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 between the plates? a) u = 0.0132 J/cm b) u = 52.0 J/cm c) u=127J/cm d) - 208 J/cm

Answer 1

Answer:
`E=2.078* 10^(-10) J/cm^3`

Step-by-step explanation:

Given

Diameter(D)=21.5 cm

distance between plate=1.75 mm

and we know capacitance is

`C=(\epsilon _0A)/(d)`

and we know charge is

Q=CV

and charge density (
`\sigma [tex])[tex]=(Q)/(A)=(\epsilon _0AV)/(Ad)=(\epsilon V)/(d)`

Electric field(E)
`=(\sigma )/(\epsilon _0)`

Energy density
`=(1)/(2)\epsilon _0E^2`

Energy desity
`=(4\pi \epsilon _0)/(8\pi )(V^2)/(d^2)`

Energy desity
`=(1)/(8\pi \cdot 9* 10^9)* \left [ (12* 10^3)/(1.75)\right ]^2`

`E=2.078* 10^(-10) J/cm^3`