Q.Two capacitors, $C_1 = 4\ \mu\text{F}$ and $C_2 = 6\ \mu\text{F}$, are connected to a $12\ \text{V}$ battery, first in series and then in parallel. Find the energy stored in the combination in each case.

Capacitor Energy Storage

The Intuition

Charging a capacitor is like piling sand onto a growing heap. The first grain of charge lands on an empty plate easily. But every later bit of positive charge must be pushed onto a plate that is already positive, and it resists. So more and more work is needed as the plate fills up. All of that work does not disappear — it is stored in the capacitor as electrostatic potential energy, ready to be released later.

Building the Formula

Suppose at some instant during charging the capacitor already holds charge $q$, so the voltage across it at that moment is $v = q/C$. Moving one more small charge $dq$ onto the plate costs work:

Adding up (integrating) all these small contributions as the charge builds from $0$ to a final value $Q$ gives the total work done:

This work is exactly the energy $U$ stored in the charged capacitor.

Three Equivalent Forms

Using $Q = CV$, the same stored energy can be written three ways — pick whichever matches the quantities you know:

• Use $\dfrac{Q^2}{2C}$ when the charge is fixed (capacitor disconnected from the source).
• Use $\dfrac{1}{2}CV^2$ when the voltage is fixed (capacitor stays connected to a battery).

Where the Energy Lives — Energy Density

The energy is stored in the electric field occupying the space between the plates, not on the plates themselves. For a parallel-plate capacitor this leads to a general result: energy stored per unit volume of field is

where $E$ is the field strength. Wherever an electric field exists, energy is stored there, with density proportional to $E^2$.

A Quick Example

A $10\ \mu\text{F}$ capacitor is charged to $100\ \text{V}$. The stored energy is:

That 0.05 J can be released almost instantly — which is exactly how a camera flash works: charge slowly, discharge fast. …