A plastic ball, of diameter $1 cm$ and carrying a uniform charge of ${10}^{-8}C$, is suspended by an insulating string with its lowest point $1cm$ above a large container of brine (salted water). As a result, the surface of the water below the ball wells up a little.
How large is the rise in water level immediately below the ball? Ignore the effect of surface tension, and take the density of salted water to be $1000kg m^{-3}.$

Brine is a good conductor, because positive and negative ions can move easily within it. When the charged plastic ball is placed close to the surface of the water, opposing charges are induced in the surface, whilst like charges are repelled from it. The resulting electric field lines above the water surface will be perpendicular to it, whilst beneath it the net electric field vanishes.

The charged ball attracts the water below it, and the surface wells up in a hump. The electrical forces exerted on the hump are balanced mainly by gravity and the effect of surface tension can be ignored. We don’t know the shape of the hump exactly, but can be sure that the rise in water level will be small and there will be only a slight deviation from a plane surface; this is why we can use the so-called method of image charges. It will be sufficient to consider the maximum effect and find the rise at the point $P$ shown in Fig.

At P the electric field due to the charge Q is

$E_1=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{{\left(3r\right)}^2} .........................................\left(1\right)$

The effect of the unknown surface charge distribution can be replaced by that due to an image charge of $-Q$ situated at a depth below the surface of $3r$3r (see Fig). The electric field at $P$ due to the image charge has the same magnitude and direction as $E_1$​, and so the net electric field is

$E=2E_1=\frac{1}{2\pi {\epsilon }_0} \frac{Q}{{\left(3r\right)}^2}..................\left(2\right)$

According to Gauss’s law the surface charge density at $P$ is

$\sigma ={\epsilon }_{0}E=\frac{1}{2\pi }\frac{Q}{{\left(3r\right)}^2}........\left(3\right)$

At the water surface the force exerted on a unit area is the product of the surface charge density $\sigma$ and the electric field $E_1$ due to the ball:

$\frac{F}{A}=\sigma E_1 ........\left(4\right)$

This is the upward force at $P$ per unit area and is balanced by the hydrostatic pressure associated with the maximum rise $h$in water level:

$\frac{F}{A}=\rho gh.....\left(5\right)$

Substituting for the electric field and surface charge gives an expression for $h$ as

$h= \frac{1}{\rho g}\frac{1}{2\pi }\frac{Q}{{\left(3r\right)}^2}\frac{1}{4\pi {\epsilon }_0}\frac{Q}{{\left(3r\right)}^2}......\left(6\right)$

Inserting the numerical data into this equation, yields $h\approx 0.29 mm,$which is very small compared to the diameter of the ball and justifies our treating the water surface as being close to flat.