derive an expression for energy stored in a parallel platecapacitor
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PHYSICS
(a) Derive the expression for the energy stored in a parallel plate capacitor. Hence obtain the expression for the energy density of the electric field.
(b) A fully charged parallel plate capacitor is connected across an uncharged identical capacitor. Show that the energy stored in the combination is less than that stored initially in the single capacitor
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ANSWER
(a)
Power across capacitor is P=VI
P=V
dt
dQ
By definition of capacitance, Q=CV
Pdt=
C
Q
dQ
Energy stored is
U=∫Pdt=
2C
Q
2
...........(i)
Electric field inside the capacitor is given by:
E=
ε
o
σ
=
Aε
o
Q
Capacitance C=
d
ε
o
A
Q=ECd.........(ii)
Substituting (ii) in (i),
U=
2C
E
2
C
2
d
2
=
2
E
2
ε
o
Ad
Energy density
Ad
U
=
2
1
ε
o
E
2
(b)
Let charge on the capacitor before connecting be Q.
After connection, each capacitor has a charge Q/2.
Energy stored before connection, U
1
=
2C
Q
2
Energy stored after connection, U
2
=2
2C
(Q/2)
2
=
4C
Q
2
Hence, energy stored in the combination is less than that of the single capacitor.