16.7 Energy Stored in a Capacitor

16.7 Energy Stored in a Capacitor

A capacitor is a device to store charge. Alternatively, it is possible to think of a capacitor as a device for storing electrical energy. After all the charge on the plates possesses electrical potential energy which arises because work is to be done to deposit charge on the plates.

A good evidence of the energy stored in a charged capacitor is that of its discharging. If we connects the plates of charged capacitor by means of a thick wire, the charge flows from one plate to the other to neutralize the opposite charge on it. As a result, the wire becomes hot. Since heat is a form of energy, so the capacitor possessed energy when it was charged, and it was this energy which appeared as heat.

Let us determine the electric potential energy stored in a charged capacitor graphically. We have already studied that the charge $Q$ on the plates increases with the increase in potential $V$ between the plates of the capacitor, i.e., $Q \propto V$

Therefore if we draw a $V$ vs $Q$ graph, it will be a straight line passing through the origin as shown in Fig. 16.10. In fact, with each small increment of charge deposited during the charging process, the potential difference between the plates increases, and a larger amount of work is needed to bring up next increment of charge.

(Fig. 16.15)

This problem can be resolved by dividing the area under the graph into very small increments of charge $\Delta Q$ each as shown in Fig. 16.10. The work required to move a small charge $\Delta Q$ through a potential difference $V$ across the plates of capacitor is

This work done in depositing charge $\Delta Q$ on the plates is stored in the capacitor as increase in its electric potential energy. But $(\Delta Q)(V_{o})$ is equal to the area of the rectangle shown red in the figure.

Therefore total work done to charge the capacitor to a final value $Q$ is equal to the sum of areas of all such rectangles under $V$ vs $Q$ graph. Fig. 16.10 shows that this sum is equal to the total area under the graph, i.e., area of the right angled triangle OAB.

Area of triangle OAB $= \tfrac{1}{2}$ (base) (height) $= \tfrac{1}{2}$ (OA) (AB)

If OA at this stage is equal to total charge $Q$ and AB is equal to the final potential difference $V$ then,

Area of the triangle OAB $= \tfrac{1}{2}QV$

Therefore, energy stored in a charged capacitor is

If we put the value of $Q = CV$ in eq. (16.39), we get

Energy stored $= W = \tfrac{1}{2}(CV)V$

or

Therefore,

$\mathrm{P.E} = \tfrac{1}{2}CV^{2}$

It is also possible to regard the energy as being stored in electric field between the plates, rather than the potential energy of the charges on the plates. Such a view point is useful when electric field strength between the plates instead of charges on the plates causing field is to be considered. This relation can be obtained by substituting $V = Ed$ and $C = \dfrac{A\epsilon_{o} \epsilon_{r}}{d}$ in Eq. (16.41).

Energy $= \tfrac{1}{2} \left( \dfrac{A\epsilon_{o} \epsilon_{r}}{d} \right) (Ed)^{2}$

As $(Ad)$ is the volume between the plates, so

$= \tfrac{1}{2} \epsilon_{o} \epsilon_{r} E^{2} \times (Ad)$

This equation is valid for any electric field strength.

Example 16.7

Two capacitors of capacitances $4\,\mu\mathrm{F}$ and $8\,\mu\mathrm{F}$ are connected in series and charged by a $100\,\mathrm{V}$ power supply. Find the total charge and energy stored in the system.

Solution

Equivalent capacitance $C_{e}$ is given by;

$C_{e} = \dfrac{8}{3} \times 10^{-6}\,\mathrm{F}$

Putting the values of $C_{1}$ and $C_{2}$, we get

and $Q = C_{e}V = \left(\dfrac{8}{3} \times 10^{-6}\,\mathrm{F}\right) \times 120\,\mathrm{V}$

$Q = 3.2 \times 10^{-4}\,\mathrm{C}$

Energy stored in the system is given by

$W = \tfrac{1}{2}QV$

$W = \tfrac{1}{2}(3.2 \times 10^{-4}\,\mathrm{C}) \times 120\,\mathrm{V}$

$= 1.92 \times 10^{-2}\,\mathrm{J}$

Example 16.8

A $16\,\mu\mathrm{F}$ capacitor is charged to a potential difference of $100\,\mathrm{V}$. Determine the energy stored in it. If the potential difference is doubled by what factor does the energy increase?

Solution

Energy stored is given by;

$W = \tfrac{1}{2}CV^{2}$

Putting the values of $C$ and $V$, we get

$W = \tfrac{1}{2}(16 \times 10^{-6}\,\mathrm{F}) \times (100)^{2}$

$= 8 \times 10^{-2}\,\mathrm{J}$

Since energy depends on $V^{2}$, therefore if $V^{2}$ is doubled, the energy increases by a factor $(2)^{2}$, i.e., energy increases by $4$ times.