Lower plate of capacitor is fixed and upper plate of capacitor is connected to a spring. In the steady position, the distance between the plates is d 0 . When the capacitor is connected with an electric source with the voltage V, a new equilibrium appears, with the distance between the plates as d 1 . Mass of the upper plate is m.
(i) Spring constant k is-
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Ans.
(i)
Sol.
Initial extension is x 0
∴ kx 0 = mg
when capacitor is charged
then,
k[x 0 + (d 0 – d 1 )] = mg + 
k(d 0 – d 1 ) = 
A
k = 
(ii)
Sol. V 2 = 
Voltage is depending on d 1
For maximum value of V
= 0
= 
[2d 1 d 0 – 
]
= 0
d 1 = 
V max = 
(iii)
Sol. Let the small displacement of the upper plate be x downwards from the equilibrium position
F = – k[l 0 + (d 0 – d) + x] + mg
+ 
F = – k(d 0 – d) – kx + 
= – k(d 0 – d) – kx + 
= – k(d 0 – d) – kx + k(d 0 – d 1 ) 
F = – kx 
Acceleration = 
= 
x
ω = 
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