# I have trouble understanding why two different capacitors should have the same amount of charge on each side of a circuit

There is a rhombus with four different capacitors. There's a voltage between the point $$a$$ and the point $$b$$, but there is no voltage between the point $$c$$ and the point $$d$$. Apparently, assuming that there is no excess of charges in the circuit, the capacitors on the left side have the same charge, and the same holds for the right side. I have trouble understanding why the charges should be equal on each side. It is true, only if the capacitors have capacitance values that are symmetric ( $$C_1 = C_2$$ and $$C_3 = C_4$$ ). Else is not true. You can do the computation using the formula of a series of capacitors $$1/C_{tot} = 1/C_1 + 1/C_2 +...+ 1/C_n$$ (here $$n = 2$$). You can compute the charge by the well known relationship $$Q = C \times V$$.

You should have trouble with it. It isn't true. For example, suppose $$C_3 = 2C_1$$ $$C_2 = 10C_1$$ $$C_4 = 20 C_1$$

You will find that

$$V_{ac} = V_{ad}$$

But

$$Q_{right} = 10 Q_{left}$$

In the above circuit $$C_1$$ and $$C_3$$ are in series and also parallel to series capacitors: $$C_3$$ and $$C_4$$. There is a voltage potential between $$a$$ and $$b$$, but none between $$c$$ and $$d$$. Capacitors in series have the same charge, and the charge on the left and right sides is the same.

For the series capacitors:

$$left \ hand \ side \ \ \frac{1}{C_{s1}} = \frac{1}{C_1} + \frac{1}{C_3} = \frac{C_3 + C_1}{C_1 C_3}$$

$$right \ hand \ side \ \ \frac{1}{C_{s2}} = \frac{1}{C_2} + \frac{1}{C_4} = \frac{C_4 + C_2}{C_2 C_4}$$

We add the capacitance on left and right sides that are in parallel. I.e. capacitance is added, because the charges are added but the voltage is the same:

$$C_p = C_{s1} + C_{s2}$$

Therefore, for the whole system:

$$Q_{tot} = C \times V = (C_{s1} + C_{s2}) \times V$$

But the charges on the left and right hand sides are the same, therefore, $$C_{s1}$$ and $$C_{s2}$$ must be the same. So, $$C_1 = C_2$$ and $$C_3 = C_4$$, or even, $$C_1 = C_2 = C_3 = C_4$$.

I.e. $$Q_{rightside} = C_{s1} V = Q_{leftside} = C_{s2} V$$