A parallel-plate capacitor was lowered into water in a horizontal position, with water filling up the gap between the plates d = 1.0 mm wide. Then a constant voltage V = 500 V was applied to the capacitor. Find the water pressure increment in the gap.
Text Solution
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Sol. F = 
(when battery remains connected)
F = – 
(when battery is disconnected)
Let at an instant a thickness x of gap between plates is filled with water shown in fig.
C = 
= 
∴ U = 
CV 2 or U = 
∴ F = 
. (since, battery remains connected)
∴ F = – 
When water is filled in the gap between plates of capacitor x = d
∴ F = 
= 
= 
∴ Excess pressure is Δ P = 
= 
For water ε = 81
On putting the values, we get Δ p = 7.17 kPa = 0.07 atm.
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