Question

A parallel plate capacitor has circular plates with diameter D 21.5 cm separated by a distance d 1.75 mm. When a potential difference of AV= 12.0 V is applied across the plates, what is the energy density u between the plates? 9. a) u 0.0132 J/cm u=52.0 J/cm3 b) c) u= 127 J/cm2 u=208 J/cm2 d) =

Answer 1

energy density is 2.08 ×
`10^(-4)` J/m³

Step-by-step explanation:

given data

diameter D = 21.5 cm

distance d = 1.75 mm

potential difference V = 12.0 V

to find out

energy density u

solution

first we will apply here formula for electric field that is

electric field = potential difference / distance ....................1

put here all value in equation 1 we get electric field

electric field = 12 / 1.75 ×
`10^(-3)`

electric field = 6.857 ×
`10^(3)` N/C

so energy density will be

energy density = 1/2 × ∈ × E² .............................2

put here all value and ∈ = 8.85 ×
`10^(-12)`

energy density = 1/2 × 8.85 ×
`10^(-12)` × (6.857 ×
`10^(3)`)²

so energy density = 2.08 ×
`10^(-4)` J/m³