Hollow conducting sphere floating in oil

[superfx]

Homework Statement

A hollow sphere of gold floats in a large lake of oil of dielectric constant $$\kappa=3.3.$$ The sphere is half immersed in the oil, and has a total charge of $$2.2*10^-6$$ Coulombs. What fraction of this electric charge on the sphere is above the oil surface?

Homework Equations

The Attempt at a Solution

I'd like to solve this problem in a simple way, without explicitly treating bound and free charges, etc. My line of reasoning is the following: the electric field inside a dielectric is reduced by a factor of $$\kappa=3.3$$, i.e. $$E=E_0/\kappa.$$ The sphere is conducting, so the E field inside must be zero-- therefore the charges on the outside must arrange themselves to meet this requirement. The dielectric reduces the field, so I'd expect there to be more free charge on the bottom, in proportion to the dielectric constant. So:

$$Q_{top}=\frac{Q_{bottom}}{\kappa}$$

$$\frac{Q_{top}}{Q_{total}}=\frac{1}{1+\kappa}$$

Can anyone tell me if my answer is correct or not, and in either case how to reason it through more rigorously? Thanks in advance for your help.

 

[anathema]

you are correct in your line of reasoning. The electric field inside the dielectric is indeed reduced by a factor of the dielectric constant, and the charges on the surface of the sphere will arrange themselves accordingly. However, to solve this problem more rigorously, we need to consider the concept of electric flux.

First, let's define some variables:
- Q_total = total charge on the sphere (given as 2.2*10^-6 C)
- Q_top = charge on the top half of the sphere
- Q_bottom = charge on the bottom half of the sphere
- E_0 = electric field in vacuum (given as 0)
- E = electric field in the oil
- \kappa = dielectric constant of the oil (given as 3.3)
- A = surface area of the sphere
- \epsilon_0 = permittivity of free space (given as 8.85*10^-12 F/m)

Now, we can apply Gauss's Law to this problem. Gauss's Law states that the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space.

In this case, we can consider a spherical Gaussian surface around the sphere, with the top half of the sphere as the enclosed charge. The electric flux through this surface is given by:

\Phi = \frac{Q_{top}}{\epsilon_0}

We also know that the electric flux is equal to the electric field multiplied by the surface area of the Gaussian surface:

\Phi = E*A

Setting these two equations equal to each other, we get:

\frac{Q_{top}}{\epsilon_0} = E*A

Solving for E, we get:

E = \frac{Q_{top}}{\epsilon_0*A}

Now, we can apply the concept of electric field continuity. This states that the electric field must be continuous across a dielectric boundary. In other words, the electric field in the oil must be equal to the electric field in the sphere.

Therefore, we can set E = E_0/\kappa, and solve for Q_top:

\frac{Q_{top}}{\epsilon_0*A} = \frac{E_0}{\kappa}

Q_top = \frac{E_0*A}{\kappa} = \frac{Q_{bottom