Question: The power supply for a pulsed nitrogen laser has a \({\bf{0}}{\bf{.050}}\;{\bf{\mu F}}\) capacitor with a maximum voltage rating of 35 kV. (a) Estimate how much energy could be stored in this capacitor. (b) If 12% of this stored electrical energy is converted to light energy in a pulse that is 6.2 microseconds long, what is the power of the laser pulse?

The energy stored in a capacitor can be calculated by examining the value of the capacitance of the capacitor and the value of the voltage. Its value is altered linearly to the value of the capacitance.

The capacitance is,\[C = 0.050\;{\rm{\mu F}}\].

The voltage rating is,\[V = 35\;{\rm{kV}}\].

The time is,\[t = 6.2\;{\rm{\mu s}}\].

The energy stored in the capacitor can be calculated as:

\[\begin{array}{c}PE = \frac{1}{2}C{V^2}\\ = \frac{1}{2}\left( {0.050\;{\rm{\mu F}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}\;{\rm{F}}}}{{{\rm{1}}\;{\rm{\mu F}}}}} \right){\left\{ {\left( {{\rm{35}}\;{\rm{kV}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{\rm{3}}}\;{\rm{V}}}}{{{\rm{1}}\;{\rm{kV}}}}} \right)} \right\}^2}\\ = 30.6\;{\rm{J}}\\ \approx 31\;{\rm{J}}\end{array}\]

Thus, the energy stored in the capacitor is \[31\;{\rm{J}}\].

If 12% of electrical energy is converted to light energy, then the value of the light energy will be:

\[\begin{array}{c}LE = 0.12\left( {PE} \right)\\ = 0.12\left( {31\;{\rm{J}}} \right)\\ = 3.7\;{\rm{J}}\end{array}\]

The power of the laser pulse can be calculated as:

\[\begin{array}{c}P = \frac{{LE}}{t}\\ = \frac{{\left( {3.7\;{\rm{J}}} \right)}}{{\left( {6.2\;{\rm{\mu s}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}\;{\rm{s}}}}{{{\rm{1}}\;{\rm{\mu s}}}}} \right)}}\\ = 5.9 \times {10^5}\;{\rm{W}}\end{array}\]

Thus, the power of the laser pulse is \[5.9 \times {10^5}\;{\rm{W}}\].