The area of each plate of parallel plate air capacitor is `150 cm^(2)`. The distance between its plates is 0.8 mm. It is charged to a pot. Diff of 1200 volt. What will be its energy ? What wil be the energy when it is filled with a medium of K = 3 and then charged. If it is charged. If it is charged first as an air capacitor and then filled with this dielectric, what will happen to energy ?

Explanation

Here, `A = 150 cm^(2) = 150xx10^(-4) m^(2)`,
`d = 0.8 mm = 8xx10^(-4) m, V_(0) = 1200 V, C_(0) = ? E_(0) = ?`
`C_(0) = (in_(0) A)/(d) = (8.85xx10^(-12)xx150xx10^(-4))/(8xx10^(-4))`
`= 1.66xx10^(-10) F`
`U_(0) = (1)/(2) C_(0) V_(0)^(2) xx 1.66xx10^(-10) xx(1200)^(2)`
`= 1.2xx10^(-4) J`
With dielectric medium, capacity becomes
`C = KC_(0) = 3xx1.66xx10^(-10) farad`
When charged to same potential , `V = 1200V`,
Energy, `U = (1)/(2) CV^(2) = (1)/(2) (KC_(0)) V_(0)^(2) = K (E_(0))`
`= 3xx1.2xx10^(-4) = 3.6xx10^(-4) J`
When capacitor charged first as air capacitor, then on filling with the dielectric, its potential
becomes `V = (V_(0))/(K) = (1200)/(3) = 400` volt, because capacity becomes 3 times whereas charge remain the same.
`:.` New energy of capacitor
`= (1)/(2) CV^(2) = (1)/(2) (KC_(0)) ((V_(0))/(K))^(2)`
`= (1)/(2) (C_(0) V_(0)^(2))/(K) = (1.2xx10^(-4))/(3) = 4xx10^(-5)` joule