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

Answer:

energy density is 2.08 ×$10^{-4}$ J/m³

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