In the given circuit, two identical parallel conducting plates A and B are connected to a 25 V battery by metal springs of spring constant 2k and k respectively. Initially, the switch ‘S’ is open and the plates are uncharged. In fact, the two plates form a capacitor. When the switch ‘S’ is closed, distance between the plates becomes 2 mm and it is one third of the initial distance between plates. Also the electric potential energy stored in the capacitor is found to be 7.5 × 10 –4 J.
(i) Initially, when the plates are uncharged capacitance of the capacitor is
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Ans.
(i)
Sol.
Energy of capacitor
= 
C = 
This is final C
As 
= 
C 2 = 
= 
C 2 = 0.8 µF
C 2 is initial capacitance
(ii)
Sol. Force between plates = 
= 
⇒ 
= 2k × x 1
put value of x 1 and get k
x 1 = 
= 
(iii)
Sol. F = 2kx 1
F = kx 2
2kx 1 = kx 2
2x 1 = x 2
Also, x 1 + x 2 = 4
= 4
3x 2 = 8
x 2 = 
= 2.67 mm
where x 2 = extension of spring connected to plate B.
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