Answer:
[tex]C=2.54\cdot 10^5\ F[/tex]
Explanation:
Energy Stored on a Capacitor
This energy is stored in the electric field present when a voltage V is applied to a capacitor C. It can be computed by the formula
[tex]\displaystyle E=\frac{CV^2}{2}[/tex]
We can compute the energy by knowing the power dissipated in the flashlamp is P=8 W in a time of [tex]7 \mu s=7\cdot 10^{-6}\ sec[/tex]
[tex]\displaystyle E=\frac{P}{t}=\frac{8\ W}{7\cdot 10^{-6}\ sec}=1.14\cdot 10^6\ J[/tex]
Solving the first equation for C
[tex]\displaystyle C=\frac{2E}{V^2}[/tex]
[tex]\displaystyle C=\frac{2\times 1.14\cdot 10^6}{3^2}[/tex]
[tex]\boxed{C=2.54\cdot 10^5\ F}[/tex]
