Question

What is the p.d of a capacitor with a capacitance of 3,300pF needs to store a 0.5mj of energy?

How to find the voltage and what is the formula for energy and also the power

Answer 1

`550.5V`, rounded to two decimal places

Explanation:

Energy `U` stored in a capacitor is given by

`U=1/2Q^2/C=1/2QV=1/2CV^2`
where `Q` is charge held, `C` its capacitance and `V` potential difference between the plates.

Using the expression of interest and inserting given values we get
`U=1/2CV^2`
`0.5xx10^-3=1/2xx(3300xx10^-12)V^2`
`=>V^2=2xx0.5xx10^-3/(3300xx10^-12)`
`=>V^2=2xx0.5xx10^-3/(3300xx10^-12)`
`=>V=550.5V`, rounded to two decimal places