Saved Bookmarks
| 1. |
The plates of parallel plate capacitor have an area of 90 cm^2 each and area separated by 2.5mm. The capacitor is charged by connecting it to a 400V supply. (a) How much electrostatic energy is stored by the capacitor ? (b) View this energy as stored in the electrostatic field between the plates , and obtain the energy per unit volume n, Hence arrive at a relation between u and the magnitude of electric field E between the plates. |
|
Answer» Solution :Here `A = 90 cm^(2) = 90 xx 10^(-4) m^(2)` `= 9 xx 10^(-3) m^(2)` `d = 2.5 MM = 2.5 xx 10^(-3) m` V = 400 volt, `E^(1) = ?` `E^(1) = (1)/(2) CV^(2) = (1)/(2) = (1)/(2) (epsilon_(0)A)/(d) V^(2)` `E^(1) = (8.85 xx 10^(-12) xx 9 xx 10^(-3) (400)^(2))/(2 xx 2.5 xx 10^(-3))` `= 2.55 xx 10^(-6) J` (b) volume of capacitor `V = A xx d` `= 90 xx 10^(-4) xx 25 xx 10^(-3) m^(3)` `= 2.25 xx 10^(-4) m^(3)`. Energy volume `= U = (2.55 xx 10^(-6))/(2.25 xx 10^(-4))` `= 0.113J//m^(3)` As `U = (E^(1))/(V) = ((1)/(2) CV^(2))/(Ad) = ((epsilon_(0)A)/(2d)V^(2))/(Ad) = (1)/(2) epsilon_(0) ((V)/(d)^(2)` But `(V)/(d)` = E, Electric intensity `:. U = (1)/(2) epsilon_(0) epsilon^(2)` |
|