Saltwater Elevation below Ball: 1.85cm

[Dinheiro]

Homework Statement

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

Homework Equations

Eletrostatic equations

The Attempt at a Solution

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...

Spoiler

0,29cm

Were my considerations that wrong?

 

[dauto]

Why are you using the volume of the sphere? it seems to play no role in the problem.

 

[dauto]

Also, please refrain from plugging in any data until the very last step of your solution. That makes your solution much easier to understand.

 

[Naz93]

The volume of the sphere is irrelevant, as dauto says. Are you familiar with Gauss' Law? The charge outside the sphere is the same as if all the charge were at the centre, as far as all points outside the sphere are located. You need to find the effect on the water of the charge 10^-8 Coulomb being concentrated at the centre of the sphere...

 

[Dinheiro]

I know, this problem was driving me crazy. It makes no sense, but it seems to be a polish olympic problem, so I was trying to get an answer out of it... This question is wrong, right?

 

[dauto]

I know, this problem was driving me crazy. It makes no sense, but it seems to be a polish olympic problem, so I was trying to get an answer out of it... This question is wrong, right?

No, the question is correct. It is possible to find the solution with the information provided.

 

[Dinheiro]

Then, how should I proceed?

 

[dauto]

Then, how should I proceed?

1st: Find the electric field at the surface field using the images method
2nd: Find the surface charge density using Gauss' law on a "Pillbox" region enclosing the surface.
3rd: Using the electric field and surface charge find the force acting on the surface. Beware of a factor of 1/2 that appears in such problem because half of the field is the field of the surface charge itself that cannot produce a force on itself.
4th: Calculate the work done by that force on the rising water.
5th: Subtract that work from the gravitational potential energy to find the total potential energy (electric plus gravitational).
6th: Minimize that energy

 

[dauto]

You may take a shortcut and use equilibrium between the electric force and gravity. That way you won't have to calculate the energy and minimize it.

 

[dauto]

I solved the problem and found an answer 10 times smaller than the one provided in the spoiler box in the OP.

Spoiler

h = 0.29 mm

 

[Dinheiro]

Thanks, guys. I tried to redo it yesterday and I got 0,56mm. Could you post your solution please, dauto?