Irodov 3.143: The Field Squeezes the Water – Amit & Deepali Sinha Foundation

Problem Irodov 3.143

A parallel-plate capacitor with horizontal plates is immersed in water (permittivity $\varepsilon = 81$ ), so that water fills the gap of width $d = 1.0$ mm between its plates. A constant voltage $V = 500$ V is then applied across the plates. By how much does the water pressure in the gap rise?

The capacitor edge on, its plates horizontal, fully under water, which fills the gap d between them; the line above is the free surface. — The capacitor edge on, its plates horizontal, fully under water, which fills the gap $d$ between them; the line above is the free surface.

Given

$d = 1.0$ mm
$V = 500$ V
water, $\varepsilon = 81$ (table value)

Find

the pressure increment

\Delta p

in the gap

Solution

Step 1Field and charge in the gap

Water has $\varepsilon = 81$ (a table value). Away from the edges the field between the plates is uniform, and the free charge density on a plate is the displacement $D = \varepsilon_0\varepsilon E$ :

E = \frac{V}{d}, \qquad \sigma = \varepsilon_0\varepsilon\,\frac{V}{d} \tag{1}

Step 2The electric force on the plate’s own charge

The force on the plate is the force on its free charge $\sigma$ . To find the field that acts on that charge, imagine a vanishingly thin vacuum slit between the plate and the water. The normal component of $\vect D$ is continuous across it, so in the slit $E_0 = \sigma/\varepsilon_0$ , while inside the metal the field is zero. The charge layer feels the average of the two:

f_{\mathrm{el}} = \sigma\cdot\frac{E_0 + 0}{2} = \frac{\sigma^2}{2\varepsilon_0} \tag{2}

This is the force per unit area pulling the plate toward the other plate.

Step 3The net force from the field energy

The net force per unit area that must hold the plate in place, field and water together, follows from the energy. At fixed charge the field energy per unit plate area is $\sigma^2x/2\varepsilon_0\varepsilon$ for a gap $x$ ; widening the gap by $dx$ lets water flow in at the outside pressure and raises this energy, so the plates attract with

f_{\mathrm{net}} = \frac{d}{dx}\left(\frac{\sigma^2 x}{2\varepsilon_0\varepsilon}\right) = \frac{\sigma^2}{2\varepsilon_0\varepsilon} \tag{3}

This is $\varepsilon$ times weaker than (2).

Step 4The pressure makes up the difference

The only other force on the plate’s inner face is the water pressure in the gap, which exceeds the pressure outside by $\Delta p$ and pushes the plate outward. The balance $f_{\mathrm{net}} = f_{\mathrm{el}} - \Delta p$ , with (2) and (3), gives

\Delta p = \frac{\sigma^2}{2\varepsilon_0} - \frac{\sigma^2}{2\varepsilon_0\varepsilon} = \frac{\sigma^2}{2\varepsilon_0}\cdot\frac{\varepsilon - 1}{\varepsilon}

Substituting $\sigma$ from (1):

\Delta p = \frac{\varepsilon_0^2\varepsilon^2V^2}{2\varepsilon_0 d^2}\cdot\frac{\varepsilon - 1}{\varepsilon}

Numerically:

\Delta p = \frac{8.85\times10^{-12}\cdot 81\cdot 80\cdot 500^2}{2\cdot(1.0\times10^{-3})^2}\ \text{Pa}

\Delta p = \dfrac{\varepsilon_0\varepsilon(\varepsilon - 1)V^2}{2d^2} \approx 7.2\text{ kPa}

(4)

That is about $0.07$ atm.

Check

With $\varepsilon = 1$ (an empty gap) (4) gives $\Delta p = 0$ : nothing to compress, and the two force expressions agree. Dimensions: $\varepsilon_0 V^2/d^2$ is $\varepsilon_0 E^2$ , an energy density, J/m $^3$ = N/m $^2$ , a pressure. Size: $\varepsilon_0E^2/2$ for $E = 5\times10^5$ V/m is about $1.1$ Pa, and the factor $\varepsilon(\varepsilon - 1) = 6480$ brings it to $7.2$ kPa.

Remark

A liquid dielectric is drawn into a region of strong field, so the field compresses the water between the plates until the extra pressure balances the pull; the pressure climbs across the fringing field at the edges of the plates, where the water is drawn in. The factor $\varepsilon(\varepsilon - 1)$ makes water, with its large permittivity, a dramatic case: the same field in oil ( $\varepsilon \approx 2$ ) would raise the pressure about three thousand times less.