The plates of a parallel plate capacitor have an area of 90 cm^(2) each and are separated by 2.5mm. The capacitance is charged by connecting it to a 400V supply. <br> (a) How much electrostatic energy is stored by the capacitor ? <br> (b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume (u). Hence arrive at a relation between U and the magnitude of electric field E between the plates.

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Answer Text

Solution :
Area of the plate of a parallel plate capacitor, `A=90cm^(2)=90xx10^(-4)m^(2)` <br> Distnace betweent he plates , d=2.5 mm =`2.5xx10^(-3) m` <br> potential difference across the plates , V=400 v <br> (a) capacitance of the capacitor is given by the relation. <br> `C=(in_(0)A)/d` <br> Electrostatic energy stored in the capacitor is given by the relation , `E_(1)=1/2 CV^(2)` <br> `=1/2 (in_(0)A)/d V^(2)` <br> where, `in_(0)`=permitivity of free space =`8.85xx10^(-12)C^(2)N^(-1)m^(-2)` <br> `:. E_(1)=(1xx8.85xx10^(-12)xx90xx(400)^(2))/(2xx2.5xx10^(-3))=2.55xx10^(-6)J` <br> Hence, the electrostatic energy stored by the capacitor is `2.55xx10^(-6)J` <br> (B) volume of the given capacitor <br> `V'=A xx d` <br> `=90xx10^(-4) xx25xx10^(-3)` <br> `=2.25xx10^(-4) m^(3)` <br> Energy stored in the capacitor per unit volume is given by <br> `u=(E_(1))/(V')` <br> `=(2.55xx10^(-6))/(2.25xx10^(-4)) =0.11 Jm^(-3)` <br> Again `u=(E_(1))/(v')` <br> `=(1/2 CV^(2))/(Ad) =((in_(0)A)/(2d)V^(2))/(Ad) =1/2 in_(0)(V/d)^(2)` <br> where, <br> `V/d`=electric intensity =E <br> `:. u=1/2 in_(0)E^(2)`

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