A plastic ball of 1cm diameter and 10^-8 Coulomb of charge is suspended by an insulating string. The lowest point of the ball is 1cm above a big container of saltwater. Indeed, the water's surface rises a bit. What is the water elevation height exactly below the ball?

Ignore the water's surface tension and consider 1000kg/m³ the saltwater density.

My english is quite rusty, I am sorry for that, but I would really appreciate the help xD

In my solution, I considered the saltwater surface an infinite conductive plane, and so an image charge would be attracted by the plastic ball as the water rises. It was, basically the idea. From now, I considered the image charge as if it was made of saltwater but with the same plastic ball volume, indeed, the water will rise until the balance of forces (electric and weight):

Let k = 9*10^9Nm²/C² be the electrical constant, d = 1000kg/m³ the density of salt water, g = 10m/s² the gravity, L the length between the ball's center and of its image, and V=4/3π(0,005)³ the plastic ball volume:

The charge image will rise until x such that

k(10^-8)²/x² = (dV)g
(9*10^9)(10^-8)²/x² = (1000)(4/3π(0,005)³)(10)

=> x² = (9*10^9)(10^-8)²/(1000)(4/3π(0,005)³)(10)

solving for x

=> x = 0,01311 m = 1,311 cm

But, geometrically, as the charge image rises until x, namely, the distance L-x, the saltwater surface would rise y = (L-x)/2

L = 0,5 + 1 +1 + 0,5 = 3cm

=> y = (3-1,311)/2 = 0,85cm

But the answer seems to be...