Q.(a) A $900\ \text{pF}$ capacitor is charged by $100\ \text{V}$ battery [Fig. 2.31(a)]. How much electrostatic energy is stored by the capacitor?

Concept: Capacitor Energy Storage — $U=\tfrac12CV^2=Q^2/2C$; connecting a charged capacitor to an identical uncharged one conserves total charge but not total energy.

(a) $C=900\ \text{pF}=9\times10^{-10}\ \text{F}$, $V=100\ \text{V}$:

(b) Disconnected, the charge $Q=CV=9\times10^{-8}\ \text{C}$ is fixed. Connected to an identical uncharged $900\ \text{pF}$ capacitor (now in parallel, $C_{\text{eq}}=1.8\times10^{-9}\ \text{F}$), it splits equally, $4.5\times10^{-8}\ \text{C}$ each: …

Part (a): energy stored while connected to the battery

Using $U=\tfrac12CV^2$:

The charge on the capacitor at this point is $Q=CV=(9\times10^{-10})(100)=9\times10^{-8}\ \text{C}$ — this is the charge carried over into part (b).

Part (b): disconnect, then connect to an identical uncharged capacitor

Once the battery is removed, the charge $Q=9\times10^{-8}\ \text{C}$ on the first capacitor is fixed (nothing left to add or remove charge). Connecting it to a second, identical, uncharged $900\ \text{pF}$ capacitor puts the two in parallel, and since there is no external source, the total charge $Q$ must simply redistribute between them.

By symmetry (identical capacitors), the charge splits equally:

The combined (parallel) capacitance is $C_{\text{eq}}=C+C=1.8\times10^{-9}\ \text{F}$, and the total stored energy afterward is …

Method: Capacitor Energy Storage — Whether the Capacitor Stays Connected to the Source or Is Isolated

This method applies to any problem comparing a capacitor's stored energy before and after some change (adding a dielectric, connecting to another capacitor, disconnecting from the battery) — the correct formula to use depends on what is held fixed during the change.

Steps

Step 1: Decide what is held constant — voltage or charge

If the capacitor stays connected to a battery throughout, the voltage across it is fixed by the battery; use $U = \tfrac12 CV^2$. If the capacitor is disconnected from every source before the change happens, its charge is fixed (nothing can add or remove charge); use $U = \dfrac{Q^2}{2C}$.

Step 2: Compute the initial energy with the appropriate formula

Use whichever of $U = \tfrac12CV^2$, $U = \tfrac12QV$, or $U = Q^2/2C$ matches the quantities actually given at that stage.

Step 3: For a "disconnect, then connect to another capacitor" scenario, apply charge conservation first …