Q.A $600\ \text{pF}$ capacitor is charged by a $200\ \text{V}$ supply. It is then disconnected from the supply and is connected to another uncharged $600\ \text{pF}$ capacitor. How much electrostatic energy is lost in the process?

Concept: Capacitor Energy Storage — energy is stored as $U = \frac{1}{2} C V^2$, and when capacitors share charge, the total energy after redistribution is less due to the loss (dissipated as heat/spark).

Step 1: Initial energy

Only the first capacitor is charged:

Step 2: Final common voltage

Charge is conserved: $Q = CV = (600 \times 10^{-12})(200) = 1.2 \times 10^{-7}\ \text{C}$.

Total capacitance after connection: $C_{\text{eq}} = 600 + 600 = 1200\ \text{pF}$.

Common voltage: $V_f = \frac{Q}{C_{\text{eq}}} = \frac{1.2 \times 10^{-7}}{1200 \times 10^{-12}} = 100\ \text{V}$.

Step 3: Final energy …

The key idea is that energy is not conserved when you connect capacitors in this way — charge is conserved, but the electrostatic energy stored in the capacitors drops. This happens because the redistribution of charge involves current flow through wires, which inevitably dissipates some energy as heat (or, in an idealised circuit, as electromagnetic radiation). The problem asks you to calculate exactly how much is lost.

Let’s walk through it.

1. Find the initial energy stored in the first capacitor. The energy stored in a capacitor is given by

Here $C = 600\ \text{pF} = 600 \times 10^{-12}\ \text{F}$ and $V = 200\ \text{V}$.

So

Compute stepwise:

$200^2 = 40\,000$, so

$600 \times 4 = 2400$, so

2. What happens when the second (uncharged) capacitor is connected? The two capacitors are identical ($C_1 = C_2 = 600\ \text{pF}$). When connected in parallel (the only way to connect them after disconnecting the supply), the total capacitance becomes

The initial charge on the first capacitor is

This charge now redistributes between the two capacitors. Since they are identical, the final voltage across each will be the same, and the total charge is conserved:

So

Notice: the voltage halves.

3. Compute the final stored energy. The final energy in the two-capacitor system is

$100^2 = 10\,000$, so …

Method: Energy Change When a Charged Capacitor Shares Charge with an Uncharged One

This method finds the energy lost when a charged capacitor is disconnected from its source and then connected to an uncharged capacitor.

Steps

Step 1: Compute the initial energy from the single charged capacitor

using only the capacitor that was actually charged; the second capacitor contributes nothing yet.

Step 2: Apply charge conservation to find the common final voltage

Once connected together (effectively in parallel, sharing the same two nodes), the total charge $Q = C_1V_1$ redistributes across the combined capacitance $C_{\text{eq}} = C_1+C_2$, settling at one common voltage:

Step 3: Compute the final total energy …