# Potential difference of a capacitor

A defibrillator is used to restart a person's heart after it stops beating. Energy is delivered to the heart by discharging a capacitor through the body tissues near the heart. If the capacitance of the defibrillator is 8.00 µF and the energy delivered is to be 100 J, to what potential difference must the capacitor be charged?

I wasn't sure how to start this problem.
Would I need to use
U = qV
but where would the capacitance fit in?

G01
Homework Helper
Gold Member

Last edited:
I used the formula.
$$q = C\Delta V$$
q = (8)(100) = 800

Then I used the U = qV equation.
U = (800)(100J) = 80000

which is wrong. (It says the answer is 5.00 kV)

G01
Homework Helper
Gold Member
This is the correct way to do this problem. The previous method I gave you was incorrect, I'm sorry.
The energy stored in a capacitor is:

$$U_c=1/2C(\Delta V)^2$$

- If you want the energy delivered to be 100J, then the capacitor must store 100J!

Also, don't forget to convert the 8microfarads to SI.

Good Luck, and sorry about that!

Last edited:
It says that I am not getting the answer.
$$q = C\Delta V$$
q = (8.0 x 10^-6)(100J) = 8 x 10^-4

$$U = q\Delta V$$
100J = (8.0 x 10^-4) V
V = 125000
???

sorry that it's taking me awhile :)

I am trying this out and not getting the correct answer.
The answer should be 5.00 kV.