Question

1. A voltage of 12 V is placed on a capacitor with C = 85 pF. (a) What is the charge on the capacitor? (b) How much energy is stored in the capacitor?

2. In open-heart surgery, a much smaller amount of energy will defibrillate the heart. (a) What voltage is applied to the 7 F capacitor of a heart defibrillator that stores 60.0 J of energy? (b) Find the amount of stored charge.

3. Calculate the energy stored in a 5 μF capacitor charged to 30 V.

Answer 1

1.

(a)

$$q=CV\\ =85*10^{-12}*12=1.02*10^{-9}\:\rm C$$

(b)

$$W=\frac{CV^2}{2}\\ =\frac{85*10^{-12}*12^2}{2}=6.12*10^{-9}\:\rm J$$

2.

$$V=\sqrt{2W/C}\\ =\sqrt{2*60.0/7}=4.14\:\rm V$$

3.

$$W=\frac{CV^2}{2}\\ =\frac{5*10^{-6}*30^2}{2}=2.25*10^{-3}\:\rm J$$