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Electric Capacitance, Energy of an Electric Field

Problem 3.143

A parallel-plate capacitor was lowered into water in a horizontal position, with water filling up the gap between the plates d=1.0 mmd=1.0 \mathrm{~mm} wide. Then a constant voltage V=500 VV=500 \mathrm{~V} was applied to the capacitor. Find the water pressure increment in the gap.

Reveal Answer
Δp=ε0ε(ε−1)V2/2d2=7kPa=0.07 atm. \Delta p=\varepsilon_{0} \varepsilon(\varepsilon-1) V^{2} / 2 d^{2}=7 \mathrm{k} \mathrm{Pa}=0.07 \text { atm. }
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