Energy Stored In A Capacitor

The energy stored in a charged capacitor is $U=\frac{1}{2}C{V}^2$, where $C$ is the capacitance and $V$ the voltage across it, a result obtained in NCERT Class 12, Chapter 2 (Electrostatic Potential and Capacitance) by summing the work done adding charge incrementally. Substituting $C=10\times 10^{-6}$ F and $V=200$ V gives $U=\frac{1}{2}\times 10\times 10^{-6}\times (200{)}^2$. First $(200{)}^2=4\times 10^4$; multiplying by $10\times 10^{-6}$ gives $0.4$; taking half yields $U=0.2$ J. The value 2 J is wrong because it omits the factor of one half and mishandles a power of ten. The value 0.1 J is wrong because it uses $V$ linearly rather than squared for part of the calculation. The value 0.4 J is wrong because it forgets the factor of one half entirely. This flash-circuit context explains the sudden bright discharge. The same energy can equally be written as $\frac{1}{2}QV$ or $\frac{Q^2}{2C}$, all three forms being equivalent once the relation $Q=CV$ is used. A magnitude and unit check confirms farads times volts squared gives joules, and a fraction of a joule is a realistic burst of stored energy for a small camera flash before it discharges.