)
U1 = C2*V/(C1+C2) = 6*10^-6 * 280/ (3*10^-6 + 6*10^-6) = 186.6 v.
So,

Vad = U1 = 186.6 v.

U3 = C4*V/(C3+C4) = 3*10^-6 * 280/ (6*10^-6 + 3*10^-6) = 93.3 v.

Vac = U3 = 93.3 v.

Vcd = Vca - Vda = -93.3 + 186.6 = 93.3 v.

2.

What is the potential difference across each capacitor after the switch S is closed?
After the switch is closed there will be two capacitors connected in series. Their capacities are equal = 3 + 6 = 9 microF.

So, U1 = U3 = V/2 = 140 v.

U2 = U4 = V/2 = 140 v.

3.

The most subtle point.

The charge flowed through the switch when it was closed will be equal =

(Q1+Q2)-(Q5+Q6), where Q5 - charge on the C1 after switch is closed, Q6 - change of C2.

Q1 + Q2 = 0. Their sum of charge were zero.

Q5 = C1*V/2; Q6 = - C2 * V/2

Pay attention, Q5 and Q6 are in opposite charges.

So

, the answer = (C2-C1)*V/2 = 3*10^-6 * 280 /2 = 4.2 * 10 ^(-4) C

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