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Common Potential | Common Potential Formula | Common Potential Of Capacitor

Common Potential | Common Potential Formula | Common Potential Of Capacitor :- When two capacitors that have been charged to different electric potentials are connected by a conducting wire, electric charge flows from the capacitor at higher potential to the capacitor at lower potential. This flow of charge continues until the electric potentials of both capacitors become equal. This equal electric potential of the two capacitors is called the common potential. During this process, charge is conserved; that is, there is no loss of charge.

Let C₁ & C₂ be the capacitances of two capacitors whose electric potentials are V₁ & V₂ respectively.

Total charge before combination :

$\displaystyle {{Q}_{net}}={{Q}_{1}}+{{Q}_{2}}={{C}_{1}}{{V}_{1}}+{{C}_{2}}{{V}_{2}}$ …..(1)

If the common potential after combination is V, then the total charge after combination :

$\displaystyle {{Q}_{net}}=Q_{1}^{'}+Q_{2}^{'}={{C}_{1}}V+{{C}_{2}}V$ …..(2)

Since charge is conserved, from equations (1) and (2),

$\displaystyle {{Q}_{1}}+{{Q}_{2}}=Q_{1}^{'}+Q_{2}^{'}$

$\displaystyle \Rightarrow {{C}_{1}}V+{{C}_{2}}V={{C}_{1}}{{V}_{1}}+{{C}_{2}}{{V}_{2}}$

$\displaystyle V=\frac{{{C}_{1}}{{V}_{1}}+{{C}_{2}}{{V}_{2}}}{{{C}_{1}}+{{C}_{2}}}$ …..(3)

That is,

$\displaystyle Common\text{ }Potential(V)\text{ = }\frac{Total\text{ }Charge}{Total\text{ }Capacity}$

Equation (3) is the required equation for the common potential.

From equation (3),

$\displaystyle {{C}_{1}}V+{{C}_{2}}V={{C}_{1}}{{V}_{1}}+{{C}_{2}}{{V}_{2}}$

$\displaystyle {{C}_{1}}{{V}_{1}}-{{C}_{1}}V={{C}_{2}}V-{{C}_{2}}{{V}_{2}}$

⇒ Loss of charge in one capacitor = gain of charge in the other capacitor

The above statement is not applicable to electric potential; that is, the decrease in the potential of one capacitor is not equal to the increase in the potential of another capacitor, because the capacitances of the two capacitors are different.

Note :-

The derivation of equation (3) has been carried out assuming that the two capacitors are connected in parallel in such a way that the positive plates of both capacitors are connected to one point and the negative plates are connected to another point. However, if the two capacitors are connected in parallel in such a manner that the positive plate of the first capacitor is connected to the negative plate of the second capacitor, and the negative plate of the first capacitor is connected to the positive plate of the second capacitor, then the charge on one capacitor will tend to neutralize the charge on the other capacitor, because the nature of the charges on them is opposite. This type of connection is shown in the figure below :-

In the above diagram, the common potential of the combination is :-

$\displaystyle Common\text{ }Potential(V)\text{ = }\frac{Total\text{ }Charge}{Total\text{ }Capacity}=\frac{{{Q}_{1}}-{{Q}_{2}}}{{{C}_{1}}+{{C}_{2}}}$

$\displaystyle \Rightarrow V=\frac{{{C}_{1}}{{V}_{1}}-{{C}_{2}}{{V}_{2}}}{{{C}_{1}}+{{C}_{2}}}$ …..(4)

That is, in this situation, the total charge will be equal to the algebraic sum of the charges on both capacitors.

Note that in equation (4), the total capacitance will be (C₁ + C₂) because this is a parallel combination of capacitors.

Example 1.

In the circuit shown in the figure, C₁ = 1 μF and C₂ = 2 μF. Capacitor C₁ is charged to 100 V and capacitor C₂ is charged to 20 V. After charging, they are connected as shown in the figure. When switches S₁ , S₂and S₃ are all closed, then –

(1) No charge will flow through S₂

(2) A charge of 80 μC will flow through S₁

(3) A charge of 40 μC will flow through S₂

(4) A charge of 60 μC will flow through S₃

Solution :

The values of charge on the capacitors before closing switches S₁ , S₂and S_{3 ,}

$\displaystyle {{Q}_{1}}={{C}_{1}}{{V}_{1}}=1\mu \times 100=100\mu C$

$\displaystyle {{Q}_{2}}={{C}_{2}}{{V}_{2}}=2\mu \times 20=40\mu C$

After closing switches S₁ , S₂and S₃, both capacitors will be in parallel combination, and the negative plate of capacitor C₁ will be connected to the positive plate of capacitor C₂. Hence, they will have a common potential,

$\displaystyle V=\frac{{{Q}_{1}}-{{Q}_{2}}}{{{C}_{1}}+{{C}_{2}}}=\frac{100\mu -40\mu }{1\mu +2\mu }=20Volt$

New values of charge on the capacitors,

$\displaystyle Q_{\text{1}}^{'}={{C}_{1}}V=1\mu \times 20=20\mu C$

$\displaystyle Q_{\text{2}}^{'}={{C}_{2}}V=2\mu \times 20=40\mu C$

Change in the polarities of charges :

Capacitors

C₁

C₂

Before closing the switches

Left Plate +100 μC

Right Plate -100 μC

Left Plate +40 μC

Right Plate -40 μC

After closing the switches

Left Plate +20 μC

Right Plate -20 μC

Left Plate -40 μC

Right Plate +40 μC

👉 C₂ reverses polarity.

Statement (1) :

No charge will flow through S₂. ❌

This statement is incorrect, because as soon as all three switches are closed, a redistribution of charges will occur between capacitors C₁ and C₂.

Statement (2) :

The charge flowing through switch S₁,

$\displaystyle Q={{Q}_{1}}-Q_{\text{1}}^{'}=100\mu C-20\mu C=80\mu C$

Therefore, statement (2) is correct.

Statement (3) :

Considering the negative plate of C₁, the charge flowing through switch S₂,

$\displaystyle Q={{Q}_{1}}-Q_{\text{1}}^{'}=-100\mu C-(-20\mu C)=-80\mu C$

Therefore, statement (3) is incorrect.

Statement (4) :

Considering the negative plate of C₂, the charge flowing through switch S₃,

$\displaystyle Q={{Q}_{2}}-Q_{2}^{'}=-40\mu C-(+40\mu C)=-80\mu C$

Therefore, statement (3) is incorrect.

Note :-

In the above question, when switches S₁ , S₂and S₃ are closed, the two capacitors are considered to be in parallel because :

The negative plate of capacitor $C_{1}$ is connected to the positive plate of capacitor $C_{2}$ through switch $S_{2}$ .
Similarly, the positive plate of capacitor $C_{1}$ is connected to the negative plate of capacitor $C_{2}$ through switch $S_{1}$ , and also to ground through switch $S_{3}$ .
This means that both plates of the two capacitors are connected between the same two points (nodes). According to the rule of parallel combination, the potential difference across all components is the same.

Common Potential | Common Potential Formula | Common Potential Of Capacitor

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