Question

Find the electric energy density between the plates of a 225-μF parallel-plate capacitor. The potential difference between the plates is 365 V , and the plate separation is 0.200 mm .

Answer 1

Energy density will be 14.73
`J/m^3`

Step-by-step explanation:

We have given capacitance
`C=225\mu F=225* 10^(-6)F`

Potential difference between the plates = 365 V

Plate separation d = 0.200 mm
`0.2* 10^(-3)m`

We know that there is relation between electric field and potential

`E=(V)/(d)`, here E is electric field, V is potential and d is separation between the plates

So
`E=(V)/(d)=(365)/(0.2* 10^(-3))=1825000N/C`

Energy density is given by
`E=(1)/(2)\varepsilon _0E^2=(1)/(2)* 8.85* 10^(-12)* (1.825* 10^6)^2=14.73J/m^3`