Energy Stored In A Capacitor

Question

Charging a 10 microfarad capacitor in a camera flash circuit up to a potential of 200 volts stores electrical energy, so how much energy is held?

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The energy stored in a charged capacitor is $U = \frac{1}{2}CV^2$, 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 $C = 10	imes10^{-6}$ F and $V = 200$ V gives $U = \frac{1}{2}	imes10	imes10^{-6}	imes(200)^2$. First $(200)^2 = 4	imes10^{4}$; multiplying by $10	imes10^{-6}$ 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.

Energy Stored In A Capacitor

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