Question

In an electronic circuit, you need a capacitor to store 6.000e-9 J of energy. You have 1.50 volts available to charge it with. What capacitance should you choose?

Answer 1

The electrical energy stored in a capacitor is given by

$E_{cap}= \frac{1}{2}CV^2$

Since we need E=6.000e-9J, and V=1.5V, we can solve this equation for C:

$2* \frac{6.000*10^{-9} }{1.5^2V}=C= 5.33*10^{-9}=5.33nF$