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Consider a parallel plate capcaitor with...

Consider a parallel plate capcaitor with plates 20 cm by 20 cm and separated by 2 mm. The dielectric constant of the material between the plates is 5. The plates are connected to a voltage source of 500 V. The energy density of the field between the plates will be close to

A

`3 2.65 J //m^3`

B

`3 1.95 J //m^3`

C

`3 1.38 J //m^3`

D

`3 0.69J //m^3`

Text Solution

Verified by Experts

The correct Answer is:
C
`A = 20 xx20 xx10^(-4) = 4xx10^(-2)m^2`
`d = 2mm = 2 xx10^(-3) m`
k = 5
`E=V/d =(500)/(2 xx10^(-3))=250 xx10^3 =25 xx10^4 ` Volt
Energy density
`U=1/2 (epsilon_0 E^2 xxk)`
`=1/2 xx 8.85 xx10^(-12) xx 5 xx(25)^2 xx10^8 = 1.38 J//m^3`
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