Energy stored in Capacitor

Most of us have seen dramatisations of medical help using a defibrillator to pass an electrical current through a patient’s heart to get it to beat normally. Often realistic in detail, the person applying the shock directs another person to “make it 400 Joules this time.” The energy delivered by the defibrillator is stored in a capacitor and can be adjusted to fit the situation.

SI units of Joules (J) are often used. Capacitors are used  in microelectronics to supply energy when batteries are charged. Capacitors are also used to supply energy for flash lights on cameras.

Energy stored in a capacitor formula

Energy stored in a capacitor is electrical potential energy, and it’s therefore related to the charge Q and voltage V on the capacitor. We must be careful when applying the equation for electrical potential energy P.E=qV to a capacitor.

Remember that P.E is the potential energy of a charge q going through a voltage V . But the capacitor starts with zero voltage and gradually comes up to its full voltage as it’s charged. The final charge placed on a capacitor endured V=V , since the capacitor now has its full voltage V on it. The average voltage on the capacitor during the charging process is  V/2 ,

and so, the average voltage endured by the full charge q is V/2 .

Therefore, the energy stored in a capacitor E_{cap} is,

E_cap=QV/2,

where Q is the charge on a capacitor with a voltage applied,V .

(Keep in mind that the energy isn’t QV , but QV/2 )

Charge and voltage are related to the capacitance C of a capacitor by Q=CV , and so the expression for E_cap can be algebraically manipulated into three identical expressions

.

E_cap=QV/2=CV²/2=Q²/2C

Energy stored in a capacitor derivation

A capacitor, as we know, is a setup of conductors with charge Q and -Q . To decide the energy stored in this configuration, suppose there are two uncharged conductors 1 and 2. Let us consider a system of moving charge from conductor 2 to conductor 1 bit by bit, in order that on the end, conductor 1 receives charge Q . By charge conservation, conductor 2 has charge -Q at the end.

In shifting positive charge from conductor 2 to conductor 1, work has to be done externally, considering that at any level conductor 1 is at a higher potential than conductor 2. To calculate the total amount of work done, we first calculate the work in a small step involving movement of a very small amount of charge. Consider the intermediate situation while the conductors 1 and 2 have charges Q’ and -Q’ respectively.

At this stage, the potential difference among conductors 1 to 2 is Q’/C , wherein C is the capacitance of the system. Now consider that a small charge ∂Q’ is transferred from conductor 2 to 1. Work done in this step (∂W), due to which charge on conductor 1 growing to Q’+Q’ , is given by<br>

W=V’∂Q’=(Q’/C)∂Q’……….(1)

Since Q’ is very small, it is made as small as we like, Eq. (1) may be written as

∂W=1/2C [ (Q’+∂Q’)²-Q’²]……….(2)

Equations (1) and (2) are similar due to the fact the time period of second order in Q’,
i.e.,Q’²/2C , is negligible, as Q’ is arbitrarily small. The overall work done (W) is the sum of the small works

(W) over the very large range of steps concerned in constructing the charge Q’ from 0 to Q .

W=∑∂W

=∑(1/2C)[(Q’+Q’)²-Q’²]………..(3)

=(1/2C)[(∂Q’²-0)+(2∂Q’)²-∂Q’²]+(3∂Q’)²+………(∂Q²-(Q-∂Q’)²)……..(4)

=(1/2C)[Q²-0]=Q²/2C…….(5)

We can get the same result from Eq. (1) using integration,br>

W=₀∫^QQ’/C Q’=₀∫^Q1/C Q’²/2=Q²/2C

We can write the Eq. (5) in multiple ways,

W=Q²/2C=1/2 CV²=1/2 QV……..(6)

As we know that electrostatic force is conservative, the work done is stored in the form of potential energy. Also, the work done is independent of the manner in which the charge setup of the capacitor is built. When the capacitor is discharged, the stored energy is released. We can easily see that the potential energy of the capacitor is stored between the two plates of the capacitor setup.

Now let us consider, a parallel plate capacitor having area A each for both the plates and the distance between the plates is d.,br>

Energy stored in the capacitor is

=(1/2)Q²/C=(Aσ)²/2d/∈₀A……….(7)

Here, σ is the surface charge density and E is the electric field between the two plates of the capacitor.

E=σ/∈₀……..(8)

From Eq. (7) and (8), we can say that

Energy stored in the capacitor is

U=(1/2)∈₀E²Ad……..(9)

Conclusion

Capacitors are used to deliver power to a lot of devices, such as defibrillators, microelectronics along with calculators, and flash lamps. The energy contained in a capacitor is the work required to charge the capacitor, starting without a charge on its plates. The electricity is saved within the electric field withinside the area in between the capacitor plates. It relies upon the quantity of electrical charge at the plates and at the potential difference between the plates. The energy contained in a capacitor system is the sum of the energies contained on each capacitor withinside the system. It may be computed because the energy contained withinside the equal capacitor of the system.