ENERGY STORED IN CAPACITOR:-
We know that capacitor is charge storing device. Its charge storing ability that is capacitance increases in the presence of dielectric medium as explained in the previous topic. Capacitor store its energy inside electric field which can be calculated as follow:
We know that Potential Difference= Work
Charge
Or V = Energy
Charge
By capacitor equation Q=CV-----(ii)
·
When key is open V1=0 & Q1=0
·
When key is closed V2=V & Q2=Q
Average P.D= V1+V2
= 0 + V
2 2
We can write Vavg=
V -------(iii)
2
Then (i) can be written as
Energy=Vavg x
Q----(iv)
Using (iii) & (ii) in (iv) we
get
Energy= ( V ) (VC)
2
Energy=CV2---------(v)
2
One other way to derive (v) is from graph method which is also shown below
One other way to derive (v) is from graph method which is also shown below
The potential difference across the plates of a capacitor is directly proportional to the charge stored on the plates. This gives a straight line through the origin on a voltage-charge graph. The area under this graph gives the energy stored in a capacitor.
As the area under the graph is a triangle,
area = ½ base x height.
so Energy = Area= 1
QV
2
Whereas Q=CV so
Energy= 1 CV2
2
Now
In case of any dielectric as
medium between between plates of capacitor the capacitance is given by
Cmed= Aεo
εr ----------------- (vi)
d
also we know tha E= V
d
or V= Ed-----(vii)
in case of energy in the presence
of dielectric (v) can be written as
Energy = 1 (Cmed)(V2)------------(vii)
2
Putting (vii) & (vi) in
(viii) we get
Energy = 1 ( Aεo
εr )(Ed)2
2 d
after cancellation of d we get
Energy = 1 εo
εr E2(Ad)
2
Where as Ad= Volume of dielectric so
Energy = 1 εo εr E2(Volume)
2
Energe = 1 εo εr E2
Volume 2
As Energy =
Energy density so
Volume
Energy density = 1 Aεo
εr E2-------(ix)
2
CONCLUSION:-
- Energy density is directly proportional to εr
- Energy density is directly proportional to E2
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