Question

15 points!

a. Calculate the electric potential energy stored in a 1.4 x 10-7 F capacitor
that stores 3.40 x 10-6 C of charge at 24.0 V.

Answer 1

`4.12* 10^(-5)\ J`.

Step-by-step explanation:

Given that,

Capacitance,
`C=1.4* 10^(-7)\ F`

Charge stored in the capacitor,
`Q=3.4* 10^(-6)\ C`

We need to find the electric potential energy stored in the capacitor. The formula for the electric potential energy stored in the capacitor is given by :

`E=(Q^2)/(2C)`

Put all the values,

`E=((3.4* 10^(-6))^2)/(2* 1.4* 10^(-7))\\\\=4.12* 10^(-5)\ J`

So, the required electric potential eenergy is equal to
`4.12* 10^(-5)\ J`.