# Electrostatic Influence and Series-Connected Capacitors

• I
• reterty
In summary, the assumption that the absolute values of the charges on the plates in a series circuit are equal is a basic assumption of circuit theory.
reterty
As you know, "in the classics" the charges of the capacitor plates are equal in absolute value and opposite in sign. However, let us consider a series connection of capacitors. In this case, a charge of the opposite sign is induced on every second plate of each capacitor due to electrostatic induction. But how can we rigorously prove the strict equality of the absolute values of these charges to the charges of the first plates?

Do you mean configuration as attached?

No, we consider the case of capacitors in series connected with battery. The first plate of the first capacitor has charge q, whereas the second plate of the last capacitor has charge -q. Intermediate plates are charged through influence and it is necessary to prove that their absolute values are also equal to q

This is one of the basic assumptions of circuit theory, that the current in a series branch is everywhere the same:

$$I(t)=\frac{dQ_1}{dt}=\frac{dQ_2}{dt}=...=\frac{dQ_n}{dt}$$ from which equality by integration you can deduce that all the charges differ by a constant, this constant being the charge each plate has at the time t=0. If all the capacitors are uncharged at t=0, then their charges are equal at any time t.

The above assumption holds only if the wavelength of the current in the circuit is big in comparison with the physical length of the series branch.

Klystron, DaveE and vanhees71
reterty said:
Intermediate plates are charged through influence and it is necessary to prove that their absolute values are also equal to q
I think it is zero because no charge can enter into or go out from the intermediate part ,i.e. the second plate charged -q, the third plate charged q and the wire between non charged.

Here's another proof:
Say you have 4 unit-area plates stacked close together vertically. 4 plates, numbered 1 to 4, left to right. 3 gaps.

Number the surfaces 1 thru 8 left to right.

Put charge Q on plate 1 and -Q on plate 4. Assume the center two plates have zero charge.

Fact: the charges on each facing pair of surfaces (e.g. #2 & #3 or #4 & #5) must be equal and opposite, otherwise the divergence in the E field is non-zero which violates Maxwell.

Thus,
## \sigma_2 = \sigma_4 = \sigma_6 = -\sigma_3 =- \sigma_5 = -\sigma_7 ##.

But also, ## \sigma_1 + \sigma_2 = Q ## and
## \sigma_7 + \sigma_8 = -\sigma_2 + \sigma_8 = -Q ##.

Now put a + test charge inside plate 1. This charge will see a rightward force due to ## \sigma_1 ## and a leftward force force due to ## \sigma_8 ##. Note that ## \sigma_2 ## thru ## \sigma_7 ## forces cancel each other out.

That leaves us with
## \sigma_1 + \sigma_2 = Q ##
## \sigma_8 - \sigma_2 = -Q ##
and since the net force on the test charge must = 0,
## \sigma_1 - \sigma_8 = 0 ##.

OK look at these equations. They have no solution!
So the assumed charge distribution is impossible.

Matter of fact, no matter what charge you put on each plate, no matter how many plates, the outside 2 surfaces must have the same charge including sign!

Last edited:

## 1. What is electrostatic influence?

Electrostatic influence is the phenomenon in which an electric field from one charged object affects the charge distribution on another object without direct contact.

## 2. How do series-connected capacitors work?

In a series-connected capacitor circuit, the capacitors are connected one after the other, with the positive end of one connected to the negative end of the next. This creates a single path for the flow of charge, and the capacitors share the same voltage. The total capacitance of the circuit is equal to the sum of the individual capacitances.

## 3. What is the effect of adding capacitors in series?

Adding capacitors in series increases the total capacitance of the circuit. This is because the capacitors share the same voltage, and the charge on each capacitor is the same. Therefore, the total charge stored in the circuit is divided among the capacitors, resulting in an increase in the overall capacitance.

## 4. How does electrostatic influence affect series-connected capacitors?

Electrostatic influence can cause the charge distribution on one capacitor to affect the charge distribution on the other capacitors in the circuit. This can result in a change in the overall capacitance of the circuit. Additionally, if one capacitor in the series becomes fully charged, it can prevent the other capacitors from reaching their full charge, reducing the overall capacitance of the circuit.

## 5. What are some practical applications of series-connected capacitors?

Series-connected capacitors are commonly used in electronic devices to store and regulate electrical energy. They are also used in power factor correction circuits, where they help to improve the efficiency of electrical systems by reducing reactive power and improving the power factor. Additionally, series-connected capacitors are used in voltage multiplier circuits to increase the output voltage of power supplies.

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